LAPLACE TRANSFORMS - BookSpar · LAPLACE TRANSFORMS . INTRODUCTION Laplace transform is an integral transform employed in solving physical problems. Many physical problems when analysed

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LAPLACE TRANSFORMS

INTRODUCTION

Laplace transform is an integral transform employed in solving physical problems.

Many physical problems when analysed assumes the form of a differential equation subjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable are specified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different values of the independent variable, the conditions are called boundary conditions.

The problem with initial conditions is referred to as the Initial value problem.

The problem with boundary conditions is referred to as the Boundary value problem.

Example 1 : The problem of solving the equation xydxdy

dxyd

=++2

2

with conditions y(0) =

y′ (0) = 1 is an initial value problem


dxyd cos23 2

2

=++ with y(1)=1,

y(2)=3 is called Boundary value problem.

Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. Besides the technique may also be employed to find certain integral values also. The transform is named after the French Mathematician P.S. de’ Laplace (1749 – 1827).

The subject is divided into the following sub topics.

LAPLACE TRANSFORMS

Definition and Properties

Transforms of some functions

Convolution theorem

Inverse transforms

Solution of differential equations

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Definition :

Let f(t) be a real-valued function defined for all t ≥ 0 and s be a parameter, real or complex.

Suppose the integral ∫∞

−

0

)( dttfe st exists (converges). Then this integral is called the Laplace

transform of f(t) and is denoted by Lf(t).

Thus,

Lf(t) = ∫∞

−

0

)( dttfe st (1)

We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t) is a function of s denoted by F(s) or )(sf .

Thus,

Lf(t) = F(s) (2)

Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denoted by L-1 [F(s)].

Thus,

L-1 [F(s)] = f(t) (3)

Suppose f(t) is defined as follows :

f1(t), 0 < t < a

f(t) = f2(t), a < t < b

f3(t), t > b

Note that f(t) is piecewise continuous. The Laplace transform of f(t) is defined as

Lf(t) = ∫∞

−

0

)(tfe st

= ∫ ∫ ∫∞

−−− ++a b

a b

ststst dttfedttfedttfe0

321 )()()(

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NOTE : In a practical situation, the variable t represents the time and s represents frequency.

Hence the Laplace transform converts the time domain into the frequency domain.

Basic properties

The following are some basic properties of Laplace transforms :

1. Linearity property : For any two functions f(t) and φ(t) (whose Laplace transforms exist)

and any two constants a and b, we have

L [a f(t) + b φ(t)] = a L f(t) + b Lφ(t)

Proof :- By definition, we have

L[af(t)+bφ(t)] = [ ]dttbtafe st∫∞

− +0

)()( φ = ∫ ∫∞ ∞

−− +0 0

)()( dttebdttfea stst φ

= a L f(t) + b Lφ(t)

This is the desired property.

In particular, for a=b=1, we have

L [ f(t) + φ(t)] = L f(t) + Lφ(t)

and for a = -b = 1, we have

L [ f(t) - φ(t)] = L f(t) - Lφ(t)

2. Change of scale property : If L f(t) = F(s), then L[f(at)] =

asF

a1 , where a is a positive

constant.


Lf(at) = ∫∞

−

0

)( dtatfe st (1)

Let us set at = x. Then expression (1) becomes,

L f(at) = ∫∞

−

0

)(1 dxxfea

xas

=

asF

a1


3. Shifting property :- Let a be any real constant. Then

L [eatf(t)] = F(s-a)


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L [eatf(t)] = [ ]∫∞

−

0

)( dttfee atst = ∫∞

−−

0

)( )( dttfe as

= F(s-a)

This is the desired property. Here we note that the Laplace transform of eat f(t) can be

written down directly by changing s to s-a in the Laplace transform of f(t).

TRANSFORMS OF SOME FUNCTIONS 1. Let a be a constant. Then

L(eat) = ∫ ∫∞ ∞

−−− =0 0

)( dtedtee tasatst

= asas

e tas

−=

−−

∞−− 1)( 0

)(

, s > a

Thus,

L(eat) = as −

1

In particular, when a=0, we get

L(1) = s1 , s > 0

By inversion formula, we have

atat es

Leas

L ==−

−− 11 11

2. L(cosh at) =

+ −

2

atat eeL = [ ]∫∞

−− +02

1 dteee atatst

= [ ]∫∞

+−−− +0

)()(

21 dtee tastas

Let s > |a| . Then,

∞+−−−

+−

+−−

=0

)()(

)()(21)(cosh

ase

aseatL

tastas

= 22 ass−

Thus,

L (cosh at) = 22 ass−

, s > |a|

and so

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atas

sL cosh221 =

−−

3. L (sinh at) = 222 asaeeL

atat

−=

− −

, s > |a|

Thus,

L (sinh at) = 22 asa−

, s > |a|

and so,

a

atas

L sinh122

1 =

−−

4. L (sin at) = ∫∞

−

0

sin ate st dt

Here we suppose that s > 0 and then integrate by using the formula

[ ]∫ −+

= bxbbxaba

ebxdxeax

ax cossinsin 22

Thus,

L (sinh at) = 22 asa+

, s > 0

and so

a

atas

L sinh122

1 =

+−

5. L (cos at) = atdte st cos0∫∞

−

Here we suppose that s>0 and integrate by using the formula

[ ]∫ ++

= bxbbxaba

ebxdxeax

ax sincoscos 22

Thus,

L (cos at) = 22 ass+

, s > 0

and so

atas

sL cos221 =

+−

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6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then

L(tn) = ∫∞

−

0

dtte nst

Let s > 0 and set st = x, then

∫ ∫∞ ∞

−+

− =

=

0 01

1)( dxxess

dxsxetL nx

n

nxn

The integral dxxe nx∫∞

−

0

is called gamma function of (n+1) denoted by )1( +Γ n . Thus

1

)1()( +

+Γ= n

n

sntL

In particular, if n is a non-negative integer then )1( +Γ n =n!. Hence

1

!)( += nn

sntL

and so

)1(

11

1

+Γ=+

−

nt

sL

n

n or !n

t n

as the case may be

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TABLE OF LAPLACE TRANSFORMS

f(t) F(s)

1 s1 , s > 0

eat as −

1 , s > a

coshat 22 ass−

, s > |a|

sinhat 22 asa−

, s > |a|

sinat 22 asa+

, s > 0

cosat 22 ass+

, s > 0

tn, n=0,1,2,….. 1!+ns

n , s > 0

tn, n > -1 1)1(

+

+Γns

n , s > 0

Application of shifting property :-

The shifting property is

If L f(t) = F(s), then L [eatf(t)] = F(s-a)

Application of this property leads to the following results :

1. [ ]ass

assat

bssbtLbteL

−→−→

−== 22)(cosh)cosh( = 22)( bas

as−−

−

Thus,

L(eatcoshbt) = 22)( basas−−

−

and

btebas

asL at cosh)( 22

1 =−−

−−

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2. 22)()sinh(

basabteL at

−−=

and

btebas

L at sinh)(1

221 =

−−−

3. 22)()cos(

basasbteL at

+−−

=

and

btebas

asL at cos)( 22

1 =+−

−−

4. 22)()sin(

basbbteL at

−−=

and

b

btebas

Lat sin

)(1

221 =

−−−

5. 1)()1()( +−

+Γ= n

nat

asnteL or 1)(

!+− nas

n as the case may be

Hence

)1()(

11

1

+Γ=

− +−

nte

asL

nat

n or 1)(!

+− nasn as the case may be

Examples :-

1. Find Lf(t) given f(t) = t, 0 < t < 3

4, t > 3

Here

Lf(t) = ∫ ∫ ∫∞ ∞

−−− +=0

3

0 3

4)( dtetdtedttfe ststst

Integrating the terms on the RHS, we get

Lf(t) = )1(11 32

3 ss es

es

−− −+

This is the desired result.

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2. Find Lf(t) given f(t) = sin2t, 0 < t ≤ π

0, t > π

Here

Lf(t) = ∫∫∞

−− +π

π

dttfedttfe stst )()(0

= ∫ −π

0

2sin tdte st

= { }π

02 2cos22sin

4

−−

+

−

ttsse st

= [ ]ses

π−−+

14

22


3. Evaluate : (i) L(sin3t sin4t)

(ii) L(cos2 4t)

(iii) L(sin32t)

(i) Here

L(sin3t sin4t) = L [ )]7cos(cos21 tt −

= [ ])7(cos)(cos21 tLtL − , by using linearity property

= )49)(1(

244912

12222 ++

=

+−

+ sss

ss

ss

(ii) Here

L(cos24t) =

++=

+

641

21)8cos1(

21

2ss

stL

(iii) We have

( )θθθ 3sinsin341sin3 −=

For θ=2t, we get

( )ttt 6sin2sin3412sin3 −=

so that

)36)(4(

4836

64

641)2(sin 2222

3

++=

+−

+=

sssstL


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4. Find L(cost cos2t cos3t)

Here

cos2t cos3t = ]cos5[cos21 tt +

so that

cost cos2t cos3t = ]coscos5[cos21 2 ttt +

= ]2cos14cos6[cos41 ttt +++

Thus

L(cost cos2t cos3t) =

+++

++

+ 41

163641

222 ss

sss

ss

5. Find L(cosh22t)

We have

2

2cosh1cosh2 θθ +=

For θ = 2t, we get

2

4cosh12cosh2 tt +=

Thus,

−+=

161

21)2(cosh 2

2

ss

stL

-------------------------------------------------------05.04.05-------------------------

6. Evaluate (i) L( t ) (ii)

t

L 1 (iii) L(t-3/2)

We have L(tn) = 1)1(

+

+Γns

n

(i) For n= 21 , we get

L(t1/2) = 2/3

)121(

s

+Γ

Since )()1( nnn Γ=+Γ , we have 22

1211

21 π

=

Γ=

+Γ

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Thus, 2

32

)(s

tL π=

(ii) For n = -21 , we get

ss

tL π=

Γ

=−

21

21 2

1

)(

(iii) For n = -23 , we get

sss

tL ππ 2221

)(2

12

12

3−=

−=

−Γ

=−−

−

7. Evaluate : (i) L(t2) (ii) L(t3)

We have,

L(tn) = 1!+ns

n

(i) For n = 2, we get

L(t2) = 332!2ss

=

(ii) For n=3, we get

L(t3) = 446!3ss

=

8. Find L[e-3t (2cos5t – 3sin5t)]

Given =

2L(e-3t cos5t) – 3L(e-3t sin5t)

= ( )25)3(

1525)3(

32 22 ++−

+++

sss , by using shifting property

= 346

922 ++

−ss

s , on simplification

9. Find L[coshat sinhat]

Here

L[coshat sinat] = ( )

+ −

ateeLatat

sin2

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=

++

++− 2222 )()(2

1aas

aaas

a

, on simplification

10. Find L(cosht sin3 2t)

Given

−

+ −

46sin2sin3

2tteeL

tt

= ( )[ ])6sin()2sin(3)6sin(2sin381 teLteLteLteL tttt −− −+−⋅

=

++

−++

++−

−+− 36)1(

64)1(

636)1(

64)1(

681

2222 ssss

=

++

−++

++−

−+− 36)1(

124)1(

136)1(

14)1(

143

2222 ssss

11. Find )( 254 −− teL t

We have

L(tn) = 1)1(

+

+Γns

n Put n= -5/2. Hence

L(t-5/2) = 2/32/3 34)2/3(

−− =−Γ

ssπ Change s to s+4.

Therefore, 2/32/54

)4(34)( −

−−

+=

steL t π

Transform of tn f(t) Here we suppose that n is a positive integer. By definition, we have

F(s) = ∫∞

−

0

)( dttfe st

Differentiating ‘n’ times on both sides w.r.t. s, we get

∫∞

−

∂∂

=0

)()( dttfes

sFdsd st

n

n

n

n

Performing differentiation under the integral sign, we get

])][()[()2(

2222

22

aasaasasa

+++−+

=

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∫∞

−−=0

)()()( dttfetsFdsd stn

n

n

Multiplying on both sides by (-1)n , we get

∫∞

− ==−0

)]([)(()()1( tftLdtetftsFdsd nstn

n

nn , by definition

Thus,

L[tnf(t)]= )()1( sFdsd

n

nn−

This is the transform of tn f(t).

Also,

)()1()(1 tftsFdsdL nn

n

n

−=

−

In particular, we have

L[t f(t)] = )(sFdsd

− , for n=1

L[t2 f(t)]= )(2

2

sFdsd , for n=2, etc.

Also,

)()(1 ttfsFdsdL −=

− and

)()( 22

21 tftsF

dsdL =

−

Transform of t

f(t)

We have , F(s) = ∫∞

−

0

)( dttfe st

Therefore,

∫∞

∫∞

∫∞ −=

s sdsdttfstedssF

0)()( = dtdsetf

s

st∫ ∫∞ ∞

−

0

)(

= ∫∞ ∞−

−0

)( dtt

etfs

st

= ∫∞

−

=

0

)()(ttfLdt

ttfe st

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Thus,

∫∞

=

s

dssFttfL )()(

This is the transform of ttf )(

Also,

∫∞

− =s t

tfdssFL )()(1

Examples : 1. Find L[te-t sin4t]

We have,

16)1(

4]4sin[ 2 ++=−

steL t

So that,

L[te-t sin4t] =

++−

17214 2 ssds

d

= 22 )172()1(8

+++ss

s

2. Find L(t2 sin3t)

We have

L(sin3t) = 9

32 +s

So that,

L(t2 sin3t) =

+ 93

22

2

sdsd = 22 )9(

6+

−s

sdsd = 32

2

)9()3(18

+−

ss

3. Find

−

tteL

t sin

We have

1)1(

1)sin( 2 ++=−

steL t

Hence

−

tteL

t sin = [ ]∫∞

∞− +=++0

12 )1(tan

1)1( sss

ds

= )1(tan2

1 +− − sπ = cot –1 (s+1)

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4. Find

ttL sin . Using this, evaluateL

tatsin

We have

L(sint) = 1

12 +s

So that

Lf(t) =

ttL sin = [ ]∞−

∞

=+∫ S

s

ss

ds 12 tan

1

= )(cottan2

11 sFss ==− −−π

Consider

L

tatsin = a L )(sin ataLf

atat

=

=

asF

aa 1 , in view of the change of scale property

=

−

as1cot

5. Find L

−

tbtat coscos

We have

L [cosat – cosbt] = 2222 bss

ass

+−

+

So that

L

−

tbtat coscos = ds

bss

ass

s∫∞

+−

+ 2222 = ∞

++

sbsas

22

22

log21

=

++

−

++

∞→ 22

22

22

22

loglog21

bsas

bsasLt

s

=

++

+ 22

22

log021

asbs =

++

22

22

log21

asbs

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6. Prove that ∫∞

− =0

3

503sin tdtte t

We have

∫∞

− =0

)sin(sin ttLtdtte st = )(sin tLdsd

− =

+−

11

2sdsd

= 22 )1(2+ss

Putting s = 3 in this result, we get

∫∞

− =0

3

503sin tdtte t

This is the result as required.

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ASSIGNMENT I Find L f(t) in each of the following cases :

1. f(t) = et , 0 < t < 2

0 , t > 2

2. f(t) = 1 , 0 ≤ t ≤ 3

t , t > 3

3. f(t) = at , 0 ≤ t < a

1 , t ≥ a

II. Find the Laplace transforms of the following functions :

4. cos(3t + 4)

5. sin3t sin5t

6. cos4t cos7t

7. sin5t cos2t

8. sint sin2t sin3t

9. sin2 5t

10. sin2(3t+5)

11. cos3 2t

12. sinh2 5t

13. t5/2

14. 3

1

+

tt

15. 3t

16. 5-t

17. e-2t cos2 2t

18. e2t sin3t sin5t

19. e-t sin4t + t cos2t

20. t2 e-3t cos2t

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21. te t21 −−

22. t

ee btat −− −

23. t

t2sin

24. t

tt 5sin2sin2

III. Evaluate the following integrals using Laplace transforms :

25. ∫∞

−

0

2 5sin tdtte t

26. ∫∞

−

0

35 cos tdtte t

27. ∫∞ −−

−

0

dtt

ee btat

28. ∫∞ −

0

sin dtt

te t

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-------------------------------------------- 28.04.05 -------------------------- Transforms of the derivatives of f(t) Consider

L )(tf ′ = ∫∞

− ′0

)( dttfe st

= [ ] ∫∞

−∞− −−0

0 )()()( dttfestfe stst , by using integration by parts

= [ ] )()0()(( tsLfftfeLt st

t+−−

∞→

= 0 - f (0) + s Lf(t) Thus L )(tf ′ = s Lf(t) – f(0) Similarly, L )(tf ′′ = s2 L f(t) – s f(0) - )0(f ′ In general, we have )0(.......)0()0()()( 121 −−− −−′−−= nnnnn ffsfstLfstLf

Transform of ∫t

0

f(t)dt

Let φ (t) = ∫t

dttf0

)( . Then φ(0) = 0 and φ′ (t) = f(t)

Now,

L φ(t) = ∫∞

−

0

)( dtte stφ

= ∫∞ −∞−

−′−

− 00

)()( dts

ets

etstst

φφ

= ∫∞

−+−0

)(1)00( dtetfs

st

Thus,

∫ =t

tLfs

dttfL0

)(1)(

Also,

∫=

−

t

dttftLfs

L0

1 )()(1

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Examples : 1. By using the Laplace transform of sinat, find the Laplace transform of cosat. Let

f(t) = sin at, then Lf(t) = 22 asa+

We note that atatf cos)( =′ Taking Laplace transforms, we get )(cos)cos()( ataLataLtfL ==′

or L(cosat) = [ ])0()(1)(1 ftsLfa

tfLa

−=′

=

−

+01

22 assa

a

Thus

L(cosat) = 22 ass+


2. Given 2/3

12s

tL =

π, show that

stL 11

=

π

Let f(t) = πt2 , given L[f(t)] = 2/3

1s

We note that, tt

tfππ1

212)( ==′

Taking Laplace transforms, we get

=′

tLtfL

π1)(

Hence

)0()()(1 ftsLftfLt

L −=′=

π

= 012/3 −

ss

Thus

st

L 11=

π


3. Find ∫

−t

dtt

btatL0

coscos

Here

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Lf(t) =

++

=

−

22

22

log21coscos

asbs

tbtatL

Using the result

L ∫ =t

tLfs

dttf0

)(1)(

We get,

∫

−t

dtt

btatL0

coscos =

++

22

22

log21

asbs

s

4. Find ∫ −t

t tdtteL0

4sin

Here

[ ] 22 )172()1(84sin

+++

=−

ssstteL t

Thus

∫ −t

t tdtteL0

4sin = 22 )172()1(8

+++sss

s

Transform of a periodic function A function f(t) is said to be a periodic function of period T > 0 if f(t) = f(t + nT) where n=1,2,3,….. The graph of the periodic function repeats itself in equal intervals. For example, sint, cost are periodic functions of period 2π since sin(t + 2nπ) = sin t, cos(t + 2nπ) = cos t.

The graph of f(t) = sin t is shown below :

Note that the graph of the function between 0 and 2π is the same as that between 2π and 4π and so on.

