Benginning Calculus Lecture notes 12 - anti derivatives indefinite and definite integrals

Beginning Calculus- Antiderivatives and The Definite Integrals -

Shahrizal Shamsuddin Norashiqin Mohd Idrus

Department of Mathematics,FSMT - UPSI

(LECTURE SLIDES SERIES)

VillaRINO DoMath, FSMT-UPSI

(I1) Antiderivatives and The Definite Integrals 1 / 37

Anti Derivatives Area Under a Curve Riemann Sums Net Area Properties of the Definite Integral

Learning Outcomes

Use substitution and advanced guessing methods to evaluate antiderivatives.

Compute Riemann Sums.

Compute areas under the curve and net areas.

State and apply properties of the definite integrals.




Anti Derivatives

G (x) =∫g (x) dx

G (x) is called the anti derivative of g , or the indefinite integralof g .

G ′ (x) = g (x)




Example

∫sin xdx = − cos x + C




Example

∫xadx =

xa+1

a+ 1+ C , for a 6= −1




Example

∫ dxx= ln |x |+ C




More Examples

∫sec2 xdx = tan x + C∫ dx√1− x2

= sin−1 x + C∫ dx1+ x2

= tan−1 x + C∫exdx = ex + C




Uniqueness of anti derivatives up to a constant

Theorem 1

If F ′ = G ′, then F (x) = G (x) + C .

Proof.

Suppose F ′ = G ′. Then,

(F − G )′ = F ′ − G ′ = 0F (x)− G (x) = C

⇒ F (x) = G (x) + C




Method of Substitution - For Differential Notation

∫x3(x4 + 2

)5dx




Example

∫ xdx√1+ x2




Example

∫e6xdx




Example - Advanced Guessing

∫xe−x

2dx




Example

∫ dxx ln x




Area Under a Curve

b

( )dxxfb

a∫

a

( )xfy =

Area under a curve =∫ ba f (x) dx




Area Under a Curve

To compute the area under a curve:

b

L

a

1 Divide into n rectangles2 Add up the areas3 Take the limit as n→ ∞ (the rectangles get thinner and thinner).




Example

f (x) = x2; a = 0, b = arbitrary

a = 0 nb/n

f(x) = x2

b/n 2b/n

f(x)L

L3b/n

divide into n rectangles

each rectangle has equal base-length =bn.




Example - continue

Base xbn

2bn

3bn

· · · b =nbn

Height f (x)(bn

)2 (2bn

)2 (3bn

)2· · · b2

The sum of the areas of the rectangles(bn

)(bn

)2+

(bn

)(2bn

)2+

(bn

)(3bn

)2+ · · ·+

(bn

)(nbn

)2=

(bn

)3 (12 + 22 + 32 + · · ·+ (n− 1)2 + n2

)=

(bn

)3 n

∑i−1

i2




Example - continuen

∑i=1

i2 =n (n+ 1) (2n+ 1)

6.

b3

n3

(n (n+ 1) (2n+ 1)

6

)=

b3(2n3 + 3n2 + n

)6n3

=2b3 +

3n+1n2

6Take the limit as n→ ∞.

limn→∞

2b3 +3n+1n2

6=b3

3So the sum of the areas of the rectangles:∫ b

0x2dx =

b3

3




Examples

f (x) = x . The area under the curve:∫ b0xdx =

b2

2

f (x) = 1. The area under the curve:∫ b01dx =

b1

1= b

In general, f (x) = xn . The area under the curve:∫ b0xndx =

bn+1

n+ 1




General Procedures for Definite Integrals

Divide the base into n intervals with equal length ∆x .

x∆

ixa b

( )xfy =

( )ixf

∆x =b− an

; xi = a+ i∆x

The Riemann sum:n∑i=1

f (xi )∆x :

∫ baf (x) dx = lim

n→∞

n

∑i=1

f (xi )4x




General Procedures for Definite Integrals - continue

Note that: ∫ baf (x) dx = lim

n→∞

n

∑i=1

f (xi )4x

can also be written as∫ baf (x) dx = lim

n→∞

(b− an

) n

∑i=1

f(a+

i (b− a)n

)




Example - continue

To evaluate∫ 10x2dx : 4x = 1− 0

n=1n; xi = 0+ i4x =

in.

So, the definite integral is∫ 10x2dx = lim

n→∞

n

∑i=1

f(in

)(1n

)= lim

n→∞

(1n

) n

∑i=1

f(in

)= lim

n→∞

(1n

) n

∑i=1

i2

n2

= limn→∞

(1n3

) n

∑i=1

i2




Example - continue

= limn→∞

(1n3

) n

∑i=1

i2

= limn→∞

(1n3

)(n (n+ 1) (2n+ 1)

6

)= lim

n→∞

2n2 + 3n+ 16n2

= limn→∞

2+3n+1n2

6=26=13




Example

f (x) = x3 − 6x is a bounded function on [0, 3] . To evaluate theRiemann sum with n = 6,

4x = 3− 06

= 0.5

x1 = 0+ 0.5 = 0.5, x2 = 1.0, x3 = 1.5, x4 = 2.0, x5 = 2.5, x6 = 3.0.

So, the Riemann sum is

n

∑i=1

f (xi )4x

=12[f (0.5) + f (1.0) + f (1.5) + f (2.0) + f (2.5) + f (3.0)]

=12(−2.875− 5− 5.625− 4+ 0.625+ 9)

= −3.9375




Example - continue

To evaluate the definite integral∫ 30

(x3 − 6x

)dx : 4x = 3− 0

n=3n; xi = 0+ i4x =

3in.

So, the definite integral is∫ 30

(x3 − 6x

)dx = lim

n→∞

n

∑i=1

f (xi )4x

= limn→∞

n

∑i=1

f(3in

)(3n

)

= limn→∞

(3n

) n

∑i=1

[(3in

)3− 6

(3in

)]




Example - continue

= limn→∞

(3n

) n

∑i=1

[27n3i3 − 18

ni]

= limn→∞

(3n

)[(27n3

) n

∑i=1

i3 −(18n

) n

∑i=1

i

]

= limn→∞

[(81n4

) n

∑i=1

i3 −(54n2

) n

∑i=1

i

]

= limn→∞

[(81n4

)(n (n+ 1)

2

)2−(54n2

)(n (n+ 1)

2

)]




Example - continue

= limn→∞

[(814

)(n4 + 2n3 + n2

n4

)−(542

)(n2 + nn2

)]= lim

n→∞

[(814

)(1+

2n+1n2

)− 27

(1+

1n

)]= lim

n→∞

[(814

)(1+

1n

)2− 27

(1+

1n

)]

=814− 27 = −27

4




Net Area

Geometrically the value of the definite integral represents the areabounded by y = f (x) , the x−axis and the ordinates at x = a andx = b only if f (x) ≥ 0.If f (x) is sometimes positive and sometimes negatives, the definiteintegral represents the algebraic sum of the area above and belowthe x−axis (the net area).




Area Under The Curve and Net Area

b

x∆

kxa

( )xfy =

y

x

If f (x) ≥ 0, the Riemann

sumn

∑k=1

f (xk ) · 4x is the

sum of the areas of rectangles.

ba

y

x

( )xfy =

If f (x) ≥ 0, the Integral∫ baf (x) dx is the area under

the curve from a to b.




Area Under The Curve and Net Area

x

y

)(xfy =+ +

ba

∫ baf (x) dx is the net area




Extend Integration to the Case f < 0 - Example

∫ 2π

0sin xdx

x

y

∫ 2π

0sin xdx = (− cos x)|2π

0

= (− cos 2π)− (− cos 0) = −1+ 1 = 0




Total Distance and Net Distance

Total distance travelled: ∫ ba|v (t)| dt

Net distance travelled: ∫ bav (t) dt




Monotonicity, Continuity and Integral

Theorem 2

Every monotonic function f on [a, b] is integrable.

Theorem 3

Every continuous function f on [a, b] is integrable.




Properties of the Definite Integral

Let f and g be integrable functions on [a, b], and c is a constant. Then,

1.∫ bacdx = c (b− a)

2.∫ aaf (x) dx = 0

3.∫ baf (x) dx = −

∫ abf (x) dx

4. cf is integrable and∫ bacf (x) dx = c

∫ baf (x) dx .

5. f ± g is integrable and∫ ba(f ± g) (x) dx =

∫ baf (x) dx ±

∫ bag (x) dx .




Properties of the Definite Integral - continue

6.∫ baf (x) dx =

∫ caf (x) dx +

∫ bcf (x) dx provided that f is integral

on [a, c ] and [c , b] . (works without ordering a, b, c )

7. (Estimation) If f (x) ≤ g (x) for x ∈ [a, b] , then∫ baf (x) dx ≤

∫ bag (x) dx . (a < b )

8. |f | is integrable and∣∣∣∣∫ ba f (x) dx

∣∣∣∣ ≤ ∫ ba |f (x)| dx .




Example - Illustration of Property (6).

ex ≥ 1, x ≥ 0∫ b0exdx ≥

∫ b01dx

∫ b0exdx = (ex )|b0 = eb − 1∫ b01dx = b

eb ≥ 1+ b, b ≥ 0




Example - continue

Repeat:ex ≥ 1+ x , x ≥ 0∫ b

0exdx ≥

∫ b0(1+ x) dx

∫ b0exdx = (ex )|b0 = eb − 1∫ b

0(1+ x) dx =

(x +

x2

2

)∣∣∣∣b0= b+

b2

2

eb ≥ 1+ b+ b2

2, b ≥ 0

Repeat: Gives a good approximation of ex .



Education

Benginning Calculus Lecture notes 12 - anti derivatives indefinite and definite integrals