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The graph of f(t) = cos t is shown below :

Note that the graph of the function between 0 and 2π is the same as that between 2π and 4π and so on. Formula : Let f(t) be a periodic function of period T. Then

∫ −−−

=T

stST dttfe

etLf

0

)(1

1)(

Proof : By definition, we have

L f(t) = ∫ ∫∞ ∞

−− =0 0

)()( duufedttfe sust

= ∞+++++ ∫∫∫+

−−− ....)(.......)()()1(2

0

Tn

nT

suT

T

suT

su duufeduufeduufe

= ∑ ∫∞

=

+−

0

)1(

)(n

Tn

nT

su duufe

Let us set u = t + nT, then

L f(t) = ∑ ∫∞

= =

+− +0 0

)( )(n

T

t

nTts dtnTtfe

Here f(t+nT) = f(t), by periodic property Hence

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∫∑ −∞

=

−=T

st

n

nsT dttfeetLf00

)()()(

= ∫ −−

−

Tst

ST dttfee 0

)(1

1 , identifying the above series as a geometric series.

Thus

L f(t) = ∫ −−

−

Tst

sT dttfee 0

)(1

1

This is the desired result. Examples:- 1. For the periodic function f(t) of period 4, defined by f(t) = 3t, 0 < t < 2 6 , 2 < t < 4 find L f(t) Here, period of f(t) = T = 4 We have,

L f(t) = ∫ −−

−

Tst

sT dttfee 0

)(1

1 = ∫ −−

−

4

04 )(

11 dttfee

sts

=

+

− ∫∫ −−−

4

2

2

04 63

11 dtedttee

ststs

=

−

+

−−

−−

−−−

− ∫4

2

2

0

2

04 6.13

11

sedt

se

set

e

ststst

s

= ( )

−−−

−−

− 2

42

4

2131

1s

seee

ss

s

Thus,

Lf(t) = )1(

)21(342

42

s

ss

essee

−

−−

−−−

2. A periodic function of period ωπ2 is defined by

Esinωt, 0 ≤ t < ωπ

f(t) =

0 , ωπ ≤ t ≤

ωπ2

where E and ω are positive constants. Show that

L f(t) = )1)(( /22 wsews

Eπ

ω−−+

Here

23 of 133



T = ωπ2 . Therefore

L f(t) = ∫ −−−

ωπ

ωπ

/2

0)/2( )(

11 dttfe

est

s = ∫ −−−

ωπ

ωπ ω/

0)/2( sin

11 tdtEe

est

s

= { }ωπ

ωπ ωωωω

/

022)/2( cossin

1

−−

+−

−

− ttss

ee

E st

s

= 22

/

)/2(

)1(1 ω

ω ωπ

ωπ ++

−

−

− se

eE s

s

= ))(1)(1(

)1(22//

/

ωω

ωπωπ

ωπ

++−+

−−

−

seeeE

ss

s

= ))(1( 22/ ω

ωωπ +− − seE

s

This is the desired result. 3. A periodic function f(t) of period 2a, a>0 is defined by E , 0 ≤ t ≤ a f(t) =

-E, a < t ≤ 2a show that

L f(t) =

2tanh as

sE

Here T = 2a. Therefore

L f(t) = ∫ −−−

ast

as dttfee

2

02 )(

11

=

−+

− ∫ ∫ −−−

a a

a

ststas dtEedtEe

e 0

2

211

= ( )[ ])(1)1(

22

asassaas eee

esE −−−

− −+−−

= ( )[ ])1)(1(

)1(1)1(

22

2 asas

asas

as eeseEe

esE

−−

−−

− +−−

=−−

=

=

+−

−

−

2tanh2/2/

2/2/ assE

eeee

sE

asas

asas

This is the result as desired.

24 of 133



----------------------------------------------- 29.04.05 -------------------------------------- Step Function : In many Engineering applications, we deal with an important discontinuous function H(t-a) defined as follows :

0, t ≤ a H (t-a) =

1, t > a

where a is a non-negative constant.

This function is known as the unit step function or the Heaviside function. The function is named after the British electrical engineer Oliver Heaviside. The function is also denoted by u(t-a). The graph of the function is shown below:

H(t-a)

1 0 a t Note that the value of the function suddenly jumps from value zero to the value 1 as at → from the left and retains the value 1 for all t>a. Hence the function H(t-a) is called the unit step function. In particular, when a=0, the function H(t-a) become H(t), where 0 , t ≤ 0 H(t) = 1 , t > 0 Transform of step function By definition, we have

L[H(t-a)] = ∫∞

− −0

)( dtatHe st

= ∫ ∫∞

−− +a

a

stst dtedte0

)1(0

= s

e as−

25 of 133



In particular, we have L H(t) = s1

Also, )(1 atHs

eLas

−=

−− and )(11 tH

sL =

−

Heaviside shift theorem Statement :- L [f(t-a) H(t-a)] = e-as Lf(t) Proof :- We have

L [f(t-a) H(t-a)] = ∫∞

−−−0

)()( dteatHatf st

= ∫∞

− −a

st dtatfe )(

Setting t-a = u, we get

L[f(t-a) H(t-a)] = duufe uas )(0

)(∫∞

+−

= e-as L f(t) This is the desired shift theorem. Also, L-1 [e-as L f(t)] = f(t-a) H(t-a) Examples : 1. Find L[et-2 + sin(t-2)] H(t-2) Let f(t-2) = [et-2 + sin(t-2)] Then f(t) = [et + sint] so that

L f(t) = 1

11

12 +

+− ss

By Heaviside shift theorem, we have L[f(t-2) H(t-2)] = e-2s Lf(t) Thus,

++

−=−−+ −−

11

11)2()]2sin([ 2

2)2(

ssetHteL st

2. Find L(3t2 +2t +3) H(t-1) Let f(t-1) = 3t2 +2t +3

26 of 133



so that f(t) = 3(t+1)2 +2(t+1) +3 = 3t2 +8t +8

Hence

sss

tLf 886)( 23 ++=

Thus L[3t2 +2t +3] H(t-1) = L[f(t-1) H(t-1)] = e-s L f(t)

=

++−

ssse s 886

23

3. Find Le-t H(t-2) Let f(t-2) = e-t , so that, f(t) = e-(t+2) Thus,

L f(t) = 1

2

+

−

se

By shift theorem, we have

1

)()]2()2([)1(2

2

+==−−

+−−

setLfetHtfL

ss

Thus

[ ]1

)2()1(2

+=−

+−−

setHeL

st

4. Let f(t) = f1 (t) , t ≤ a f2 (t), t > a Verify that f(t) = f1(t) + [f2(t) – f1(t)]H(t-a) Consider f1(t) + [f2(t) – f1(t)]H(t-a) = f1(t) + f2 (t) – f1(t), t > a 0 , t ≤ a = f2 (t), t > a f1(t), t ≤ a = f(t), given Thus the required result is verified. 5. Express the following functions in terms of unit step function and hence find their Laplace transforms. 1. t2 , 1 < t ≤ 2 f(t) = 4t , t > 2 2. cost, 0 < t < π

27 of 133



f(t) = sint, t > π 1. Here, f(t) = t2 + (4t-t2) H(t-2) Hence,

L f(t) = )2()4(2 23 −−+ tHttL

s (i)

Let φ(t-2) = 4t – t2 so that φ(t) = 4(t+2) – (t+2)2 = -t2 + 4 Now,

ss

tL 42)( 3 +−=φ

Expression (i) reads as

L f(t) = [ ])2()2(23 −−+ tHtL

sφ

= )(2 23 tLe

ss φ−+

=

−+ −

32

3

242ss

es

s

This is the desired result. 2. Here f(t) = cost + (sint-cost)H(t-π) Hence,

L f(t) = )()cos(sin12 π−−+

+tHttL

ss (ii)

Let φ (t-π) = sint – cost Then φ(t) = sin(t + π) – cos(t + π) = -sint + cost so that

L φ(t) = 11

122 +

++

−s

ss

Expression (ii) reads as

L f(t) = [ ])()(12 ππφ −−+

+tHtL

ss

= )(12 tLe

ss s φπ−++

=

+−

++

−

11

1 22 sse

ss sπ

28 of 133



---------------------------------- 3.05.04 --------------------------- CONVOLUTION

The convolution of two functions f(t) and g(t) denoted by f(t) ∗ g(t) is defined as

f(t) ∗ g(t) = ∫ −t

duugutf0

)()(

Property : f(t) ∗ g(t) = g(t) ∗ f(t) Proof :- By definition, we have

f(t) ∗ g(t) = ∫ −t

duugutf0

)()(

Setting t-u = x, we get

f(t) ∗ g(t) = ∫ −−0

))(()(t

dxxtgxf

= ∫ ∗=−t

tftgdxxfxtg0

)()()()(

This is the desired property. Note that the operation ∗ is commutative. Convolution theorem :- L[f(t) ∗ g(t)] = L f(t) . L g(t) Proof :- Let us denote

f(t) ∗ g(t) = φ(t) = ∫ −t

duugutf0

)()(

Consider

∫ ∫∞

− −=0 0

)]()([)]([t

st dtugutfetL φ

= ∫ ∫∞

− −0 0

)()(t

st duugutfe (1)

We note that the region for this double integral is the entire area lying between the lines u =0 and u = t. On changing the order of integration, we find that t varies from u to ∞ and u varies from 0 to ∞. u u=t t=u t = ∞ 0 u=0 t Hence (1) becomes

L[φ(t)] = ∫ ∫∞

=

∞

=

− −0

)()(u ut

st dtduugutfe

29 of 133



= ∫ ∫∞ ∞

−−−

−0

)( )()( dudtutfeugeu

utssu

= ∫ ∫∞ ∞

−−

0 0

)()( dudvvfeuge svsu , where v = t-u

= ∫ ∫∞ ∞

−−

0 0

)()( dvvfeduuge svsu

= L g(t) . L f(t) Thus L f(t) . L g(t) = L[f(t) ∗ g(t)] This is desired property. Examples : 1. Verify Convolution theorem for the functions f(t) and g(t) in the following cases : (i) f(t) = t, g(t) = sint (ii) f(t) =t, g(t) = et (i) Here,

f ∗ g = ∫ −t

duutguf0

)()( = ∫ −t

duutu0

)sin(

Employing integration by parts, we get f ∗ g = t – sint so that

L [f ∗ g] = )1(

11

112222 +

=+

−ssss

(1)

Next consider

L f(t) . L g(t) = )1(

11

112222 +

=+

⋅ssss

(2)

From (1) and (2), we find that L [f ∗ g] = L f(t) . L g(t) Thus convolution theorem is verified. (ii) Here

f ∗ g = ∫ −t

ut duue0

Employing integration by parts, we get f ∗ g = et – t – 1 so that

L[f ∗ g] = )1(

1111

122 −

=−−− sssss

(3)

Next

L f(t) . L g(t) = )1(

11

1122 −

=−

⋅ssss

(4)

From (3) and (4) we find that

30 of 133



L[f ∗ g] = L f(t) . L g(t) Thus convolution theorem is verified. 2. By using the Convolution theorem, prove that

∫ =t

tLfs

dttfL0

)(1)(

Let us define g(t) = 1, so that g(t-u) = 1 Then

∫ ∫ ∗=−=t t

gfLdtutgtfLdttfL0 0

][)()()(

= L f(t) . L g(t) = L f(t) . s1

Thus

∫ =t

tLfs

dttfL0

)(1)(

This is the result as desired. 3. Using Convolution theorem, prove that

∫ ++=−−

tu

ssduuteL

02 )1)(1(

1)sin(

Let us denote, f(t) = e-t g(t) = sin t, then

∫ ∫ −=−−t t

u duutgufLduuteL0 0

)()()sin( = L f(t) . L g(t)

= )1(

1)1(

12 +

⋅+ ss

= )1)(1(

12 ++ ss

This is the result as desired.

31 of 133



ASSIGNMENT 1. By using the Laplace transform of coshat, find the Laplace transform of sinhat.

2. Find (i) ∫t

dtt

tL0

sin (ii) ∫ −t

t tdteL0

cos (iii) ∫ −t

t tdteL0

cos

(iv) ∫

t

t dtet

tL0

sin (v) ∫t

atdttL0

2 sin

3. If f(t) = t2, 0 < t < 2 and f(t+2) = f(t) for t > 2, find L f(t) 4. Find L f(t) given f(t) = t , 0 ≤ t ≤ a f(2a+t) = f(t) 2a – t, a < t ≤ 2a

5. Find L f(t) given f(t) = 1, 0 < t < 2a f(a+t) = f(t)

-1, 2a < t < a

6. Find the Laplace transform of the following functions : (i) et-1 H(t-2) (ii) t2 H(t-2) (iii) (t2 + t + 1) H(t +2) (iv) (e-t sint) H(t - π) 7. Express the following functions in terms of unit step function and hence find their Laplace transforms : (i) 2t, 0 < t ≤ π (ii) t2 , 2 < t ≤ 3 f(t) = f(t) = 1 , t > π 3t , t > 3 (iii) sin2t, 0 < t ≤ π (iv) sint, 0 < t ≤ π/2 f(t) = f(t) = 0 , t > π cost, t > π/2 8. Let f1 (t) , t ≤ a f(t) = f2 (t) , a<t ≤ b f3 (t) , t > b Verify that f(t) = f1(t) + [f2(t) – f1(t)]H(t-a) + [f3(t) - f2(t)] H(t-b) 9. Express the following function in terms of unit step function and hence find its Laplace transform. Sint, 0 < t ≤ π f(t) = Sin2t, π < t ≤ 2π Sin3t, t > 2π

32 of 133



10. Verify convolution theorem for the following pair of functions: (i) f(t) = cosat, g(t) = cosbt (ii) f(t) = t, g(t) = t e-t (iii) f(t) = et g(t) = sint 11. Using the convolution theorem, prove the following:

(i) ∫ ++=− −

tu

sssudueutL

022

1

)1()1(cos)(

(ii) 220 )(

1)(ass

duueutLt

au

+=−∫ −

33 of 133



INVERSE LAPLACE TRANSFORMS Let L f(t) = F(s). Then f(t) is defined as the inverse Laplace transform of F(s) and is denoted by L-1 F(s). Thus L-1 F(s) = f(t).

Linearity Property Let L-1 F(s) = f(t) and L-1 G(s) = g(t) and a and b be any two constants. Then

L-1 [a F(s) + b G(s)] = a L-1 F(s) + b L-1 G(s)

Table of Inverse Laplace Transforms

F(s) )()( 1 sFLtf −=

0,1>s

s

1

asas

>−

,1 ate

0,22 >+

sas

s

Cos at

0,122 >

+s

as aatSin

asas

>−

,122

aath Sin

as

ass

>−

,22

ath Cos

0,11 >+ s

sn

n = 0, 1, 2, 3, . . . !n

tn

0,11 >+ s

sn

n > -1 ( )1+Γ ntn

Examples 1. Find the inverse Laplace transforms of the following:

22)(

asbsii

++

22 994

25452)(

ss

ssiii

−−

++−

Here

521)(−s

i

34 of 133



25

11

21

25

121

521)(

t

es

Ls

Li =−

=−

−−

at

abat

asLb

assL

asbsLii sincos1)( 22

122

122

1 +=+

++

=++ −−−

92

94

4252

5

42

984

25452)( 2

1

2

122

1

−

−−

+

−=

−−

++− −−−

s

sL

s

sL

ss

ssLiii

−−

−= ththtt 3sin

233cos4

25sin

25cos

21

Evaluation of L-1 F(s – a) We have, if L f(t) = F(s), then L[eat f(t)] = F(s – a), and so L-1 F(s – a) = eat f(t) = e at L-1 F(s)

Examples

( )41

113: Evaluate 1.

++−

ssL

( )( ) ( ) ( )4

13

14

1-

112

113

111-1s3 L Given

+−

+=

+++

= −−

sL

sL

s

41

31 1213

sLe

sLe tt −−−− −=

Using the formula

getweandntakingandnt

sL

n

n ,32!

11

1 ==+−

323 32 teteGiven

tt −−

−=

52s-s2sL : Evaluate 2. 2

1-

++

35 of 133



( )( )

( ) ( ) ( ) 4113

411

4131

412 L Given 2

12

12

12

1-

+−+

+−

−=

+−

+−=

+−

+= −−−

sL

ssL

ssL

ss

413

4 21

21

++

+= −

sLe

ss Le t-t

tet e tt 2sin

232cos +=

1312: Evaluate 2

1

+++−

sssL

( )

( )( )

( ) ( )

−+−

−+

+=

−+

−+= −−

45

1

45

24

51

2L Given 2

23

12

23

23

12

23

23

1-

sL

s

sL

s

s

−−

−= −

−−

−

45

1

452

212

3

212

3

sLe

ssLe

tt

=

−

ththet

25sin

52

25cos2 2

3

sssss

2452 : Evaluate 4. 23

2

−+−+

havewe

( ) ( )( ) 1212452

2452

2452 2

2

2

23

2

−+

++=

−+−+

=−+−+

=−+−+

sC

sB

sA

sssss

sssss

sssss

Then

2s2+5s-4 = A(s+2) (s-1) + Bs (s-1) + Cs (s+2) For s = 0, we get A = 2, for s = 1, we get C = 1 and for s = -2, we get B = -1. Using these values in (1), we get

11

212

2452

23

2

−+

+−=

−+−+

ssssssss

Hence

tt ee

ssssL +−=−+−+ −− 2

22

21 2

25452

36 of 133



( ) ( )2154: Evaluate 5. 2

1

++++−

sssL

Let us take

( ) ( ) ( ) 2112154

22 ++

++

+=

++++

sC

sB

sA

sss

Then

4s + 5 = A(s + 2) + B(s + 1) (s + 2) + C (s + 1)2

For s = -1, we get A = 1, for s = -2, we get C = -3 Comparing the coefficients of s2, we get B + C = 0, so that B = 3. Using these values in (1), we get

( ) ( ) ( ) ( ) 23

13

11

2154

22 +−

++

+=

++++

ssssss

Hence

( ) ( ) sLe

sLe

sLe

sssL ttt 13131

2154 121

21

21 −−−−−−− −+=

++++

ttt eete 233 −−− −+=

44

31: Evaluate 5.

assL−

−

Let

)1(2244

3

asDCs

asB

asA

ass

++

++

+−

=−

Hence

s3 = A(s + a) (s2 + a2) + B (s-a)(s2+a2)+(Cs + D) (s2 – a2)

For s = a, we get A = ¼; for s = -a, we get B = ¼; comparing the constant terms, we get D = a(A-B) = 0; comparing the coefficients of s3, we get 1 = A + B + C and so C = ½. Using these values in (1), we get

2244

3

2111

41

ass

asasass

++

++

−=

− Taking inverse transforms, we get

37 of 133



[ ] ateeas

sL atat cos21

41

44

31 ++=

−−−

[ ]athat coscos21

+=

1: Evaluate 6. 24

1

++−

sssL

Consider

( ) ( ) ( )( )

+−++=

+−+++=

++ 112

21

111 222224 sssss

sssss

sss

( ) ( )( )( ) ( ) ( )

( ) ( )

+−

+=

++

++−

+−=

++−

+−=

+−+++−−++

=

−−−−

43

1

43

121

1

43

1

43

121

11

11

21

1111

21

2

121

2

121

241

2

212

21

2222

22

sLe

sLe

sssL

Therefore

ss

ssssssssssss

tt

−=−

2323sin

2323sin

21 2

121 t

et

ett

=

2sin

23sin

32 tht

Evaluation of L-1[e-as F(s)] We have, if Lf(t) = F(s), then L[f(t-a) H(t-a) = e-as F(s), and so

L-1[e-as F(s)] = f(t-a) H(t-a)

38 of 133



Examples

( )41

2:)1(

−

−−

seLEvaluate

ss

Here

( )

( )

( )( )( ) ( )5

65

)()(2

61

21)()(

21)(,5

352

4

51

32

412

411

4

−−

=

−−=−

==−

==

−==

−

−−

−−−

tHte

atHatfseL

Thus

tes

Les

LsFLtfTherefore

ssFa

t

s

tt

++

+

−−−

41:)2( 2

2

21

sse

seLEvaluate

ss ππ

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( )ππ

ππππ

ππππ

22coscos

222cossin

)1(

2cos4

)(

sin1

1)(

)1(22

21

2

21

1

21

−+−−=

−−+−−=

=+

=

=+

=

−−+−−=

−

−

tHttHt

tHttHtGiven

asreadsrelationNow

ts

sLtf

ts

LtfHere

tHtftHtfGiven

39 of 133



Inverse transform of logarithmic and inverse functions

( )[ ] ( ) .ttfL then F(s), f(t) L if have, We HencesFdsd

−==

( ) )(1 ttfsFdsdL =

−−

Examples

++−

bsasLEvaluate log:)1( 1

( ) ( )

( )

( ) [ ]

( )

( )b

eetfThus

eetftor

eesFdsdLthatSo

bsassF

dsdThen

bsasbsassFLet

atbt

atbt

btat

−−

−−

−−−

−=

−=

−−=

−

+−

+−=−

+−+=

++

=

1

11

logloglog)(

40 of 133



(2) Evaluate 1−L

−

sa1tan

( )

( )

( )

( )

( )

( ) ( )

( ) ( )∫

∫

=

=

=

=

=

−

+=−

=

−

−

−

t

t

dttfssFL

havewessFdttfSinceL

aattf

attftor

thatsoatsFdsdLor

asasF

dsdThen

sasFLet

0

1

0

1

22

1

,,

sin

sin

sin

tan)(

ssFoftransformInverse

Examples

( )

+

−22

1 1:)1(ass

LEvaluate

( )

( )

( )( )

( )2

0

122

1

1

22

cos1

sin1

sin)(

1

aat

dta

atssFL

assLThen

aatsFLtf

thatsoas

sFdenoteusLet

t

−=

==+

=

==+

=

∫−−

−

41 of 133



( )

( )

( )

( )[ ]

( )( )[ ]

( ) ( )[ ]

[ ]

[ ] ( ) ( )∫

∫

∫

−=∗=

=⋅=∗

==

−++=

+−=+

+−=

=+

=+

+

−

−−

−

−

−−

−−

−

t

atat

tat-

at

tat

at

duugutftgtfsGsFL

soandsGsFtLgtLf

eeata

dtateaass

L

atea

dtteass

LHence

teas

Lhavewe

assLEvaluate

0

1

3

0222

1

2

02

1

21

221

)()()()(

)()()()(g(t) f(t)L

thenG(s), Lg(t) and F(s) L(t) if have, We

1211

1111

get we this,Using

parts.by n integratioon,111

1

1

1:)2(

theorem nconvolutio using by F(s) of transform Inverse

transformLaplace inversefor n theoremconvolutio thecalled is expression This

NEXT CLASS 24.05.05

42 of 133



Examples

Employ convolution theorem to evaluate the following :

( )( )bsasL

++− 1)1( 1

( )( )( ) ( )

( )

baee

baee

dueedueebsas

L

e, g(t) ef(t)

bssG

as

atbttbaat

tubaat

tbuuta-

-bt-at

−−

=

−

−=

==++

==

+=

+=

−−−−

−−−−− ∫∫

1

1

n theorem,convolutioby Therefore,

get weinverse, theTaking

1)(,1 F(s) denote usLet

00

1

( )2221)2(

as

sL+

−

( ) ( )

( )

( )a

atta

auatatu

duauatata

duauutaaas

sL

at , g(t) at f(t)

asssG

asF(s)

t

t

t-

2sin

22cossin

2a1

formula angle compound usingby ,2

2sinsin1

cossin1

n theorem,convolutioby Hence

cosa

sin

Then)(,1 denote usLet

0

0

0222

1

2222

=

−−

−=

−+=

−=+

==

+=

+=

∫

∫

43 of 133



( )( )11)3( 2

1

+−−

sssL

Here

1)(,

11 2 +

=−

=s

ssGs

F(s)

Therefore

( )( ) ( )

( ) ( )[ ] [ ]ttettee

uueeduuess

L

t , g(t) ef(t)

ttt

tutut-

t

cossin211cossin

2

cossin2

sin11

1

have wen theorem,convolutioBy

sin

02

1

−−=−−−−=

−−==

+−

==

−

−−∫

Assignment

By employing convolution theorem, evaluate the following :

( )( ) ( )( )

( ) ( ) ( )

( ) ( ) ( )2154)6(1)3(

11)5(

11)2(

,)4(11

1)1(

21

2221

221

221

2222

21

21

+−+

+

+++

≠++++

−−

−−

−−

sssL

asL

ssL

sssL

babsas

sLss

L

44 of 133



LAPLACE TRANSFORM METHOD FOR DIFFERENTIAL EQUATIONS

As noted earlier, Laplace transform technique is employed to solve initial-value problems. The solution of such a problem is obtained by using the Laplace Transform of the derivatives of function and then the inverse Laplace Transform. The following are the expressions for the derivatives derived earlier.

(o)f - (o)f s - f(o) s - f(t) L s (t)f[ L

(o)f - f(o) s - f(t) L s (t)fL[

f(o) - f(t) L s (t)]fL[

23

2

′′′=′′′

′=′′

=′

Examples

1) Solve by using Laplace transform method 2 y(o) t y y , ==+′ −te

45 of 133



Taking the Laplace transform of the given equation, we get

( ) ( )[ ] ( )( )

( ) ( )( )

( )( )

( )( )

( ) ( )( )

( )

( )421

11

12

13114112

1342

get we, transformLaplace inverse theTaking

1342

1121s

becomes thiscondition,given theUsing

11

2

31

3

21

3

21

3

2

2

2

+=

++

+=

++−++−+

=

+++

=

+++

=

+=−+

+=+−

−

−

−

−

te

ssL

sssL

sssLtY

ssstyL

thatso

styL

styLoytysL

t

This is the solution of the given equation. Solve by using Laplace transform method :

0)()(,sin32)2( =′==−′+′′ oyoytyyy Taking the Laplace transform of the given equation, we get

[ ] [ ]1

1)(3)()(2)()()( 22

+=−−+′−−

styLoytLysoyosytLys

Using the given conditions, we get

46 of 133



[ ]

( )( )( )

( )( )( )

( )

equation.given theofsolution required theis This

sin2cos101

401

81

sums, partial of method theusingby 151

103

1401

11

81

131

1311)(

1311)(

1132)(

3

21

21

21

2

22

ttee

s

s

ssL

sDCs

sB

sAL

sssLty

orsss

tyLor

ssstyL

tt +−−=

+

−−+

+−

−=

++

++

+−

=

++−

=

++−=

+=−+

−

−

−

−

equation. integral thesolve tomethod Transform LaplaceEmploy 3)

( ) ( )∫ −+=t

0

sinlf(t) duutuf

( ) ( )

equation. integralgiven theofsolution theis This

211)(1)(

1)(1sin)(1

get wehere, n theorem,convolutio usingBy

sin1

get weequation,given theof transformLaplace Taking

2

3

21

3

2

2

t

0

ts

sLtfors

stfL

Thus

stfL

stLtLf

sL f(t)

duutufLs

L f(t)

+=

+=

+=

++=⋅+=

−+=

−

∫

47 of 133



s. transformLaplace using t any timeat particle theofnt displaceme thefind 10, is speed initial theand 20at x is particle theofposition initial theIf t.at time particle theofnt displaceme thedenotesx

where025dtdx

6dt

xdequation the,satisfyingpath a along moving is particleA 4)(

2

2

=

=++ x

get weequation, theof transformLaplace theTaking10. (o) x'20, x(o)are conditions initial theHere

025x(t)(t)6x'(t)'x'

asrewritten bemay equation Given

==

=++

[ ]

( )( )

( )

( ) ( )

problem.given theofsolution desired theis This

24sin354cos20

163170

163320

16370320

16313020x(t)

25613020Lx(t)

013020256ssLx(t)

33

21

21

21

21

2

2

tete

sL

ssL

ssL

ssL

thatso

sss

ors

tt

−−

−−

−−

+=

+++

+++

=

++++

=

+++

=

+++

=

=−−++

−

=

−− LRt

at eeaL-R

E is t any timeat current

that theShow R. resistance and L inductance ofcircuit a to0at t applied is Ee A voltage (5) -at

The circuit is an LR circuit. The differential equation with respect to the circuit is

)(tERidtdiL =+

Here L denotes the inductance, i denotes current at any time t and E(t) denotes the E.M.F. It is given that E(t) = E e-at. With this, we have Thus, we have

48 of 133



at

at

EetiRtLi

orEeRidtdiL

−

−

=+

=+

)()('

[ ] [ ] ( ) orER at−=+ eL(t)i'L(t)i'LL TTT

get wesides,both on )( transformLaplace Taking TL

[ ] [ ]as

EtiLRoitiLsL TT +=+−

1)()()(

[ ]

( ) ( )

))(()(get we,L transforminverse Taking

)(

)(get weo, i(o) Since

11-T RsLas

ELti

RsLasEtiL

oras

ERsLtiL

T

T

T

++=

+++=

+=+=

−

desired. asresult theis This

)(

11 11

−

−=

+−

+−=

−−

−−

LRt

at

TT

eeaLR

Eti

Thus

RsLLL

asL

aLRE

1. y(o) 2, with x(o)09dtdy

,04dtdxgiven t of in termsy andfor x equations ussimultaneo theSolve )6(

===−

=+

x

y

Taking Laplace transforms of the given equations, we get

[ ][ ] 0)()()(9

0)(4 x(o)- Lx(t) s

=−+−

=+

oytLystxL

tLy

get we,conditions initialgiven theUsing

1)(5)(9

2)(4)(=+−

=+tyLtxL

tyLtxLs

get weLy(t),for equations theseSolving

3618

2 ++

=ss L y(t)

thatso

49 of 133



tt

sss Ly(t)

6sin36cos

3618

36 221

+=

++

+= −

(1)

get we,09dtdyin thisUsing =− x

[ ]tt 6cos186sin6

91x(t) +−=

or

[ ]tt 6sin6cos3

32x(t) −=

(2) (1) and (2) together represents the solution of the given equations.

Assignment

Employ Laplace transform method to solve the following initial – value problems

( )

( )

( )∫

∫

=−+=

−+=

=′=−=+′′

−=−==+′+′′

−=

==+′′

=′=++=+′+′′

′===+′+′′

=′==+′+′′

=′′=′==−′−′′+′′′

−

−

t

tu

t

t

fduuutfttf

dueutftf

yytHyy

yytyyy

yytyy

yyttyyy

yyeyyy

yyeyyy

yyygivenyyyy

0

0

2

2

2

4)0(,cos)(')9

21)()8

1)0(,0)0(,1)7

1)1(,3)0(,2)6

12

,1)0(,2cos9)5

0)0(,2)0(,22223)4

)0(1)0(,34)3

1)0(,2)0(,65)2

6)0(,0)0()0(022)1

π

any time.at current thefind initially, zero iscurrent theIf 0. at t applied issin voltagea circuit, R-Lan In 11)

t.any timeat nt displaceme the

find rest,at initially is particle theIf .5sin805dtdx4

dtxd

equation by the governed is t any timeat point fixed a fromnt x displaceme its that so line a along moves particleA )10

2

2

=

=++

t E

tx

ω

50 of 133



3)0(,8)0(,2dtdy,23

dtdx)15

3)0(,8)0(,2dtdy,32

dtdx)14

0)0(,1)0(,sindtdy,

dtdx)13

0)0(,2)0(,cosdtdy,sin

dtdx 12)

:

===+=+

==−=−=

===+=−

===+=+

yxyxxy

yxxyyx

yxtxey

yxtxty

t

t of terms in y andx for equations aldifferenti ussimultaneo following the Solve

LAPLACE TRANSFORMS

INTRODUCTION

Laplace transform is an integral transform employed in solving physical problems.

Many physical problems when analysed assumes the form of a differential equation subjected to a set of initial conditions or boundary conditions.

By initial conditions we mean that the conditions on the dependent variable are specified at a single value of the independent variable.

If the conditions of the dependent variable are specified at two different values of the independent variable, the conditions are called boundary conditions.

The problem with initial conditions is referred to as the Initial value problem.

The problem with boundary conditions is referred to as the Boundary value problem.


dxyd

=++2

2

with conditions y(0) =

y′ (0) = 1 is an initial value problem

51 of 133




dxyd cos23 2

2

=++ with y(1)=1,

y(2)=3 is called Boundary value problem.

Laplace transform is essentially employed to solve initial value problems. This technique is of great utility in applications dealing with mechanical systems and electric circuits. Besides the technique may also be employed to find certain integral values also. The transform is named after the French Mathematician P.S. de’ Laplace (1749 – 1827).

The subject is divided into the following sub topics.

Definition :

Let f(t) be a real-valued function defined for all t ≥ 0 and s be a parameter, real or complex.

Suppose the integral ∫∞

−

0

)( dttfe st exists (converges). Then this integral is called the Laplace

transform of f(t) and is denoted by Lf(t).

Thus,

Lf(t) = ∫∞

−

0

)( dttfe st (1)

We note that the value of the integral on the right hand side of (1) depends on s. Hence Lf(t) is a function of s denoted by F(s) or )(sf .

LAPLACE TRANSFORMS

Definition and Properties

Transforms of some functions

Convolution theorem

Inverse transforms

Solution of differential equations

52 of 133



Thus,

Lf(t) = F(s) (2)

Consider relation (2). Here f(t) is called the Inverse Laplace transform of F(s) and is denoted by L-1 [F(s)].

Thus,

L-1 [F(s)] = f(t) (3)

Suppose f(t) is defined as follows :

f1(t), 0 < t < a

f(t) = f2(t), a < t < b

f3(t), t > b

Note that f(t) is piecewise continuous. The Laplace transform of f(t) is defined as

Lf(t) = ∫∞

−

0

)(tfe st

= ∫ ∫ ∫∞

−−− ++a b

a b

ststst dttfedttfedttfe0

321 )()()(

NOTE : In a practical situation, the variable t represents the time and s represents frequency.

Hence the Laplace transform converts the time domain into the frequency domain.

Basic properties

The following are some basic properties of Laplace transforms :

1. Linearity property : For any two functions f(t) and φ(t) (whose Laplace transforms exist)

and any two constants a and b, we have

L [a f(t) + b φ(t)] = a L f(t) + b Lφ(t)


L[af(t)+bφ(t)] = [ ]dttbtafe st∫∞

− +0

)()( φ = ∫ ∫∞ ∞

−− +0 0

)()( dttebdttfea stst φ

53 of 133



= a L f(t) + b Lφ(t)


In particular, for a=b=1, we have

L [ f(t) + φ(t)] = L f(t) + Lφ(t)

and for a = -b = 1, we have

L [ f(t) - φ(t)] = L f(t) - Lφ(t)

2. Change of scale property : If L f(t) = F(s), then L[f(at)] =

asF

a1 , where a is a positive

constant.


Lf(at) = ∫∞

−

0

)( dtatfe st (1)

Let us set at = x. Then expression (1) becomes,

L f(at) = ∫∞

−

0

)(1 dxxfea

xas

=

asF

a1


3. Shifting property :- Let a be any real constant. Then

L [eatf(t)] = F(s-a)


L [eatf(t)] = [ ]∫∞

−

0

)( dttfee atst = ∫∞

−−

0

)( )( dttfe as

= F(s-a)

This is the desired property. Here we note that the Laplace transform of eat f(t) can be

written down directly by changing s to s-a in the Laplace transform of f(t).

TRANSFORMS OF SOME FUNCTIONS 3. Let a be a constant. Then

L(eat) = ∫ ∫∞ ∞

−−− =0 0

)( dtedtee tasatst

54 of 133



= asas

e tas

−=

−−

∞−− 1)( 0

)(

, s > a

Thus,

L(eat) = as −

1

In particular, when a=0, we get

L(1) = s1 , s > 0

By inversion formula, we have

atat es

Leas

L ==−

−− 11 11

4. L(cosh at) =

+ −

2

atat eeL = [ ]∫∞

−− +02

1 dteee atatst

= [ ]∫∞

+−−− +0

)()(

21 dtee tastas

Let s > |a| . Then,

∞+−−−

+−

+−−

=0

)()(

)()(21)(cosh

ase

aseatL

tastas

= 22 ass−

Thus,

L (cosh at) = 22 ass−

, s > |a|

and so

atas

sL cosh221 =

−−

3. L (sinh at) = 222 asaeeL

atat

−=

− −

, s > |a|

Thus,

L (sinh at) = 22 asa−

, s > |a|

and so,

a

atas

L sinh122

1 =

−−

55 of 133



4. L (sin at) = ∫∞

−

0

sin ate st dt

Here we suppose that s > 0 and then integrate by using the formula

[ ]∫ −+

= bxbbxaba

ebxdxeax

ax cossinsin 22

Thus,

L (sinh at) = 22 asa+

, s > 0

and so

a

atas

L sinh122

1 =

+−

5. L (cos at) = atdte st cos0∫∞

−

Here we suppose that s>0 and integrate by using the formula

[ ]∫ ++

= bxbbxaba

ebxdxeax

ax sincoscos 22

Thus,

L (cos at) = 22 ass+

, s > 0

and so

atas

sL cos221 =

+−

6. Let n be a constant, which is a non-negative real number or a negative non-integer. Then

L(tn) = ∫∞

−

0

dtte nst

Let s > 0 and set st = x, then

∫ ∫∞ ∞

−+

− =

=

0 01

1)( dxxess

dxsxetL nx

n

nxn

The integral dxxe nx∫∞

−

0

is called gamma function of (n+1) denoted by )1( +Γ n . Thus

56 of 133



1

)1()( +

+Γ= n

n

sntL

In particular, if n is a non-negative integer then )1( +Γ n =n!. Hence

1

!)( += nn

sntL

and so

)1(

11

1

+Γ=+

−

nt

sL

n

n or !n

t n

as the case may be

57 of 133



TABLE OF LAPLACE TRANSFORMS

f(t) F(s)

1 s1 , s > 0

eat as −

1 , s > a

coshat 22 ass−

, s > |a|

sinhat 22 asa−

, s > |a|

sinat 22 asa+

, s > 0

cosat 22 ass+

, s > 0

tn, n=0,1,2,….. 1!+ns

n , s > 0

tn, n > -1 1)1(

+

+Γns

n , s > 0

Application of shifting property :-

The shifting property is

If L f(t) = F(s), then L [eatf(t)] = F(s-a)

Application of this property leads to the following results :

1. [ ]ass

assat

bssbtLbteL

−→−→

−== 22)(cosh)cosh( = 22)( bas

as−−

−

Thus,

L(eatcoshbt) = 22)( basas−−

−

and

btebas

asL at cosh)( 22

1 =−−

−−

58 of 133



2. 22)()sinh(

basabteL at

−−=

and

btebas

L at sinh)(1

221 =

−−−

3. 22)()cos(

basasbteL at

+−−

=

and

btebas

asL at cos)( 22

1 =+−

−−

4. 22)()sin(

basbbteL at

−−=

and

b

btebas

Lat sin

)(1

221 =

−−−

5. 1)()1()( +−

+Γ= n

nat

asnteL or 1)(

!+− nas

n as the case may be

Hence

)1()(

11

1

+Γ=

− +−

nte

asL

nat

n or 1)(!

+− nasn as the case may be

Examples :-

1. Find Lf(t) given f(t) = t, 0 < t < 3

4, t > 3

Here

Lf(t) = ∫ ∫ ∫∞ ∞

−−− +=0

3

0 3

4)( dtetdtedttfe ststst

Integrating the terms on the RHS, we get

Lf(t) = )1(11 32

3 ss es

es

−− −+

59 of 133




2. Find Lf(t) given f(t) = sin2t, 0 < t ≤ π

0, t > π

Here

Lf(t) = ∫∫∞

−− +π

π

dttfedttfe stst )()(0

= ∫ −π

0

2sin tdte st

= { }π

02 2cos22sin

4

−−

+

−

ttsse st

= [ ]ses

π−−+

14

22


3. Evaluate : (i) L(sin3t sin4t)

(iv) L(cos2 4t)

(v) L(sin32t)

(i) Here

L(sin3t sin4t) = L [ )]7cos(cos21 tt −

= [ ])7(cos)(cos21 tLtL − , by using linearity property

= )49)(1(

244912

12222 ++

=

+−

+ sss

ss

ss

(ii) Here

L(cos24t) =

++=

+

641

21)8cos1(

21

2ss

stL

(iii) We have

( )θθθ 3sinsin341sin3 −=

For θ=2t, we get

( )ttt 6sin2sin3412sin3 −=

so that

)36)(4(

4836

64

641)2(sin 2222

3

++=

+−

+=

sssstL


60 of 133



4. Find L(cost cos2t cos3t)

Here

cos2t cos3t = ]cos5[cos21 tt +

so that

cost cos2t cos3t = ]coscos5[cos21 2 ttt +

= ]2cos14cos6[cos41 ttt +++

Thus

L(cost cos2t cos3t) =

+++

++

+ 41

163641

222 ss

sss

ss

5. Find L(cosh22t)

We have

2

2cosh1cosh2 θθ +=

For θ = 2t, we get

2

4cosh12cosh2 tt +=

Thus,

−+=

161

21)2(cosh 2

2

ss

stL

6. Evaluate (i) L( t ) (ii)

t

L 1 (iii) L(t-3/2)

We have L(tn) = 1)1(

+

+Γns

n

(i) For n= 21 , we get

61 of 133



L(t1/2) = 2/3

)121(

s

+Γ

Since )()1( nnn Γ=+Γ , we have 22

1211

21 π

=

Γ=

+Γ

Thus, 2

32

)(s

tL π=

(ii) For n = -21 , we get

ss

tL π=

Γ

=−

21

21 2

1

)(

(iii) For n = -23 , we get

sss

tL ππ 2221

)(2

12

12

3−=

−=

−Γ

=−−

−

7. Evaluate : (i) L(t2) (ii) L(t3)

We have,

L(tn) = 1!+ns

n

(i) For n = 2, we get

L(t2) = 332!2ss

=

(ii) For n=3, we get

L(t3) = 446!3ss

=

8. Find L[e-3t (2cos5t – 3sin5t)]

Given =

2L(e-3t cos5t) – 3L(e-3t sin5t)

= ( )25)3(

1525)3(

32 22 ++−

+++

sss , by using shifting property

= 346

922 ++

−ss

s , on simplification

62 of 133



9. Find L[coshat sinhat]

Here

L[coshat sinat] = ( )

+ −

ateeLatat

sin2

=

++

++− 2222 )()(2

1aas

aaas

a

, on simplification

10. Find L(cosht sin3 2t)

Given

−

+ −

46sin2sin3

2tteeL

tt

= ( )[ ])6sin()2sin(3)6sin(2sin381 teLteLteLteL tttt −− −+−⋅

=

++

−++

++−

−+− 36)1(

64)1(

636)1(

64)1(

681

2222 ssss

=

++

−++

++−

−+− 36)1(

124)1(

136)1(

14)1(

143

2222 ssss

11. Find )( 254 −− teL t

We have

L(tn) = 1)1(

+

+Γns

n Put n= -5/2. Hence

L(t-5/2) = 2/32/3 34)2/3(

−− =−Γ

ssπ Change s to s+4.

Therefore, 2/32/54

)4(34)( −

−−

+=

steL t π

Transform of tn f(t) Here we suppose that n is a positive integer. By definition, we have

F(s) = ∫∞

−

0

)( dttfe st

Differentiating ‘n’ times on both sides w.r.t. s, we get

])][()[()2(

2222

22

aasaasasa

+++−+

=

63 of 133



∫∞

−

∂∂

=0

)()( dttfes

sFdsd st

n

n

n

n

Performing differentiation under the integral sign, we get

∫∞

−−=0

)()()( dttfetsFdsd stn

n

n

Multiplying on both sides by (-1)n , we get

∫∞

− ==−0

)]([)(()()1( tftLdtetftsFdsd nstn

n

nn , by definition

Thus,

L[tnf(t)]= )()1( sFdsd

n

nn−

This is the transform of tn f(t).

Also,

)()1()(1 tftsFdsdL nn

n

n

−=

−

In particular, we have

L[t f(t)] = )(sFdsd

− , for n=1

L[t2 f(t)]= )(2

2

sFdsd , for n=2, etc.

Also,

)()(1 ttfsFdsdL −=

− and

)()( 22

21 tftsF

dsdL =

−

Transform of t

f(t)

We have , F(s) = ∫∞

−

0

)( dttfe st

Therefore,

∫∞

∫∞

∫∞ −=

s sdsdttfstedssF

0)()( = dtdsetf

s

st∫ ∫∞ ∞

−

0

)(

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= ∫∞ ∞−

−0

)( dtt

etfs

st

= ∫∞

−

=

0

)()(ttfLdt

ttfe st

Thus,

∫∞

=

s

dssFttfL )()(

This is the transform of ttf )(

Also,

∫∞

− =s t

tfdssFL )()(1

Examples : 1. Find L[te-t sin4t]

We have,

16)1(

4]4sin[ 2 ++=−

steL t

So that,

L[te-t sin4t] =

++−

17214 2 ssds

d

= 22 )172()1(8

+++ss

s

2. Find L(t2 sin3t)

We have

L(sin3t) = 9

32 +s

So that,

L(t2 sin3t) =

+ 93

22

2

sdsd = 22 )9(

6+

−s

sdsd = 32

2

)9()3(18

+−

ss

3. Find

−

tteL

t sin

We have

1)1(

1)sin( 2 ++=−

steL t

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Hence

−

tteL

t sin = [ ]∫∞

∞− +=++0

12 )1(tan

1)1( sss

ds

= )1(tan2

1 +− − sπ = cot –1 (s+1)

4. Find

ttL sin . Using this, evaluateL

tatsin

We have

L(sint) = 1

12 +s

So that

Lf(t) =

ttL sin = [ ]∞−

∞

=+∫ S

s

ss

ds 12 tan

1

= )(cottan2

11 sFss ==− −−π

Consider

L

tatsin = a L )(sin ataLf

atat

=

=

asF

aa 1 , in view of the change of scale property

=

−

as1cot

5. Find L

−

tbtat coscos

We have

L [cosat – cosbt] = 2222 bss

ass

+−

+

So that

L

−

tbtat coscos = ds

bss

ass

s∫∞

+−

+ 2222 = ∞

++

sbsas

22

22

log21

=

++

−

++

∞→ 22

22

22

22

loglog21

bsas

bsasLt

s

=

++

+ 22

22

log021

asbs =

++

22

22

log21

asbs

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6. Prove that ∫∞

− =0

3

503sin tdtte t

We have

∫∞

− =0

)sin(sin ttLtdtte st = )(sin tLdsd

− =

+−

11

2sdsd

= 22 )1(2+ss

Putting s = 3 in this result, we get

∫∞

− =0

3

503sin tdtte t


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ASSIGNMENT I Find L f(t) in each of the following cases :

1. f(t) = et , 0 < t < 2

0 , t > 2

2. f(t) = 1 , 0 ≤ t ≤ 3

t , t > 3

3. f(t) = at , 0 ≤ t < a

2 , t ≥ a

II. Find the Laplace transforms of the following functions :

4. cos(3t + 4)

5. sin3t sin5t

6. cos4t cos7t

7. sin5t cos2t

8. sint sin2t sin3t

9. sin2 5t

10. sin2(3t+5)

11. cos3 2t

12. sinh2 5t

13. t5/2

14. 3

1

+

tt

15. 3t

16. 5-t

17. e-2t cos2 2t

18. e2t sin3t sin5t

19. e-t sin4t + t cos2t

20. t2 e-3t cos2t

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21. te t21 −−

22. t

ee btat −− −

23. t

t2sin

24. t

tt 5sin2sin2

Dr.G.N.Shekar

DERIVATIVES OF ARC LENGTH Consider a curve C in the XY plane. Let A be a fixed point on it. Let P and Q be two

neighboring positions of a variable point on the curve C. If ‘s’ is the distance of P from A

measured along the curve then ‘s’ is called the arc length of P. Let the tangent to C at P make

an angle ψ with X-axis. Then (s,ψ) are called the intrinsic co-ordinates of the point P. Let

the arc length AQ be s + δs. Then the distance between P and Q measured along the curve C

is δs. If the actual distance between P and Q is δC. Then δs=δC in the limit Q → P along C.

. . 1Q P

si e LtC

δδ→

=

ψ

0

y

x

P(s,ψ )

Q

A

s δs

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Cartesian Form:

Let ( )y f x= be the Cartesian equation of the curve C and let ( , ) ( , )P x y and Q x x y yδ δ+ +

be any two neighboring points on it as in fig.

Let the arc length PQ sδ= and the chord length PQ Cδ= . Using distance between two

points formula we have 2 2 2 2PQ = ( C) =( x) +( y)δ δ δ

222

11

+=

+=

∴

xy

xCor

xy

xC

δδ

δδ

δδ

δδ

2

1s s C s yx C x C x

δ δ δ δ δδ δ δ δ δ

⇒ = ⋅ = +

We note that δx → 0 as Q → P along C, also that when , 1sQ PC

δδ

→ =

∴ When Q → P i.e. when δx → 0, from (1) we get

2

1 (1)ds dydx dx

= + →

Similarly we may also write

2

1

+=⋅=

yx

Cs

yC

Cs

ys

δδ

δδ

δδ

δδ

δδ

and hence when Q → P this leads to

2

1 (2)ds dxdy dy

= + →

Parametric Form: Suppose ( ) ( )x x t and y y t= = is the parametric form of the curve C.

Then from (1)

( , )Q x x y yδ δ+ +

δC

( , )P x y

δs

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222

11

+

=

+=

dtdy

dtdx

dtdx

dtdx

dtdy

dxds

2 2

(3)ds ds dx dx dydt dx dt dt dt

∴ = ⋅ = + →

Note: Since ψ is the angle between the tangent at P and the X-axis,

we have tandydx

ψ=

( )2 21 1 tands y Secdx

ψ ψ′⇒ = + = + =

Similarly

( )

22 2

1 11 1 1 cot sectan

ds Cody y

ψ ψψ

= + = + = + =′

i.e. Cos dx dyand Sinds ds

ψ ψ= =

( ) ( ) ( )2 2

2 2 21dx dy ds dx dyds ds

∴ + = ⇒ = +

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We can use the following figure to observe the above geometrical connections

among , , .dx dy ds and ψ

Polar Curves :

Suppose ( )r f θ= is the polar equation of the curve C and ( , ) ( , )P r and Q r rθ δ θ δθ+ + be

two neighboring points on it as in figure:

Consider PN ⊥ OQ.

In the right-angled triangle OPN, We have PN PNSin PN r Sin rOP r

δθ δθ δθ= = ⇒ = =

since Sin = when . is very smallδθ δθ δθ

From the figure we see that, (1)ON ONCos ON r Cos r rOP r

δθ δθ= = ⇒ = = =

cos 1 0whenδθ δθ= →

( )NQ OQ ON r r r rδ δ∴ = − = + − =

dx

ds dy

ψ

C

N

O θ

δθ r

P(r,θ)

δs

x

( , )Q r rδ θ δθ+ +

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2 2 2 2 2 2, . , ( ) ( ) ( )From PNQ PQ PN NQ i e C r rδ δθ δ= + = +

22C rrδ δ

δθ δθ ⇒ = +

2

2S S C S rrC C

δ δ δ δ δδθ δ δθ δ δθ

∴ = = +

We note that , 0 1Swhen Q P along the curve alsoC

δδθδ

→ → =

22, (4)dS drwhen Q P r

d dθ θ ∴ → = + →

22 2 2 2, ( ) ( ) ( ) 1CSimilarly C r r r

r rδ δθδ δθ δδ δ

= + ⇒ = +

221S S C Sand r

r C r C rδ δ δ δ δθδ δ δ δ δ

= = +

22, 1 (5)dS dwhen Q P we get r

dr drθ ∴ → = + →

Note:

dWe know that Tan rdrθφ =

2

2 2 2 2 21 secds drr r r Cot r Cot rCod d

φ φ φθ θ

∴ = + = + = + =

Similarly

2

2 21 1ds dr Tan Secdr dr

θ φ φ = + = + =

1dr dCos and Sinds ds r

θφ φ∴ = =

The following figure shows the geometrical connections among ds, dr, dθ and φ

73 of 133



Thus we have :

22 2 2

1 , 1 ,ds dy ds dx ds dx dydx dx dy dy dt dt dt

= + = + = +

2 22 21ds d ds drr and r

dr dr d dθ

θ θ = + = +

Example 1: 2/3 2/3 2/3ds ds and for the curve xdx dy

y a+ =

2/3 2/3 2/3 -1/3 -1/32 2x x ' 03 3

y a y y+ = ⇒ + =

-1 13 3

-13

x'y

yyx

⇒ = = −

2 1/32 /3 2 /3 2 /3 2 /3

2 /3 2 /3 2 /3

dsHence 1 1dx

dy y x y a adx x x x x

+ = + = + = = =

Similarly

2 1/32 /3 2 /3 2 /3 2 /3

2 /3 2 /3 2 /3

ds 1 1dy

dx x x y a ady y y y y

+= + = + = = =

Example 2: 2

2 2

ds a for the curve y a log dx a -x

Find

=

( )2 2 22 2 2 2

2 2log log dy x axy a a a a x adx a x a x

− = − − ⇒ = − = − −

dr

ds r dθ

φ

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( )

( )( )

( )( )

22 2 2 2 2 22 2 2 2 2

2 2 2 22 2 2 2 2 2

441 1a x a x a xds dy a x a x

dx dx a xa x a x a x

− + + + ∴ = + = + = = = −− − −

Example 3: t tIf x ae Sint, y ae Cost, find dsdt

= =

t t tx ae Sint ae Sint ae Costdxdt

= ⇒ = +

t t ty ae Cost ae Cost ae Sintdydt

= ⇒ = −

( ) ( )2 2

2 22 2 2 2t tds dx dy a e Cos t Sin t a e Cos t Sin tdt dt dt

∴ = + = + + −

( )2 22 2t tae Cos t Sin t a e= + = ( ) ( ) ( )2 2 2 22a b a b a b+ + − = +

Example 4: tIf x a Cos t log Tan , y a Sin t, find 2

dsdt

= + =

2

122 22 2 2

tSecdx a Sin t a Sin tt t tdt Tan Sin Cos

= − + = − + ⋅

( )2 211 Sin t a Cos ta Sin t a a Cost Cot t

Sin t Sin t Sin t−

= − + = = = ⋅

dy a Cos tdt

=

2 2

2 2 2 2 2ds dx dy a Cos t Cot t a Cos tdt dt dt

∴ = + = +

( )2 2 2 1a Cos t Cot t= +

2 2 2 2 2seca Cos t Co t a Cot t a Cot t= ⋅ = =

Example 5: 3 3If x a Cos t, y Sin t, find dsdt

= =

2 23 , 3dx dya Cos t Sin t a Sin t Cos tdt dt

= − =

2 2ds dx dy

dt dt dt ∴ = +

2 4 2 2 4 29 9a Cos t Sin t a Sin t Cos t= +

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( )2 2 2 2 29 3a Cos t Sin t Cos t Sin t a Sin t Cos t= + =

Exercise:

Find dsdt

for the following curves:

(i) x=a(t + Sint), y=a(1-Cost) (ii) x=a(Cost + t Sint), y= aSint(iii) x= a log(Sect + tant), y= a Sect

76 of 133



Example 6: 2 2If r a Cos 2 , Show that r is constantdsd

θθ

=

22 2 22 2 2 2 2dr dr ar a Cos r a Sin Sin

d d rθ θ θ

θ θ−

= ⇒ = − ⇒ =

2 42 2 2 4 4 2

2

12 2ds dr ar r Sin r a Sind d r r

θ θθ θ

= + = + = +

4 4 2 4 2 4 22 2 2dsr r a Sin a Cos a Sind

θ θ θθ

∴ = + = +

2 2 2 22 Constanta Cos Sin aθ θ= + = = 2 2r is constant for r a Cos2dsd

θθ

∴ =

Example 7: 2 2

-1 r dsFor the curve Cos , Show that r is constant.k dr

k rr

θ − = −

2 2

2 2

22

2

2 (1)1 1 2

1

rr k rd k rdr k rr

k

θ −

− − − − = ⋅ −

−

( )2 2 2

2 2 2 2 2

1 r k r

k r r k r

+ −−= +

− −

2 2 2

2 2 2 2 2 2 2 2

1 k r kk r r k r r k r

− − += + =

− − −

2 2k rr−

=

( )2 2 2 2 22 2

21 2 1k rds d r k r kr r

dr dr r r rθ − + − ∴ = + = + = =

dsHence r k (constant)dr

=

Example 8: ( )2

2 2

ds dsFor a polar curve r f show that ,dr d

r rpr p

θθ

= = =−

We know that dr d 1ds ds r

Cos and Sinθφ φ= =

2 22

22

dr 1 1ds

r ppCos Sin p rSinr r

ϕ φ ϕ−

∴ = = − = − = =

2 2

dsdr

rr p

∴ =−

2ds

dr r rAlso pSin p

rθ φ

= = =

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Exercise:

Find ds dsanddr dθ

for the following curves:

m

2

(i) r=a(1+ Cos ) (ii) =a Cosm

(iii) aSec ( )2

mr

r

θ

θθ

=

Mean Value Theorems

Recapitulation Closed interval: An interval of the form a x b≤ ≤ , that includes every point between a and b and also the end points, is called a closed interval and is denoted by [ ]ba, . Open Interval: An interval of the form a x b< < , that includes every point between a and b but not the end points, is called an open interval and is denoted by ( )ba, Continuity: A real valued function )(xf is said to be continuous at a point 0x if

0

0lim ( ) ( )x x

f x f x→

=

The function )(xf is said to be continuous in an interval if it is continuous at every point in the interval. Roughly speaking, if we can draw a curve without lifting the pen, then it is a continuous curve otherwise it is discontinuous, having discontinuities at those points at which the curve will have breaks or jumps. We note that all elementary functions such as algebraic, exponential, trigonometric, logarithmic, hyperbolic functions are continuous functions. Also the sum, difference, product of continuous functions is continuous. The quotient of continuous functions is continuous at all those points at which the denominator does not become zero. Differentiability: A real valued function )(xf is said to be differentiable at point 0x if

0

0

0

( ) ( )limx x

f x f xx x→

−−

exists uniquely and it is denoted by )(' 0xf .

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A real valued function f(x) is said to be differentiable in an interval if it is differentiable at

every in the interval or if 0

( ) ( )limh

f x h f xh→

+ − exists uniquely. This is denoted by )(' xf .

We say that either )(' xf exists or f(x) is differentiable. Geometrically, it means that the curve is a smooth curve. In other words a curve is said to be smooth if there exists a unique tangent to the curve at every point on it. For example a circle is a smooth curve. Triangle, rectangle, square etc are not smooth, since we can draw more number of tangents at every corner point. We note that if a function is differentiable in an interval then it is necessarily continuous in that interval. The converse of this need not be true. That means a function is continuous need not imply that it is differentiable. Rolle’s Theorem: (French Mathematician Michelle Rolle 1652-1679)

Suppose a function )(xf satisfies the following three conditions: (i) )(xf is continuous in a closed interval [ ]ba, (ii) )(xf is differentiable in the open interval ( )ba, (iii) )()( bfaf =

Then there exists at least one point c in the open interval ( )ba, such that 0)(' =cf Geometrical Meaning of Rolle’s Theorem: Consider a curve )(xf that satisfies the conditions of the Rolle’s Theorem as shown in figure:

As we see the curve )(xf is continuous in the closed interval [ ]ba, , the curve is smooth i.e. there can be a unique tangent to the curve at any point in the open interval ( )ba, and also )()( bfaf = . Hence by Rolle’s Theorem there exist at least one point c belonging to ( )ba, such that 0)(' =cf . In other words there exists at least one point at which the tangent drawn to the curve will have its slope zero or lies parallel to x-axis. Notation: Often we use the statement: there exists a point c belonging to ( )ba, such that ……. We present the same in mathematical symbols as: ∃ c ( )ba,∈ : 0)(' =cf From now onwards let us start using the symbolic notation.

[b,f(b)] [a,f(a)]

x

y

a c b

79 of 133



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Example 1: Verify Rolle’s Theorem for 2)( xxf = in [ ]1,1− First we check whether the conditions of Rolle’s theorem hold good for the given function:

(i) 2)( xxf = is an elementary algebraic function, hence it is continuous every where and so also in [ ]1,1− .

(ii) xxf 2)(' = exists in the interval (-1, 1) i.e. the function is differentiable in (-1, 1).

(iii) Also we see that 1)1()1( 2 =−=−f and 11)1( 2 ==f i.e., )1()1( ff =− Hence the three conditions of the Rolle’s Theorem hold good. ∴By Rolle’s Theorem ∃ c ( )1,1−∈ : 0)(' =cf that means 2c = 0 ⇒ c= 0 ( )1,1−∈

Hence Rolle’s Theorem is verified.

Example 2: Verify Rolle’s Theorem for 2/)3()( xexxxf −+= in [-3, 0]

)(xf is a product of elementary algebraic and exponential functions which are continuous and hence it is continuous in [ 3,0]−

2/22/ )3(21)32()( xx exxexxf −− +−+=′

2/2 )]3(21)32[( xexxx −+−+=

2/2/2 )2)(3(21]6[

21 xx exxexx −− +−−=−−−= exists in (-3, 0)

0)3( =−f and also 0)0( =f )0()3( ff =−∴ That is, the three conditions of the Rolle’s Theorem hold good. ∴ ∃ c ( )0,3−∈ : 0)(' =cf

∞−=⇒=+−−∴ − ,2,30)2)(3(21 2/ cecc c

Out of these values of c, since 2 ( 3,0)− ∈ − , the Rolle’s Theorem is verified.

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Example 3: Verify Rolle’s Theorem for nm bxaxxf )()()( −−= in [a, b] where a <b and a, b>0.

)(xf is a product of elementary algebraic functions which are continuous and hence it is continuous in [a, b]

11 )()()()()( −− −−+−−=′ nmnm bxaxnbxaxmxf

11 )())](()([ −− −−−+−= nm bxaxaxnbxm

11 )())](()([ −− −−+−+= nm bxaxnambnmx exists in ( )ba,

0)( =af and also 0)( =bf )()( bfaf =∴ Hence the three conditions of the Rolle’s Theorem hold good. ∴ ∃ ( ),c a b∈ : 0)(' =cf

banmnambcbcacnambnmc nm ,,0)())](()([ 11

++

=⇒=−−+−+∴ −−

Out of these values of c, sincenmnambc

++

= ),( ba∈ , the Rolle’s Theorem is verified.

Example 4: Verify Rolle’s Theorem for )(

log)(2

baxabxxf

++

= in [a, b]

)log(log)log()( 2 baxabxxf +−−+= is the sum of elementary logarithmic functions which are continuous and hence it is continuous in [a, b]

xabxxxf 12)( 2 −

+=′ exists in ( )ba,

01loglog)(

log)( 2

22==

+

+=

++

=abaaba

baaabaaf

and similarly 01loglog)(

log)( 2

22==

+

+=

++

=abbabb

bababbbf

Hence the three conditions of the Rolle’s Theorem hold good. ∴ ∃ c ( )ba,∈ : 0)(' =cf

abcabcabcc

abcccabc

ccf =⇒=−⇒=+

−−⇒=−

+=′ 00

)(2012)( 2

2

22

2

Since abc = ),( ba∈ , the Rolle’s Theorem is verified. Exercise: Verify Rolle’s Theorem for

(i) )cos(sin)( xxexf x −= in

45,

4ππ ,

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(ii) 2/)2()( xexxxf −= in [ ]2,0 ,

(iii) 2sin 2( ) x

xf xe

= in

45,

4ππ .

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Lagrange’s Mean Value Theorem (LMVT): (Also known as the First Mean Value Theorem) (Italian-French Mathematician J. L. Lagrange 1736-1813)

Suppose a function )(xf satisfies the following two conditions: (i) )(xf is continuous in a closed interval [ ]ba,

(ii) )(xf is differentiable in the open interval ( )ba,

Then there exists at least one point c in the open interval ( )ba, such that ab

afbfcf−−

=)()()('

Proof: Consider the function )(xφ defined by kxxfx −= )()(φ where k is a constant to be found such that ( ) ( )a bϕ φ= . Since )(xf is continuous in the closed interval [ ]ba, , )(xφ which is a sum of continuous functions is also continuous in the closed interval [ ]ba,

' '( ) ( ) (1)x f x kxφ = − → exists in the interval ( )ba, as )(xf is differentiable in ( )ba, . We have ( ) ( )a f a kaφ = − and ( ) ( )b f b kbφ = −

( ) ( )( ) ( ) ( ) ( ) (2)f b f aa b f a ka f b kb kb a

φ φ −∴ = ⇒ − = − ⇒ = →

−

This means that when k is chosen as in (2) we will have ( ) ( )a bφ φ= Hence the conditions of Rolle’s Theorem hold good for )(xφ in [ ]ba, ∴By Rolle’s Theorem ∃ c ( ),a b∈ : '( ) 0cφ =

''( ) 0 ( ) 0 ( ) (3)c f c k k f cφ ′= ⇒ − = ⇒ = →

From (2) and (3), we infer that ∃ c ( ),a b∈ : ab

afbfcf−−

=)()()('

Thus the Lagrange’s Mean Value Theorem (LMVT) is proved.

84 of 133



Geometrical Meaning of Lagrange’s Mean Value Theorem: Consider a curve )(xf that satisfies the conditions of the LMVT as shown in figure:

From the figure, we observe that the curve )(xf is continuous in the closed interval [ ]ba, ; the curve is smooth i.e. there can be a unique tangent to the curve at any point in the open interval ( )ba, . Hence by LMVT there exist at least one point c belonging to ( )ba, such

that ab

afbfcf−−

=)()()(' . In other words there exists at least one point at which the tangent

drawn to the curve lies parallel to the chord joining the points [ ], ( )a f a and[ ], ( )b f b . Other form of LMVT: The Lagrange’s Mean Value Theorem can also be stated as follows: Suppose )(xf is continuous in the closed interval [ ],a a h+ and is differentiable in the open

interval ( ), )a a h+ then (0,1) : ( ) ( ) ( )f a h f a hf a hθ θ′∃ ∈ + = + + When (0,1)θ ∈ we see that ( , )a h a a hθ+ ∈ + i.e., here c a hθ= +

Using the earlier form of LMVT we may write that ( ) ( )( )( )

f a h f af a ha h a

θ + −′ + =+ −

On simplification this becomes ( ) ( ) ( )f a h f a hf a hθ′+ = + + .

[b,f(b)]

[a,f(a)]

x

y

a c b

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Example 5: Verify LMVT for ( ) ( 1)( 2)( 3)f x x x x= − − − in [0, 4]

3 2( ) ( 1)( 2)( 3) 6 11 6f x x x x x x x= − − − = − + − is an algebraic function hence it is continuous in [0, 4]

2( ) 3 12 11f x x x′ = − + exists in (0, 4) i.e., ( )f x is differentiable in (0, 4) . i.e., both the conditions of LMVT hold good for ( )f x in[0,4] .

Hence (4) (0)(0,4) : ( )4 0

f fc f c −′∃ ∈ =−

i.e., 2 (3)(2)(1) ( 1)( 2)( 3)3 1 114 0

c c − − − −− + =

−

i.e, 2 2 23 12 11 3 3 12 8 0 2 (0,4)3

c c c c c− + = ⇒ − + = ⇒ = ± ∈

Hence the LMVT is verified.

Example 6: Verify LMVT for ( ) logef x x= in [1, ]e

( ) logef x x= is an elementary logarithmic function hence continuous in [1, ]e .

1( )f xx

′ = exists in ( )1,e or ( )f x is differentiable in ( )1,e .

i.e., both the conditions of LMVT hold good for ( )f x in [1, ]e

Hence ( ) (1) 1 log log1(1, ) : ( )1 1

f e f ec e f ce c e

− −′∃ ∈ = ⇒ =− −

1 1 1 (1, )

1c e e

c e⇒ = ⇒ = − ∈

−


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Example 7: Verify LMVT for 1( ) sinf x x−= in [0,1]

We find 2

1( )1

f xx

′ =−

exists in ( )0,1 , and hence ( )f x is continuous in [0,1] .

i.e., both the conditions of LMVT hold good for ( )f x in [0,1]

Hence 1 1

2(1) (0) 1 sin 1 sin 0(0,1) : ( )

1 0 11

f fc f cc

− −− −′∃ ∈ = ⇒ =− −

2

2 222

1 2 4121

c cc

π ππ π

−⇒ = ⇒ − = ⇒ =

−

2 4 0.7712 (0,1)c ππ

−∴ = = ∈


Example 8: Find θ of LMVT for 2( )f x ax bx c= + + or

Verify LMVT by finding θ for 2( )f x ax bx c= + + 2( )f x ax bx c= + + is an algebraic function hence continuous in [ , ]a a h+ .

( ) 2f x ax b′ = + exists in ( ),a a h+

i.e., both the conditions of LMVT hold good for ( )f x in [ , ]a a h+ Then the second form of LMVT (0,1) : ( ) ( ) ( ) (1)f a h f a hf a hθ θ′∃ ∈ + = + + →

Here ; 2 3 2 2( ) ( ) ( ) 2 (2)f a h a a h b a h c a ah a h ab bh c+ = + + + + = + + + + + → 3( ) (3)f a a ab c= + + → ; 2( ) 2 ( ) 2 2 (4)f a h a a h b a a h bθ θ θ′ + = + + = + + →

Using (2), (3) and (4) in (1) we have:

3 2 2 3 2 12 (2 2 ) (0,1)2

a ah a h ab bh c a ab c h a a h bθ θ+ + + + + = + + + + + ⇒ = ∈


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Example 9: Find θ of LMVT for ( ) logef x x= in 2,e e

( ) logef x x= is an elementary logarithmic function hence continuous in 2,e e

1( )f xx

′ = exists in ( )2, .e e

Hence by the second form of LMVT

( )

[ ] [ ]

( )

2 2 2

22

2

0,1 : ( ) ( ) ( ) ( )

( )log log( )

( 1) ( 1). ., 2 log log 2 11 ( 1) 1 ( 1)

21 ( 1) 1 0,11

e e

f e f e e e f e e e

e ee ee e e

e e ei e e ee e e

ee ee

θ θ

θ

θ θ

θ θ

′∃ ∈ = + − + −

−⇒ = +

+ −− −

= + ⇒ − =+ − + −

−⇒ + − = − ⇒ = ∈

−


Example 10: If 0x > , Apply Mean Value Theorem to show that

( ) log (1 )1 e

xi x xx

< + <+

and 1 10 1log (1 )e x x

< − <+

Consider ( ) log (1 )ef x x= + in [0, ]x . We see that ( )f x is an elementary logarithmic function

hence continuous in [0, ]x . Also 1( )1

f xx

′ =+

exists in ( )0, .x

∴ by LMVT (0,1) : ( ) (0) (0 )f x f xf xθ θ′∃ ∈ = + +

. ., log(1 ) log1 log(1 ) (1)1 1

x xi e x xx xθ θ

+ = + ⇒ + = →+ +

Since 0x > and 0 1θ< < , we have 0 1 1 1x x x xθ θ< < ⇒ < + < +

1 1 1 11 1 (2)1 1 1 1 1 1

x x xx x x x x xθ θ θ

⇒ > > ⇒ < < ⇒ < < →+ + + + + +

From (1) and (2) we get log (1 )1 e

x x xx

< + <+

Also from (1) we have, 1 1 1 1log(1 ) log(1 )

xx x x x

θ θ+= ⇒ − =

+ +

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Since 0 1θ< < , we can write 1 10 1.log(1 )x x

< − <+

Example 11: Apply Mean Value Theorem to show that

1 12 2

sin sin , 01 1

b a b ab a where a ba b

− −− −< − < < <

− −

Consider 1( ) sinf x x−= in [ , ]a b

We see that 2

1( )1

f xx

′ =−

exists in ( ),a b

Hence ( )f x is differentiable in ( ),a b and also it is continuous in [ , ]a b .

Therefore, by LMVT ( )1 1

2( ) ( ) 1 sin sin( , ) : ( ) 1

1

f b f a b ac a b f cb a b ac

− −− −′∃ ∈ = ⇒ = →− −−

Since 2 2 2 2 2 2 2 2 21 1 1a c b a c b a c b a c b< < ⇒ < < ⇒ − > − > − ⇒ − > − > −

( )2 2 2 2 2 21 1 1 1 1 1 2

1 1 1 1 1 1or

a c b a c b∴ < < < < →

− − − − − −

From (1) and (2), we have

1 11 1

2 2 2 21 sin sin 1 sin sin

1 1 1 1

b a b a b ab ab aa b a b

− −− −− − −

< < ⇒ < − <−− − − −

Exercise: Verify the Lagrange’s Mean Value Theorem for

(i) ( ) ( 1)( 2)f x x x x= − − in 10,2

(ii) 1( )f x Tan x−= in [ ]0,1

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Mean Value Theorems Cauchy’s Mean Value Theorem (CMVT): (French Mathematician A. L. Cauchy 1789-1857)

Suppose two functions )(xf and ( )g x satisfies the following conditions:

(i) )(xf and ( )g x are continuous in a closed interval [ ]ba, (ii) )(xf and ( )g x are differentiable in the open interval ( )ba,

(iii) ( ) 0 .[ ]g x for all x for all can be denoted by the symbol′ ≠ ∀

Then there exists at least one point c in the open interval ( )ba,

such that '( ) ( ) ( )'( ) ( ) ( )

f c f b f ag c g b g a

−=

−

Proof: Consider the function )(xφ defined by ( ) ( ) ( )x f x k g xφ = − where k is a constant to be found such that ( ) ( )a bϕ φ= . Since )(xf and ( )g x are continuous in the closed interval [ ]ba, , )(xφ which is a sum of continuous functions is also continuous in the closed interval [ ]ba,

' '( ) ( ) ( ) (1)x f x k g xφ ′= − → exists in the interval ( )ba, as )(xf and ( )g x are differentiable in ( )ba, . We have ( ) ( ) ( )a f a kg aφ = − and ( ) ( ) ( )b f b k g bφ = −

( ) ( )( ) ( ) ( ) ( ) ( ) ( ) (2)( ) ( )

f b f aa b f a kg a f b k g b kg b g a

φ φ −∴ = ⇒ − = − ⇒ = →

−

This means that when k is chosen as in (2) we will have ( ) ( )a bφ φ= Hence the conditions of Rolle’s Theorem hold good for )(xφ in [ ]ba, ∴By Rolle’s Theorem ∃ c ( ),a b∈ : '( ) 0cφ =

' ( )'( ) 0 ( ) ( ) 0 (3)( )

f cc f c k g c kg c

φ′

′= ⇒ − = ⇒ = →′

From (2) and (3), we infer that ∃ c ( ),a b∈ : '( ) ( ) ( )'( ) ( ) ( )


−=

−

Thus the Cauchy’s Mean Value Theorem (CMVT) is proved.

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Example 12: Verify the Cauchy’s MVT for 2( )f x x= and 4( )g x x= in [ , ]a b

2( )f x x= and 4( )g x x= are algebraic polynomials hence continuous in [ , ]a b

( ) 2f x x′ = and 3( ) 4g x x′ = exist in ( ),a b also we see that ( ) 0 ( , )g x for all x a b′ ≠ ∈ since 0 a b< < i.e., the conditions of CMVT hold good for ( )f x and ( )g x in[ , ]a b .

Hence ∃ c ( ),a b∈ : '( ) ( ) ( )'( ) ( ) ( )


−=

−

i.e., 2 2 2 2

3 4 4 2 2 22 1 1 ( , )

24 2c b a b ac a bc b a c b a

− += ⇒ = ⇒ = ∈

− +

Hence the CMVT is verified.

Example 13: Verify the Cauchy’s MVT for ( ) logf x x= and 1( )g xx

= in [1, ]e

( ) logf x x= and 1( )g xx

= are elementary logarithmic and rational algebraic functions that

are continuous in [1, ]e

1( )f xx

′ = and 21( )g x

x−′ = exist in ( )1,e

also we see that ( ) 0 (1, )g x for all x e′ ≠ ∈ i.e., the conditions of CMVT hold good for ( )f x and ( )g x in ( )1,e .

Hence ( )1,c e∃ ∈ : '( ) ( ) (1)'( ) ( ) (1)

f c f e fg c g e g

−=

−

i.e., 2

1 log log1 1 (1, )1 1 1 11 1

e ec c c ee

c e e

−= ⇒ − = ⇒ = ∈

− −− −

Hence the CMVT is verified. Example 14: Verify the Cauchy’s MVT for ( ) xf x e= and ( ) xg x e−= in [ , ]a b

( ) xf x e= and ( ) xg x e−= are elementary exponential functions that are continuous in [ , ]a b

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( ) xf x e′ = and ( ) xg x e−′ = − exist in ( ),a b also we see that ( ) 0 ( , )g x for all x a b′ ≠ ∈ , since 0 a b< < . i.e., the conditions of CMVT hold good for ( )f x and ( )g x in ( ),a b .

Hence ( ),c a b∃ ∈ : '( ) ( ) ( )'( ) ( ) ( )


−=

−

i.e., 2 21 1

c b a b a b ac c

c b a a bb a a b

e e e e e e ee ee e e e e

e e e

− − −

+

− − −= ⇒ − = ⇒ − =

− − −−

2 ( , )

2c a b a be e c a b+ +

= ⇒ = ∈


Example 15: Verify the Cauchy’s MVT for ( )f x Sin x= and ( )g x Cos x= in 0,2π

( )f x Sin x= and ( )g x Cos x= are elementary trignometric functions

that are continuous in 0,2π

.

( )f x Cos x′ = and ( )g x Sin x′ = − exist in 0,2π

also we see that ( ) 0 0,2

g x for all x π ′ ≠ ∈

.

i.e., the conditions of CMVT hold good for ( )f x and ( )g x in 0,2π

.

Hence 0,2

c π ∃ ∈

:( ) (0)'( ) 2

'( ) ( ) (0)2

f ff cg c g g

π

π

−=

−

( ) (0)2 1 0,

4 2( ) (0)2

Sin SinCos c Cot c cSin c Cos Cos

ππ π

π

− ⇒ = ⇒ = ⇒ = ∈ − −


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Exercise: Verify the Cauchy’s Mean Value Theorem for

(i) ( )f x x= and 1( )g xx

= in 1 ,14

(ii) 21( )f xx

= and 1( )g xx

= in [ ],a b

(iii) ( )f x Sin x= and ( )g x Cos x= in [ ],a b

Taylor’s Mean Value Theorem: (Generalised Mean Value Theorem): (English Mathematician Brook Taylor 1685-1731)

Suppose a function )(xf satisfies the following two conditions:

(i) )(xf and it’s first (n-1) derivatives are continuous in a closed interval [ ]ba,

(ii) ( 1) ( )nf x− is differentiable in the open interval ( )ba,

Then there exists at least one point c in the open interval ( )ba, such that

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2 3( ) ( )( ) ( ) ( ) ( ) ( ) ( )2 3

b a b af b f a b a f a f a f a− −′ ′′ ′′′= + − + + + …..

1

( 1) ( )( ) ( )..... ( ) ( ) (1)1

n nn nb a b af a f c

n n

−−− −

+ + →−

Takingb a h= + and for 0 1θ< < , the above expression (1) can be rewritten as

2 3 1( 1) ( )( ) ( ) ( ) ( ) ( ) .... ( ) ( ) (2)

2 3 1

n nn nh h h hf a h f a hf a f a f a f a f a h

n nθ

−−′ ′′ ′′′+ = + + + + + + + →

− Taking b=x in (1) we may write

2 3 1( 1)( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ... ( ) (3)

2 3 1

nn

nx a x a x af x f a x a f a f a f a f a R

n

−−− − −′ ′′ ′′′= + − + + + + + →

−

( )( ) ( ) Ren

nn

x aWhere R f c mainder term after n termsn

−= →

When ,n → ∞ we can show that 0nR → , thus we can write the Taylor’s series as

2 1( 1)

( )

1

( ) ( )( ) ( ) ( ) ( ) ( ) ... ( ) ....2 1

( )( ) ( ) (4)

nn

nn

n

x a x af x f a x a f a f a f an

x af a f an

−−

∞

=

− −′ ′′= + − + + + +−

−= + →∑

Using (4) we can write a Taylor’s series expansion for the given function f(x) in powers of (x-a) or about the point ‘a’. Maclaurin’s series: (Scottish Mathematician Colin Maclaurin 1698-1746) When a=0, expression (4) reduces to a Maclaurin’s expansion given by

2 1( 1)

( )

1

( ) (0) (0) (0) ... (0) ....2 1

(0) (0) (5)

nn

nn

n

x xf x f xf f fn

xf fn

−−

∞

=

′ ′′= + + + + +−

= + →∑

Example 16: Obtain a Taylor’s expansion for ( )f x Sin x= in the ascending powers of

4x π −

up to the fourth degree term.

The Taylor’s expansion for )(xf about 4π is

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2 3 4(4)

( ) ( ) ( )4 4 4( ) ( ) ( ) ( ) ( ) ( ) ( ) .... (1)

4 4 4 2 4 3 4 4 4

x x xf x f x f f f f

π π ππ π π π π π− − −

′ ′′ ′′′= + − + + + →

1( )4 4 2

f x Sin x f Sinπ π = ⇒ = =

; 1( )4 4 2

f x Cos x f Cosπ π ′ ′= ⇒ = =

1( )4 4 2

f x Sin x f Sinπ π ′′ ′′= − ⇒ = − = −

1( )4 4 2

f x Cos x f Cosπ π ′′′ ′′′= − ⇒ = − = −

(4) (4) 1( )4 4 2

f x Sin x f Sinπ π = ⇒ = =

Substituting these in (1) we obtain the required Taylor’s series in the form

2 3 4( ) ( ) ( )1 1 1 1 14 4 4( ) ( )( ) ( ) ( ) ( ) ....4 2 3 42 2 2 2 2

x x xf x x

π π ππ − − −

= + − + − + − +

2 3 4( ) ( ) ( )1 4 4 4( ) 1 ( ) ...

4 2 3 42

x x xf x x

π π ππ

− − − = + − − + − +

Example 17: Obtain a Taylor’s expansion for ( ) logef x x= up to the term containing

( )41x − and hence find loge(1.1). The Taylor’s series for )(xf about the point 1 is

2 3 4(4)( 1) ( 1) ( 1)( ) (1) ( 1) (1) (1) (1) (1) .... (1)

2 3 4x x xf x f x f f f f− − −′ ′′ ′′′= + − + + + →

Here ( ) log (1) log 1 0ef x x f= ⇒ = = ; 1( ) (1) 1f x fx

′ ′= ⇒ =

21( ) (1) 1f x fx

′′ ′′= − ⇒ = − ; 32( ) (1) 2f x fx

′′′ ′′′= ⇒ =

(4) (4)

46( ) (1) 6f x fx

= − ⇒ = − etc.,

Using all these values in (1) we get

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2 3 4( 1) ( 1) ( 1)( ) log 0 ( 1)(1) ( 1) (2) ( 6) ....

2 3 4ex x xf x x x − − −

= = + − + − + + −

2 3 4( 1) ( 1) ( 1)log ( 1) ....

2 3 4ex x xx x − − −

⇒ = − − + −

Taking x=1.1 in the above expansion we get

2 3 4(0.1) (0.1) (0.1)log (1.1) (0.1) .... 0.09532 3 4e⇒ = − + − =

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Example 18: Using Taylor’s theorem Show that

2 3

log (1 ) 0 1, 02 3ex xx x for xθ+ < − + < < >

Taking n=3 in the statement of Taylor’s theorem, we can write

2 3( ) ( ) ( ) ( ) ( ) (1)

2 3x xf a x f a xf a f a f a xθ′ ′′ ′′′+ = + + + + →

Consider ( ) logef x x= 2 31 1 2( ) ; ( ) ( )f x f x and f xx x x

′ ′′ ′′′⇒ = = − =

Using these in (1), we can write,

2 3

2 31 1 2log( ) log (2)

2 3 ( )x xa x a x

a a a xθ

+ = + + − + → +

For a=1 in (2) we write,

2 3 2 3

3 31 1log(1 ) log1

2 3 2 3(1 ) (1 )x x x xx x x

x xθ θ+ = + − + = − +

+ +

Since 33

10 0, (1 ) 1 1(1 )

x and x and thereforex

θ θθ

> > + > <+

2 3log(1 )

2 3x xx x∴ + < − +

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Example 19: Obtain a Maclaurin’s series for ( )f x Sin x= up to the term containing 5.x The Maclaurin’s series for f(x) is

2 3 4 5(4) (5)( ) (0) (0) (0) (0) (0) (0) .... (1)

2 3 4 5x x x xf x f x f f f f f′ ′′ ′′′= + + + + + →

Here ( ) (0) 0 0f x Sin x f Sin= ⇒ = = ( ) (0) 0 1f x Cos x f Cos′ ′= ⇒ = =

( ) (0) 0 0f x Sin x f Sin′′ ′′= − ⇒ = − = ( ) (0) 0 1f x Cos x f Cos′′′ ′′′= − ⇒ = − = −

(4) (4)( ) (0) 0 0f x Sin x f Sin= ⇒ = = (5) (5)( ) (0) 0 1f x Cos x f Cos= ⇒ = = Substituting these values in (1), we get the Maclaurin’s series for ( )f x Sin x= as

2 3 4 5 3 5( ) 0 (1) (0) ( 1) (0) (1) .... ....

2 3 4 5 3 5x x x x x xf x Sin x x Sin x x= = + + + − + + ⇒ = − +

Note: As done in the above example, we can find the Maclaurin’s series for various functions, for ex:

2 4 6( ) 1 ......

2 4 6x x xi Cosx = − + −

2 3 4 5 6( ) 1 ......

2 3 4 5 6x x x x x xii e x= + + + + + +

2 3 4( 1) ( 1)( 2) ( 1)( 2)( 3)( ) (1 ) 1 ......

2 3 4m m m m m m m m m miii x mx x x x− − − − − −

+ = + + + +

Taking 1 ( )m in iii= − we can get 1 2 3 4 5 6(1 ) 1 ......x x x x x x x−+ = + + + + + +

Replacing x by (-x) in this we get 1 2 3 4 5 6(1 ) 1 ......x x x x x x x−− = − + − + − +

We use these expansions in the study of various topics.

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Indeterminate Forms: While evaluating certain limits, we come across expressions of the form

0 00 , , 0 , , 0 , 10

and ∞∞×∞ ∞ − ∞ ∞

∞which do not represent any value. Such expressions are

called Indeterminate Forms. We can evaluate such limits that lead to indeterminate forms by using L’Hospital’s Rule (French Mathematician 1661-1704). L’Hospital’s Rule: If f(x) and g(x) are two functions such that (i) lim ( ) 0 lim ( ) 0

x a x af x and g x

→ →= =

(ii) ( ) ( ) ( ) 0f x andg x exist and g a′ ′ ′ ≠

Then ( ) ( )lim lim( ) ( )x a x a

f x f xg x g x→ →

′=

′

The above rule can be extended, i.e, if ( ) ( ) ( )( ) 0 ( ) 0 lim lim lim .....( ) ( ) ( )x ax a x a

f x f x f xf a and g a theng x g x g x→→ →

′ ′′′ ′= = = = =

′ ′′

Note:

1. We apply L’Hospital’s Rule only to evaluate the limits that in 0 ,

0∞∞

forms. Here we

differentiate the numerator and denominator separately to write ( )( )

f xg x

′′

and apply the

limit to see whether it is a finite value. If it is still in 00

or ∞∞

form we continue to

differentiate the numerator and denominator and write further ( )( )

f xg x

′′′′

and apply the

limit to see whether it is a finite value. We can continue the above procedure till we get a definite value of the limit.

2. To evaluate the indeterminate forms of the form 0 , ,×∞ ∞ − ∞ we rewrite the

functions involved or take L.C.M. to arrange the expression in either 00

or ∞∞

and then

apply L’Hospital’s Rule.

3. To evaluate the limits of the form 0 00 , 1and ∞∞ i.e, where function to the power of function exists, call such an expression as some constant, then take logarithm on both

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sides and rewrite the expressions to get 00

or ∞∞

form and then apply the L’Hospital’s

Rule. 4. We can use the values of the standard limits like

0 0 0 0 0

sin tanlim 1; lim 1; lim 1; lim 1; lim cos 1;sin tanx x x x x

x x x x x etcx x x x→ → → → →

= = = = =

Evaluate the following limits:

Example 1: Evaluate 30

sinlimtanx

x xx→

−

3 2 2 4 3 20 0 0

sin 0 cos 1 0 sin 0lim lim limtan 0 3tan sec 0 6 tan sec 6 tan sec 0x x x

x x x xx x x x x x x→ → →

− − − = = +

6 2 4 2 4 4 20

cos 1lim6sec 24 tan sec 18 tan sec 12 tan sec 6x

xx x x x x x x→

−= = −

+ + +

Method 2:

3

33 30 0 0 0

sinsin 0 0 sin 0 tanlim lim lim lim 1

tan 0 0 0tanx x x x

x xx x x x xx

x x xxx

→ → → →

−− − = = =

20 0 0

cos 1 0 sin 0 cos 1lim lim lim3 0 6 0 6 6x x x

x x xx x→ → →

− − − = = = = −


limx x

x

a bx→

−

0 0

0 log loglim lim log log log0 1

x x x x

x x

a b a a b b aa bx b→ →

− − = = − =

Example 3: Evaluate ( )20

sinlim1x x

x x

e→ −

( ) ( )20 0 0

sin 0 sin cos 0 cos cos sin 1 1 0 2lim lim lim 10 0 2[1 0] 22 . ( 1)2 11 x x x xx xx x xx

x x x x x x x x xe e e ee ee→ → →

+ + − + − = = = = = + + −− −

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log(1 )limx

x

x e xx→

− +

( )2

20 0 0

111log(1 ) 0 0 1 1 0 1 31lim lim lim

0 2 0 2 2 2

x x xx x

x

x x x

e e x ee x e xx e x xx x→ → →

+ + ++ − +− + + + + += = = =


cosh coslimsinx

x xx x→

−

0 0 0

cosh cos 0 sinh sin 0 cosh cos 1 1 2lim lim lim 1sin 0 sin cos 0 cos cos sin 1 1 0 2x x x

x x x x x xx x x x x x x x x→ → →

− + + + = = = = = + + − + −


cos log(1 ) 1limsinx

x x xx→

− + − +

2

20 0 0

11 cossin 1cos log(1 ) 1 0 0 1 1(1 )1lim lim lim 0sin 0 sin 2 0 2 cos 2 2x x x

xxx x x xxx x x→ → →

− +− − +− + − + − ++ += = = =


lim1 log

x

x

x xx x→

−− −

2 1

1 1 1

2

0 (1 log ) 1 0 (1 log ) 1 1lim lim lim 21 11 log 0 0 11

x x x x

x x x

x x x x x x xx x

x x

−

→ → →

− + − + + + = = = = − − −

1log log 1 log (1 log )

( ) (1 log )

x

x x

Since y x y x x y x y y xy

dand then x x xdx

′ ′= ⇒ = ⇒ = + ⇒ = +

= +


4

sec 2 tanlim1 cos 4x

x xxπ

→

−+

2 2 2 4 2 2 2

4 4 4

sec 2 tan 0 2sec tan 2sec 0 2sec 4sec tan 4sec tanlim lim lim1 cos 4 0 4 sin 4 0 16 cos 4x x x

x x x x x x x x x xx x xπ π π

→ → →

− − + − = = + − −

( ) ( ) ( )4 2 2

22 2 4 2 (1) 4 2 8 116 16 2

+ −= = =

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Example 9: Evaluate log(sin . cos )limlog(cos .sec )x a

x ec aa x→

2

cos cossin . coslog(sin . cos ) 0 cotlim lim lim cot

sec tan .coslog(cos .sec ) 0 tancos .sec

x a x a x a

x ec ax ec ax ec a x ax x aa x x

a x→ → →

= = =


2coslimsin

x x

x

e e xx x

−

→

+ −

0 0 0

2cos 0 2sin 0 2cos 1 1 2lim lim lim 2sin 0 sin cos 0 cos cos sin 1 1 0

x x x x x x

x x x

e e x e e x e e xx x x x x x x x x

− − −

→ → →

+ − − + + + + + = = = = + + − + −


cos log(1 )limx

x x xx→

− +

20 0

1cos sincos log(1 ) 0 01lim lim0 2 0x x

x x xx x x xx x→ →

− −− + +=

2

0

1sin sin cos0 0 0 1 1(1 )lim

2 2 2x

x x x xx

→

− − − +− − − ++= = =


0

log(1 )limlog cosx

xx→

−

2 2

2 20 0 0 0

2log(1 ) 0 2 cos 0 2 cos 2 sin 2 01lim lim lim lim 2

sinlog cos 0 (1 )sin 0 (1 )cos 2 sin 1 0cos

x x x x

xx x x x x xx

xx x x x x x xx

→ → → →

− − − −− = = = = = − − − − −


tan sinlimsinx

x xx→

−

2 2

3 2 2 30 0 0

tan sin 0 sec cos 0 2sec tan sin 0lim lim limsin 0 3sin cos 0 6sin cos 3sin 0x x x

x x x x x x xx x x x x x→ → →

− − + = = −

2 2 4

3 2 20

4sec tan 2sec cos 0 2 1 3 1lim6cos 12sin cos 9 sin cos 6 0 0 6 2x

x x x xx x x x x→

+ + + += = = =

− − − −

Method 2:

3

33 30 0 0 0

tan sintan sin tan sin 0 sinlim lim lim lim 1

sin 0sinx x x x

x xx x x x xx

x x xxx

→ → → →

−− − = = =

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2 2

20 0

sec cos 0 2sec tan sin 0lim lim3 0 6 0x x

x x x x xx x→ →

− + = =

2 2 4

0

4sec tan 2sec cos 0 2 1 3 1lim6 6 6 2x

x x x x→

+ + + += = = =


tanlimtanx

x xx x→

−

3

2 30 0 0 0

tantan tan 0 tanlim lim lim lim 1

tantan 0x x x x

x xx x x x xx

xx x x xx

→ → → →

−− − = = =

2 2 2 2 4

20 0 0

sec 1 0 2sec tan 0 4sec tan 2sec 0 2 1lim lim lim3 0 6 0 6 6 3x x x

x x x x x xx x→ → →

− + + = = = = =


limlog(1 )

ax ax

x

e ebx

−

→

−+

0 0

0 2lim limlog(1 ) 0 /(1 )

ax ax ax ax

x x

e e ae ae a a abx b bx b b

− −

→ →

− + + = = = + +


1 loglimx

x

a x ax→

− −

22

20 0 0

1 log 0 log log 0 (log ) 1lim lim lim (log )0 2 0 2 2

x x x

x x x

a x a a a a a a ax x→ → →

− − − = = =


log( )limx

x

e e exx→

− +

2 2 20 0 0

log( ) log (1 ) log log(1 ) 0lim lim lim0

x x x

x x x

e e ex e e x e e xx x x→ → →

− + − + − − + = =

2

0 0

110 1 1(1 )1lim lim 1

2 0 2 2

xx

x x

ee xxx→ →

+− ++ += = = =

Exercise: Evaluate the following limits.

3

30 0

2 log(1 ) log(1 )( ) lim ( ) limsin sin

x x

x x

e e x xi iix x x

−

→ →

− − + +

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20 22

1 sin cos log(1 ) logsin( ) lim ( ) limtan ( )

2x x

x x x xiii ivx x x

π π→ →

+ − + −

−

1

2 202

cosh log(1 ) 1 sin sin( ) lim ( ) limx x

x x x x xv vix xπ

−

→ →

+ − − +

2 2

0

(1 )( ) limlog(1 )

x

x

e xviix x→

− ++

Limits of the form ∞ ∞

:


log(sin 2 )limlog(sin )x

xx→

2

20 0 0 0 0

log(sin 2 ) (2cos 2 / sin 2 ) 2cot 2 2 tan 0 2sec 2lim lim lim lim lim 1log(sin ) (cos / sin ) cot tan 2 0 2sec 2 2x x x x x

x x x x x xx x x x x x→ → → → →

∞ = = = = = = ∞


loglimcosx

xecx→

2

0 0 0 0

log 1\ sin 0 2sin 2 0lim lim lim lim 0cos cos .cot cos 0 cos sin 1 0x x x x

x x x xecx ec x x x x x x x→ → → →

∞ − − = = = = = ∞ − − −


log coslimtanx

xxπ

→

2 2

2 2 2 2

log cos tan tan sin cos 0lim lim lim lim 0tan sec sec 1 1x x x x

x x x x xx x xπ π π π

→ → → →

∞ − − − − = = − = = = ∞


log(1 )limcotx

xxπ→

−

2

21 1 1 1

log(1 ) 1/(1 ) sin 0 2 sin cos 0lim lim lim lim 0cot cos (1 ) 0x x x x

x x x x xx ec x x

π π π ππ π π π π π→ → → →

− ∞ − − = = = = = ∞ − − − −

Example 22: Evaluate tan 2

0lim log tan 3xx

x→

tan 2 00

loglog tan 3lim log tan 3 lim loglog tan 2 log

ex bxx e

axx ax b→→

∞ = = ∞

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2

20 0 0

3sec 3 / tan 3 3/ sin 3 .cos3 3/ sin 3 .cos3lim lim lim2sec 2 / tan 2 2 / sin 2 .cos 2 2 / sin 2 .cos 2x x x

x x x x x xx x x x x x→ → →

= = =

0 0 0

6 / sin 6 6sin 4 0 24cos 4 24lim lim lim 14 / sin 4 4sin 6 0 24cos 6 24x x x

x x xx x x→ → →

= = = = =

Example 23: Evaluate log( )limlog( )x ax a

x ae e→

−−

log( ) 1/( ) ( ) 0lim lim lim lim 1

log( ) /( ) ( ) 0 ( )

x a x a

x a x x a x x x ax a x a x a x a

x a x a e e e ee e e e e e x a e x a e e→ → → →

− ∞ − − = = = = = − ∞ − − − +


0 0

log tan logsin( ) lim ( ) limlog cotx x

x xi iix x→ →

sin0 02

cot 2 sec( ) lim ( ) lim ( ) lim log sin 2cot 3 tan 3 xx xx

x xiii iv v xx xπ→ →→

Limits of the form ( )0×∞ : To evaluate the limits of the form ( )0×∞ , we rewrite the

given expression to obtain either 00

or ∞ ∞

form and then apply the L’Hospital’s Rule.


lim( 1)xx

a x→∞

−

( )

11

1 2

2

1(log )( 1) 0lim( 1) 0 lim lim

1 10

xx

xx x x

a aa xa x form

x x→∞ →∞ →∞

− − − ×∞ = = −

1

0lim (log ) log logxx

a a a a a→∞

= = =

Example 25: Evaluate

2

lim(1 sin ) tanx

x xπ

→−

( ) 2

2 2 2

(1 sin ) 0 cos 0lim(1 sin ) tan 0 lim lim 0cot 0 cos 1x x x

x xx x formx ec xπ π π

→ → →

− − − ×∞ = = = = −


limsec .log2x

xx

π→

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( )1 1

log 0limsec .log 0 lim2 0cos

2x x

xx formx

x

ππ→ →

∞× =

1

2 2

1/ 2 2lim lim1 sinsin

22 2xx

x x

xx xπ ππ π ππ

→→= = =

− −


0lim log tanx

x x→

( )0 0

log tanlim log tan 0 lim(1/ )x x

xx x formx→ →

∞ ×∞ = ∞

2 2

0 0

2

sec / tanlim lim1 sin .cosx x

x x xx x

x→ →

−= =

−

2

0 0

2 0 4 0lim lim 0sin 2 0 2cos 2 2x x

x xx x→ →

− − = = = =


1lim (1 ) tan

2x

xx π→

−

( )2

2

1 1 1 2

1 0 2 2 4lim (1 ) tan 0 lim lim2 0cot cos

2 2 2 2x x x

x x xx form x xec

ππ π π π π→ → →

− − − ×∞ = = = = −


0lim tan .logx

x x→

( )0 0

loglim tan .log 0 limcotx x

xx x formx→ →

∞ ×∞ = ∞

2

20 0 0

1/ sin 0 sin 2 0lim lim lim 0cos 0 1 1x x x

x x xec x x→ → →

− − = = = = = −

Limits of the form ( )∞ − ∞ : To evaluate the limits of the form ( )∞ − ∞ , we take L.C.M.

and rewrite the given expression to obtain either 00

or ∞ ∞

form and then apply the

L’Hospital’s Rule.


1lim cotx

xx→

−

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( )0 0

1 1 coslim cot limsinx x

xx formx x x→ →

− = − ∞ − ∞

0 0

sin cos 0 cos cos sinlim limsin 0 sin cosx x

x x x x x x xx x x x x→ →

− − + = = +

0 0

sin 0 sin cos 0 0lim lim 0sin cos 0 cos cos sin 1 1 0x x

x x x x xx x x x x x x→ →

+ + = = = = + + − + −

Example 31: Evaluate [ ]

2

lim sec tanx

x xπ

→−

[ ] ( )2 2

1 sinlim sec tan limcos cosx x

xx x formx xπ π

→ →

− = − ∞ − ∞

2 2

1 sin 0 cos 0lim lim 0cos 0 sin 1x x

x xx xπ π

→ →

− − = = = = −


1limlog 1x

xx x→

− −

( )1 1

1 ( 1) log 0lim limlog 1 ( 1) log 0x x

x x x xformx x x x→ →

− − − ∞ − ∞ = − −

1 1 1

2

1 1 log log 0 l/ 1 1lim lim lim1 1 1 10 1 1 2log 1 logx x x

x x xx x x

x x x x→ → →

− − − − − − = = = = = − + + − + +


1 1lim1xx x e→

− −

( )0 0

1 1 ( 1) 0lim lim1 ( 1) 0

x

x xx x

e xformx e x e→ →

− − − ∞ − ∞ = − −

0 0

1 0 1 1lim lim( 1) 0 1 1 0 2

x x

x x x x xx x

e ee xe e e xe→ →

− = = = = − + + + + +


1 1limsinx x x→

−

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( )0 0

1 1 sin 0lim limsin sin 0x x

x xformx x x x→ →

− − ∞ − ∞ =

0 0

1 cos 0 sin 0lim lim 0sin cos 0 cos cos sin 1 1x x

x xx x x x x x x→ →

− = == = = + + − +


1 log(1 )limx

xx x→

+ −

2

2 20 0 0 0

1111 log(1 ) log(1 ) 0 0 1(1 )1lim lim lim lim0 2 0 2 2x x x x

x x x xxx x x x→ → → →

− + − + + +− = = = =


lim cotx

a xx a→

−

( )0 0 0

cos cos 0lim cot lim lim0sin sin

x x x

x xa x a aa aformx xx a x x

a a→ → →

− = − ∞ − ∞ = −

0 0

1sin cos . cos cos sin0lim lim0sin sin cos

x x

x x x x x xa x aa a a a a a a

x x x xxa a a a

→ →

− − + = = +

0 0

2

1 1sin sin . .cos0 0 0lim lim 01 1 1 10sin cos cos cos sin 0x x

x x x x xa a a a a a ax x x x x x xa a a a a a a a a a a

→ →

+ + = = = = + + − + −


( )0

( ) lim 2 cot( ) ( ) lim cos cotx a x

xi x a ii ecx xa→ →

− − −

2

202

1( ) lim tan sec ( ) lim cot2 xx

iii x x x iv xxπ

π→→

− −

[ ]202

1 1( ) lim ( ) lim 2 tan sectanx x

v vi x x xx x x π

π→ →

− −

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Example 37: Find the value of ‘a’ such that 30

sin 2 sinlimx

x a xx→

+ is finite. Also find the

value of the limit.

Let A= 3 20 0

sin 2 sin 0 2cos 2 cos 2lim lim0 3 0x x

x a x x a x a finitex x→ →

+ + + = = ≠

We can continue to apply the L’Hospital’s Rule, if 2+a=0 i.e., a= -2 .

2For a = − ,

20 0

2cos 2 2cos 0 4sin 2 2sin 0lim lim3 0 6 0x x

x x x xAx x→ →

− − + = =

0

8cos 2 2cos 8 2lim 16 6x

x x→

− + − += = = −

lim 2 1.The given it will have a finite value when a and it is∴ = − −

Example 38: Find the values of ‘a’ and ‘b’ such that 30

(1 cos ) sin 1lim3x

x a x b xx→

− += .

3 20 0

(1 cos ) sin 0 (1 cos ) sin cos 1lim lim0 3 0x x

x a x b x a x ax x b x a bLet A finitex x→ →

− + − + + − + = = = ≠

We can continue to apply the L’Hospital’s Rule, if 1-a+b=0 i.e., a-b = 1 .

1For a b− = ,

20 0

(1 cos ) sin cos 0 2 sin cos sin 0lim lim3 0 6 0x x

a x ax x b x a x ax x b xAx x→ →

− + + + − = =

0

3 cos sin cos 3lim6 6x

a x ax x b x a b finite→

− − −= = =

1 3 1. . ., 3 23 6 3

a bThis finite value is given as i e a b−= ⇒ − =

1 11 3 2 .2 2

Solving the equations a b and a b we obtain a and b− = − = = = −

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Example 39: Find the values of ‘a’ and ‘b’ such that 20

cosh coslim 1x

a x b xx→

−= .

20

cosh coslim0x

a x b x a bLet A finitex→

− −= = ≠

We can continue to apply the L’Hospital’s Rule, if a-b=0, since the denominator=0 .

0For a b− = ,

20 0 0

cosh cos 0 sinh sin 0 cosh coslim lim lim0 2 0 2 2x x x

a x b x a x b x a x b x a bAx x→ → →

− + + + = = = =

1. 2But thisis given as a b∴ + =

0 2 1 1.Solving the equations a b and a b we obtain a and b− = + = = = Limits of the form 0 00 , 1and ∞∞ : To evaluate such limits, where function to the power of function exists, we call such an expression as some constant, then take logarithm on both

sides and rewrite the expressions to get 00

or ∞∞

form and then apply the L’Hospital’s Rule.


0lim x

xx

→

0

0lim (0 )x

xLet A x form

→=

Take log on both sides to write

0 0 0

loglog lim log lim .log (0 ) lim1/

xe x x x

xA x x x formx→ → →

∞ = = ×∞ = ∞

20 0

1/ 0lim lim 0( 1/ ) 1 1x x

x xx→ →

−= = = =

−

0

0log 0 1 lim 1x

e xA A e x

→= ⇒ = = ∴ =


0lim(cos ) xx

x→

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21

0lim(cos ) (1 )xx

Let A x form∞

→=


21

2 20 00

1 log cos 0log lim log(cos ) lim log cos ( 0 ) lim0

xe x xx

xA x x formx x→ →→

= = ∞× =

2

0 0

tan 0 sec 1lim lim2 0 2 2x x

x xx→ →

− − − = = =

211

20

1 1 1log lim(cos )2

xe x

A A e xe e

−

→= − ⇒ = = ∴ =

Example 42: Evaluate cos

2

lim(tan ) x

xx

π→

cos 0

2

lim(tan ) ( )x

xLet A x form

π→

= ∞

Take log on both sides to write cos

2 2

log lim log(tan ) lim cos log(tan ) (0 )xe

x xA x x x form

π π→ →

= = ×∞

2

2

2 2 2

log tan sec / tan cos 0lim lim lim 0sec sec . tan sin 1x x x

x x x xx x x xπ π π

→ → →

∞ = = = = = ∞

0 cos

2

log 0 1 lim(tan ) 1xe

xA A e x

π→

= ⇒ = = ∴ =


0

tanlim( ) xx

xx→

21

0

tanlim( ) (1 )xx

xLet A formx

∞

→=


21

20 0

tan 1 tanlog lim log( ) lim log( ) ( 0 )xe x x

x xA formx x x→ →

= = ∞ ×

2

20 0

tan sec 1log( ) 0 tanlim lim0 2x x

x xx x x

x x→ →

− = =

20 0 0

1 1 2 12 sin 2 0sin .cos sin 2lim lim lim

2 2 2 sin 2 0x x x

x xx x x x xx x x x→ → →

− − − = = =

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2 20 0

2 2cos 2 0 4sin 2 0lim lim4 sin 2 4 cos 2 0 4sin 2 16 cos 2 8 sin 2 0x x

x xx x x x x x x x x→ →

− − = == + + −

20

8cos 2 8 1lim24cos 2 48 sin 2 16 cos 2 24 3x

xx x x x x→

− − −= = =

− −

211 1

3 30

1 tanlog lim( )3

xe x

xA A e ex

− −

→= − ⇒ = ∴ =


0lim( )x xx

a x→

+

1

0lim( ) (1 )x xx

Let A a x form∞

→= +


1

0 0

1log lim log( ) lim log( ) ( 0 )x xxe x x

A a x a x formx→ →

= + = + ∞ ×

0 0

log( ) 0 ( log 1) /( )lim lim0 1

x x x

x x

a x a a a xx→ →

+ + + = =

log 1 log log loga a e ae= + = + =

1

0log log lim( ) .x x

e xA ea A ea Hence a x ea

→∴ = ⇒ = + =

Example 45: Evaluate tan

2lim(2 )xa

x a

xa

π

→−

tan

2lim(2 ) (1 )xa

x a

xLet A forma

π∞

→= −


( )tan

2log lim log(2 ) lim tan .log(2 ) 02

xa

e x a x a

x x xA forma a a

π π→ →

= − = − ∞×

2

2

( 1/ )

log(2 ) 2 sin0 2 22lim lim lim .0cot cos 2

2 2 2x a x a x a

ax x xa a a

x x xeca a a a

π

π π π π π→ → →

−

− − = = = = − −

2 2tan

22log lim(2 ) .xa

e x a

xA A e Hence ea

ππ π

π →∴ = ⇒ = − =

112 of 133




22

1

0 0

1 cos( ) lim(cos ) ( ) lim2

b xx

x x

xi ax ii→ →

+

2

11

2 log(1 )

1 0

sin( ) lim(1 ) ( ) limx

x

x x

xiii x ivx

−

→ →

−

( )cottan

0 0( ) lim(sin ) ( ) lim 1 sin xx

x xv x iv x

→ →+

( )2 tan 2cos

04

( ) lim(cos ) ( ) lim tan xec x

x xvii x viii x

π→ →

1

0

1( ) lim( ) ( ) lim1 3

x x x xx

x x

ax a b cix xax→∞ →

+ + + −

CURVATURE

Consider a curve C in XY-plane and let P, Q be any two neighboring points on it. Let arc

AP=s and arc PQ=δs. Let the tangents drawn to the curve at P, Q respectively make angles ψ

and ψ+δψ with X-axis i.e., the angle between the tangents at P and Q is δψ. While moving

from P to Q through a distance‘δs’, the tangent has turned through the angle ‘δψ’. This is

called the bending of the arc PQ. Geometrically, a change in ψ represents the bending of the

x

y

C

P

ψ ψ+δψ

δs Q

δψ

o

113 of 133



curve C and the ratio s

δψδ

represents the ratio of bending of C between the point P & Q and

the arc length between them.

∴Rate of bending of Curve at P is

Q P

d Ltds sψ δψ

δ→=

This rate of bending is called the curvature of the curve C at the point P and is denoted by κ

(kappa). Thus ddsψκ =

We note that the curvature of a straight line is zero since there exist no bending i.e. κ=0, and

that the curvature of a circle is a constant and it is not equal to zero since a circle bends

uniformly at every point on it.

ψκρ

ρκ

κ

dds1

letter).Greek - (rhoby denoted is and curvature of radius thecalled is 1 then 0, If

==∴

≠

Radius of curvature in Cartesian form :

Suppose y = f(x) is the Cartesian equation of the curve considered in figure.

we have ( )2

2 22

dy d y d d ds1y Tan y Sec Tandx dx dx ds dx

ψ ψψ ψ ψ′ ′′= = ⇒ = = ⋅ = + ⋅

2dsBut we know that 1dx

dydx

= +

ψ

0

y

x

c

114 of 133



32 2

2 22

22

2

1ds1 1d

dydxd y dy d dy

d ydx dx ds dxdx

ψψ

+ ∴ = + ⋅ ⋅ + ⇒ =

( )

32 21 yds

d yρ

ψ

′+ ∴ = =′′

This is the expression for radius of curvature in Cartesian form.

NOTE: We note that when y’=∞, we find ρ using the formula

32 2

2

2

1 dxdy

d xdy

ρ

+

=

Example 9: Find the radius of curvature of the curve x3+y3 = 2a3 at the point (a, a).

3 3 3 2 22 3 3 0x y a x y y′+ = ⇒ + ⋅ = ( )2

2 , , 1xy hence at a a yy

′ ′⇒ = − = −

( ) ( ) ( )2 2 3 3

4 4

2 2 2 2 4, , ,y x x y y a ay hence at a a y

y a a′ − +′′ ′′∴ = − = − = −

( ) ( )

3 32 22 21 1 1

. ., 2 24 4 2

y a ai ey a

ρ ρ ′+ + − ∴ = = = ⋅ =

′′ −

Example 10: Find the radius of curvature for x y a+ = at the point where it meets

the line y=x.

On the line y x, x x a . 2 x a4ai e or x= + = = =

i.e., We need to find at ,4 4a aρ

1 1x y a 0 . , , , 14 42 x 2 y

y a ay i e y hence at yx

′ ′ ′+ = ⇒ + = = − = −

115 of 133



1 12 2

,x y y

y xAlso y

x

′⋅ − ′′ = −

1 1( 1)4 4 1 12 2 ( ) ( 1) 44 4 2 2, ,

4 44 4 4

a aa a

a aat y a a a a

⋅ − −

− − − ′′∴ = − = − = − =

( ) ( )3 3

2 22 21 1 12 24 4 2

y a ay a

ρ ′+ + − ∴ = = = =

′′

Example 11: Show that the radius of curvature for the curve y 4 Sin x - Sin 2x =

5 5at x is 2 4π=

y 4 Sin x - Sin 2x 4 2 2y Cos x Cos x′= ⇒ = −

when x , y 4 2 0 2( 1) 22 2Cos Cosπ π π′∴ = = − = − − =

Also, y -4 Sin x 4 Sin 2x and when x , y -4 Sin 4 Sin 42 2π π π′′ ′′= + = = + = −

( )

3 32 2 2 21 y 1 2 5 54 4y

ρ ρ ′+ + ∴ = = ⇒ =

′′ −

Example 12: 2 3 3Find the radius of curvature for xy = a - x at (a, 0).

2 3 3 2 22 3xy a x y xy y x′= − ⇒ + = −

2 23 ( ,0),

2x yy and at a y

xy− −′ ′∴ = = ∞

2 2

dx 2 dxIn such cases we write ( , 0), 0dy 3 dy

xy and at ax y

= =− −

116 of 133



( )

( )

2 22

22 2 2 2 2

dx dx3 2 2 2 6 2dy dydx 2 d xAlso

dy 3 dy 3x y

x y y x xy x yxy

x y

+ + − + − = ⇒ =

+ +

( ) ( )( )( )

22 3

22 42

3 0 0 2 0d x 6 2, 0 ,dy 9 33 0

a a aAt aa aa

+ + − − − ∴ = = = +

32 2

32 2

22

11 3

2 23

dxdy o aor

d x ady

ρ ρ

+ + ∴ = = =

−

Exercise: Find the radius of curvature for the following curves:

3 3

22

2 3 3

2 2

2 2

2 3 3

2 2

3 3(i) x 3 ,2 2

( )(ii) ( ,0)

(iii) int

( ) 1 .

( ) int ( ,0).( ) 2 (3 ) int

a ay axy at

a a xy at ax

a y x a at the po where it crosses x axisx yiv at the ends of the major axisa b

v a y x a at the po avi y x x at the po where t

+ =

−=

= − −

+ =

= −

= −2 2 2

2 2 23

tan .( ) ( ) ( 2 , 2 ).

2( )

he gents are parallel to x axisvii x y a x y at a a

ax x yviii For the curve y show thata x a y x

ρ

−

= + −

= = + +

117 of 133



M.G.Geetha

INTEGRAL CALCULUS Gamma and Beta Functions

Definition:

The improper integral ( ) dxxen 1n

0

x −∞

−∫=Γ is defined as the gamma function.

Here n is a real number called the parameter of the function. Γ(n) exists for all real values of

n except 0, -1, -2., ………..the graph of which is shown below :

Recurrence formula

( ) 0 n ,1

0

>=Γ −∞

−∫ dxxen nx

∫∞

−∞

− +

=0

xn

0

xn

,dxen

xen

x integrating by parts.

Applying the definition of Gamma function and using ∞→→ xas 0x n

xe

( ) ( ) )2.......(..........1 nnnor Γ=+Γ

-4 -3 -2 -1 0 n

r(n)

( ) ( ) )1.....(....................1n

nn +Γ=Γ

118 of 133



Note:

(1) Γ(n) is not convergent when n = 0, -1, -2, …….

(2) If Γ(n) is known for 0 < n < 1, then its value for 1 < n < 2 can be found using equation (2).

Also, its values for -1 < n < 0 can be got using equation (1)

(3) If n is a +ve integer, using the recurrence relation and Γ(1) = 1, we get

Γ(n) = (n - 1) Γ(n - 1)

= (n - 1) (n - 2) Γ(n - 2) and so on

= (n - 1) (n - 2) (n - 3) …….. 1

= (n - 1) !

Problems:

(1) Prove that π=

Γ

21

By definition, ( ) ∫∞

−−=Γ0

1nx dxxen

t xsin 212

0

2

== −∞

−∫ gudtte nt

( ) ∫∞

−−=Γ∴0

1n22x dxxe2n

∫∞

−−=0

1n22y dyye2

∫∞

−=

Γ⇒

0

2x dxe21 ∫

∞−=

0

2y dye

∫ ∫∞ ∞

+−

=

Γ∴

0 0

2y2x2dxdye 4

21

Converting Cartesian (x, y) to polar (r, θ) system, we get

∫ ∫=

∞

=

−=

Γ

2

0 0

2

)2( 221 2

π

θ

θr

r drdre

119 of 133



[ ]∫ ∫ −∞− ==2

0

r-0

222

2re- ,π

θ rr edrde

∞→→θ= ∫π

r as 0e ,d22r-

2

0

= π

Taking square roots on both sides, we get π=

Γ

21

(2) Show that

Γ=∫

∞−

41

41dxe

0

4x

dyy41dxdydxx4y xLet 4

334 −=⇒=⇒=

∫∫∞

−−∞

− =∴0

43

0 414

dyyedxe yx .definition theusing ,41

41

Γ=

(3) Find ( )∫1

0

5dx xlnx

using ln x = -y, we get x = e - y , dx = -e - y dy

and y ranges from ∞ to 0 when x = 0 to 1

( ) ∫∫∞

−∞

−=∴0

y65

0

5 ,dyeydxln x x interchanging the lower and the upper limits

,6

dtte0

6

5t∫

∞−= choosing 6y = t

( ) 66 6!56

61

−=Γ−=

(4) Prove that πa

xa

dxa

=

∫0 ln

using tewetxa

==

xaget ,ln

dx = -ae-t dt and t = ∞ to 0 when x = 0 to a

120 of 133



∴ ∫∫∞

−−=

0

t21a

0dte.at

xaln

dx πaa =

Γ=

21

(5) Show that ( ) ( )( )∫ ++

−=

1

01n

nnm

1m!n1dxx logx .

Hence show that 4log1

0

−=∫ dxxx when 'n' is a positive integer and m > -1.

Let logx = - t ⇒ dx = -e-t dt and t = ∞ to 0 when x = 0 to 1.

( ) ( ) ( ) ( )∫ ∫ ∫∞ ∞

+−−− −=−=∴1

0 0 0

11 xlog dtetdtetedxx tmnntnmtnm

using ( ) ( ) dxytm n xlogx ,11

0

m∫=+ ( )( )

1- m ,1m

dye1m

y1 y

0n

nn >

++−= −

∞

∫

( ) ( )( ) ∫∫

∞−

++

−=∴

0

ny1n

nn

1

0

m dyye1m1dx xlogx

( )( )

( ) !n1n ,!n1m1

1n

n=+Γ

+

−=

+

choosing get we,21 m and 1 −==n the required result.

Beta function Definition: Beta function, denoted by B(m,n) is defined by

( ) ( ) dxx1xn,mB 1n1

0

1m −− −= ∫ , where m and n are positive real numbers.

By a property of definite integral, ( ) ( ) dxxxdxxx nmnm 11

0

111

0

1 11 −−−− ∫∫ −=−

Therefore we have B(m,n) = B(n,m)

Alternate Expressions for B(m,n)

( ) ( )∫ −− −=1

0

11 )1.......(....................1, dxxxnmB nm

121 of 133



( ) t1

tx-1 ,1

1t1

1xLet )( 2 +=

+−=⇒

+= dt

tdxi

and t = ∞ to 0 when x = 0 to 1.

( )( )

dtttt

nm

∫∞ −−

+

+

+=∴

02

11

11

11

11 n)B(m, toreduces 1

( )

( ) ( )mnBdtt

tnm

m

,nm,B using 1

also is n)B(m, 0

1

=+

= ∫∞

+

−

( ) θθθθ ddxii cossin2sin Let x 2 =⇒= 1 to0 when x 2

to0 and =π

=θ

∫π

− θθθ2

0

1m21-2n dcossin2 toreduces (1) Then

∫ −=∴2

0

121-2m cossin2 n)B(m, π

θθθ dn

Relation between Gamma and Beta functions.

Prove that ( ) ( ) ( )( )nm

nmn,mB+ΓΓΓ

=

Proof:

We know that ( ) dxxe2n 1n2

0

2x −∞

−∫=Γ ( ) dyxeand my 12

0

2

2m −∞

−∫=Γ

( ) ( ) ( ) dxdyyxen mnyx 12

0

12

0

22

4mThen −∞

−∞

+−∫ ∫=ΓΓ

( ) θθθ= ∫∫π

=θ

−−−+∞

− dcossin drre42

0

1n21m21nm2

0

2r in polar coordinates

= Γ(m+n) B(m,n) by definitions.

Note:

1 n 0 ,nsin

dxx1

x gsinU0

1-n<<

ππ

=+∫

∞

We get Γ(n) Γ(1-n) = Γ(1) B (1-n,n) using the above relation

122 of 133



( ) dxx1

xnn,-1BBut 0

1n

∫∞ −

+= using definition of Beta function.

Therefore Γ(n) Γ(1-n) = π /sin nπ, 0 < n < 1

Problems

(1) Show that

=

−∫ 2

1,52Bdx

x1

x51

05

dt54t

51dx,tget x we,t xLetting 5

15 −=== and t = 0 to 1 when x = 0 to 1

( ) dtt51t1t5dx

x1

x5 54

211

0

511

05

−−−=−

∴ ∫∫

( ) dtt1t 211

0

53 −−

−= ∫

.definitionby 21,

52

= B

(2) Evaluate ( )( ) ( )

dxx1

xdxx1

x1x

030

18

030

810

∫∫∞∞

+−

+

−

= B (11,19) - B(19,11) = 0, using B(m,n) = B(n,m).

(3) Evaluate ∫π

θθ2

0dtan

∫∫−=

2

0

21

212

0

cossintanππ

θθθθθ dd

definitionby ,41,

43

21

= B

(4) Show that ( ) 1 n 0, t ,1n,tBndxxnx1 t1t

n

0

n>>+=

− −∫

n to0 when x 1 to0 y and ndydxynx Put ===⇒=

123 of 133



( ) ndyynydxxnx ttnt

n n11

1

0

1

0

11 −−− ∫∫ −=

−∴

( ) ( ) .definitionby 1,1 11

0

+=−= −∫ ntBndyyyn ttnt

(5) Prove that ( )

= − 2

1,2

1, 12 nnnB n B

∫ −=2

0

121-2n cossin2 n)B(n, that know Weπ

θθθ dn

θθ=θθθ

= ∫π

−

−cos2sin sin2 ,d

22sin2

2

01n2

1n2

∫π

−

−φ=θφ

φ=

01n2

1n22 putting ,d

2sin

( ) φφππφφφπ

sin-sin as )(0,in even is sin ,2

sin22

012

12

== ∫ −

−

dn

n

( ) 21,

21nn,B ,function Beta of definition theApplying 12

= − ngetwe n B

(6) Prove the Duplication formula

( ) ( )nnn n 222

112 Γ=

+ΓΓ −

π

Using the previous result,

+=

++

21,

21nB

21

21n,

21nB n2 …………(1)

( )

( ))2..(....................

21

2112

121

21

21,

21

21,

21

Also

Γ

+Γ+Γ

+Γ

+Γ

+Γ

=

+

++

nn

nnn

nB

nnB

( )

( ) nnn

nnFrom 22

12221n

get we(2) and (1) =Γ

Γ

+Γ

π

∴ ( ) ( )nnn n 222

1. 12 Γ=

+ΓΓ −

π

124 of 133



(7) Prove that ( )( )

dxx1

xxn,mB1

0nm

1n1m

∫ +

−−

+

+=

By one of the definitions of Beta function, we have

( )( )∫

∞

+

−

+=

0nm

1ndx

x1xn,mB

( ) ( )

dxx1

xdxx1

x

1nm

1n1

0nm

1n

∫∫∞

+

−

+

−

++

+=

= I1 + I2 (say)

Substituting get we,Iin y1x 2=

( ) ( )

dyy1

ydyy1

y1yyI

1

0nm

1m0

12nm1n

nm2 ∫∫ +

−

+−

+

+=

−

+=

( )( )

dxx1

xxnm,B Hence1

0nm

1n1m

∫ +

−−

+

+=

Conclusion: We evaluated double integrals using change of order of integration. We used

change of variables for double and triple integrals to simplify the integrals in certain cases.

We saw how Gamma and Beta functions were useful in evaluating complicated integrals.

Multiple Choice:

1. The value of ∫ ∫1

0

2

1

2 is dydxxy

(a) 1/2

(b) 7/3

(c) 7/6

(d) 7/2

2. The value of ∫ ∫1

0

2x

xisdy xdx

(a) -1/12

(b) -1/4

(c) 0

(d) -1/15

125 of 133



3. The value of ( )∫ ∫ ∫−

+

−

++1

1

z

0

zx

zxis dxdydzzyx

(a) 1

(b) 0

(c) 2

(d) 8

4. The value of Γ(-10) is

(a) 10!

(b) 9!

(c) -9!

(d) Not defined

5. The value of Γ(-5/2) is

(a) π− 2

(b) π34

(c) π−158

(d) Not defined

6. ( )( )

tosimplifies x1

dxx1x

026

89

∫∞

+

−

(a) 0

(b) ( )!28

18! !10

(c) ( )!27

17! !9

(d) ( )!2717! !9 ,2

7. ∫π

θθ2

0 toequal is dcot

126 of 133



(a) 2π

(b) 2π

(c) 2 π

(d) π / 2

8. The value of B(1/2, 1/2) is

(a) π

(b) π

(c) π2

(d) 1

Questions & Answers

(1) Evaluate ( )∫ ∫ ∫

− −

+++

1

0

x1

0

yx

03zyx1

dxdydz

Ans: Let this integral be I, then

( )∫ ∫∫−

−−

−

−

−

+++=

1

0x

yx1

0z3

x1

0ydxdy

zyx1dzI

( )∫ ∫−

−

−

−−

−

+++

−=

1

0x

x1

0y

yx1

0z3 dydx

zyx121

( )∫ ∫−

−

−

+++

−=

1

0x

x1

0y2 dx

zyx121

−=

852ln

21

(2) Change the order of integration and hence evaluate the integral

∫ ∫−1

0

x2

xdxdy

yx

Ans: The region of integration is the shaded portion shown in the figure below

Y y = x

x = 0

B

127 of 133



To get the limits for x first, we need to divide the shaded area into two parts AMB and AMO

where the x values are 0 to 2-y and 0 to y respectively. The respective y values are 1 to 2 and

0 to 1.

The given integral is now

∫ ∫ ∫ ∫− − −

−

−

+=1

0y

y

0x

2

1y

y2

0xdxdy

yxdxdy

yx

dy22y

y2dy

2y 2

1y

1

0y∫∫−−

−++=

= 2 ln 2 -1.

(3) Change to polar coordinates and hence evaluate

∫ ∫+

a

0

a

y22

dxdyyx

x

Ans: The region of integration is as shown below

X

Y

0

x = y

x = a

y = a

y = 0

128 of 133



We see that θ ranges from 0 to π/4 in the shaded region. R ranges from 0 to x = a i.e., a sec θ.

The given integral

∫ ∫θ

π

θθ=seca4

0 0

22 drdcosr

( )[ ] 40

34

0

3tansecln

3adsec

3a π

π

θ+θ=θθ= ∫

( )12ln3

a3+=

129 of 133



(4) Find the area between the parabola y = 4x - x2 and the line y = x, using double integration.

Ans: The required area is ∫∫r

dxdy

( )∫∫ ∫−−

−

−

−==3

0x

23

0x

2xx4

xydxxx3dxdy

29

3xx

23 3

23

0

=

−=

Practice Questions 1. Evaluate the following integrals

( ) ( )∫ ∫ ++3

0

2

1dxdyyx1x i

( ) ∫ ∫π

θθ2

0

a

0

2 drdsinr ii

( ) ( )∫ ∫ +1

0

x

x

22 dydxyx iii

( ) ∫ ∫a

1

b

1dydx

xy1 iv

( ) ∫ ∫ ∫ ++1

0

1

0

1

0

zyx dxdydze v

(2,4) y = x

(3,3)

130 of 133



( ) ∫ ∫ ∫3

1

1

x1

xy

0dxdy dz xyz vi

( ) ∫ ∫ ∫π

θ −θ

2a

0

cos

0

2r2a

0ddr rdz vii

2. Change the order of integration and hence evaluate

( ) ∫ ∫1

0

y-2

y

xydxdy i

( ) ∫ ∫ 4a

0

ax2

a42x

dydx ii

( ) ∫ ∫a

0

x-2a

a2x

xydydx iii

( )( )( )∫ ∫ −−

a

0

x

0dydx

yaxaycos iv

3. Change to respective coordinate systems and hence evaluate

( ) ∫ ∫+

2

0

2x-2x

022

systempolar todydxyx

x i

( ) ∫ ∫ +a

0

2x-2a

0

222 systempolar todydxyxy ii

( ) ∫ ∫ +1

0

2x-2x

x

22 systempolar todydx yx iii

( ) polar spherical tozyx1

dzdydx iv1

0

2x-1

0

2y2x1

0222∫ ∫ ∫

−−

−−−

131 of 133



( ) polar spherical tozyx

dzdydx v1

0

2x-1

0

1

2y2x222∫ ∫ ∫

−++

( ) polar lcylindrica toxyzdxdydz vi1

0x

1

0y

2

2y2xz

∫ ∫ ∫= = +=

4. Find the area between y = 2 - x and circle x2 + y2 = 4

5. Find the area bounded between the parabola y2 = 4ax and x2 = 4ay

6. Show that the area of one loop of the lemniscates r2 = a2 cos2θ is a2/2.

7. Find the area of one petal of the rose r = a sin3θ.

8. Find the area of the circle r = a sinθ outside the cardioide r = a (1 - cosθ)

9. Find the volume of the paraboloid of revolution x2 + y2 = 4z cut off by the plane z =4.

10. Find the volume of the region bounded by the paraboloid az = x2 + y2 and the cylinder

x2+ y2 = R2

11. Find the volume of the portion of the sphere x2 + y2 + z2 = a2 lying inside the cylinder

x2 + y2 = ax.

12. Find the volume cut off the sphere x2 + y2 + z2 = a2 by the cone x2 + y2 = z2

13. dxex Evaluate0

4x2∫∞

−

14. ( )∫1

0

3x

1log Find dx

15. ( ) 313

2

0

4x8

163x Evaluate

−−∫ dx

16. ∫ <<2

0

p 1 p 0 ,tan π

θθ dFind

17. ( )ndtet2a that Show0

2at1n2n Γ=∫∞

−−

18. a log2

1 is 0 a ,dxe that Show0

alog2x- π>∫

∞

132 of 133



19. If m and n are real constants > -1, prove that

( )( ) 1n

n1

0

m

1m1ndx

x1logx

++

+Γ=

∫

20. Show that ( )2

cos1cos0

1 πnna

dxaxx nn Γ=∫

∞−

Hint: Write cos ax = Real part of eiax

21. Show that

=

−∫ 2

1,n1B

n1

x1

dx1

0n

22. Show that ( ) ( ) ( ) dxx1x12

1n,mB 1n1

1

1m1nm

−

−

−−+

−+= ∫

Hence deduce that ∫− −

+1

1 11

xx dx = π

23. Show that 216

dxexdxxe0

4x2

0

8x π=∫∫

∞−

∞−

24. Show that ( ) ( ) ( ) ( )n,mBabdxxbaxb

a

1nm1n1m∫ −+−− −=−−

133 of 133

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LAPLACE TRANSFORMS - BookSpar · LAPLACE TRANSFORMS . INTRODUCTION Laplace transform is an integral transform employed in solving physical problems. Many physical problems when analysed