mathematics

© D.J.Dunn www.freestudy.co.uk 1

EDEXCEL NATIONAL CERTIFICATE

UNIT 28 – FURTHER MATHEMATICS FOR TECHNICIANS

OUTCOME 4 - CALCULUS

TUTORIAL 2 - INTEGRATION

CONTENTS

4 Be able to apply calculus

Differentiation: review of standard derivatives, differentiation of a sum, function of a function,

product and quotient rules, numerical values of differential coefficients, second derivatives,

turning points (maximum and minimum) e.g. volume of a rectangular box

Integration: review of standard integrals, indefinite integrals, definite integrals e.g. area under a

curve, mean and RMS values; numerical e.g. trapezoidal, mid-ordinate and Simpson’s rule

It is assumed that the student has completed the module MATHEMATICS FOR TECHNICIANS.

In this tutorial you will learn how to differentiate more complicated expressions.


1. REVISION

You should have covered integration in the core unit where you learned that integration is the reverse of

differentiation. The standard integrals are:

C1 -ln x x dx ln x

C x-tan x dxx tan

C2xsin 4

1 x

2

1 dxx cos

C2xsin 4

1 -x

2

1dxx sin

C-ln xdxtan x

Csin xa

1 dxax cos

C xcosa

1dxax sin

Cea

1dxe

Cln x dx x

C1n

axdxax

2

2

2

axax

1

1nn

FINDING AREAS

Lets consider how to find the area under the

graph of y = f(x) =3 + x2. A graph of this

function looks like this.

If we solve the area by use of calculus the

area would be precisely solved as follows.

Over the range x = 0 to x = 10 the area is

expressed as follows.

4x

0x

2 )dxx(3A

Carrying out the integration gives the following.

4

0

34x

0x

2

3

x3x)dxx(3A

Evaluating between limits we get the following.

units 33.33021.3312A

3

00 x 3

3

44 x 3

3

x3x)dxx(3A

334

0

34x

0x

2

This is a precise answer and we will compare it with the results found in the following work.


GRAPHICAL METHODS

Consider the same function again and this time more grid lines are shown.

COUNTING RECTANGLES

A simple but crude way to find the area under the graph is to count the rectangles. Each rectangle

on the graph above has an area of 1 unit. Count them up judging the divided ones to the nearest half.

You should get an answer of about 34 units depending on how good you are at doing it.

MID-ORDINATE RULE

The values of y corresponding to x = 0, x = 1, x = 2 and so on are called the ordinates. The values of

y corresponding to x = 0.5, x = 1.5, x = 2.5 and so on are called the mid-ordinates.

Each column is approximately a rectangle w wide and h high. The

area is approximately w h.

The area under the whole graph is approximately

A = w h1 + w h2 + w h3 +w h4

A = w(h1 + h2 + h3 + h4)

Usually, as in this case, w =1

Putting in the mid-ordinate values we find the following.

A = 1(3.25 + 5.25 + 9.25 + 15.25) = 33 units

Clearly if we took more strips by say making w = 0.5, we would get a more accurate answer and in

the limits as w becomes very small the answer will be the same as found by integrating.


TRAPEZOIDAL RULE

Consider that each strip has a straight line joining the top corners as shown. The height at the

middle is not quite the same as the mid-ordinate and is the average of the two ordinates. If h is the

average then

h1 = (A+B)/2 h2 = (B+ C)/2 h3 = (C+ D)/2 h4 = (D+E)/2

The area of each strip is

wh1 = w(A+B)/2 wh2 = w(B+ C)/2 wh3 = w(C+ D)/2

wh4 = w(D+E)/2

The total area is

A = (w/2)[(A+B) + (B+C) + (C+D) + (D+E)]

A = (w/2)[(A+B + B+C + C+D + D+E]

A = (w/2)[(A + E) + 2(B+C+D)]

Hence in our example

A = (1/2)[(3+19) +2(4+7+12)] = (1/2)(22+46) = 34

This is slightly larger than the correct figure but again, if smaller strips are used, the answer will be

accurate. The above rule may be written as follows.

rest theof sum x 2Last First 2

wA

WORKED EXAMPLE No.1

Find the area under the graph of the function y = sin between the limits 0 and radians using

integration and the trapezoidal rule.

SOLUTION

Evaluating the ordinates and mid-ordinates at intervals of /8 produces the table and graph shown.

Integration

211 0 cos πcoscosθ-dθ sinθAπ0

π

0


Mid-ordinates w = /4

2.0680.393)0.9240.924(0.3934

π )hhhw(hA 4321

Trapezoidal Rule w = /4

units 1.896707.00.1707.02002

π/4A


wA

Note one answer is slightly large and the other slightly small.

SIMPSON'S RULE

The area is divided into an even number of strips. The ordinates are h1, h2 .... The area is calculated

on the assumption that the curve joining neighbouring ordinates are a quadratic that passes through

the mid ordinate. It follows that if the curve is a parabola, the area will be exact. The derivation is

not given here as it is quite complicated but the result is as follows.

ordinates odd remaining theof sum2ordinateseven theof sum4last first 3

w

I

WORKED EXAMPLE No.2

Find the area under the curve f(x) = 2x2 + 4x + 8 between x = 0 and x = 4 using Simpson's Rule

with eight strips and determine the error.

SOLUTION

4

0

2 8)dx4x(2xA 106.67 0)4(8)4(23

)4(282

3

2 23

4

0

23

xx

x

SIMPSON'S RULE

ordinates odd remaining theof sum2ordinateseven theof sum4last first 3

wI

106.676403

0.515242464

3

0.5762106464

3

0.5I

382414246.530.518.510.5456 83

0.5I

The error is zero and it always is when the function is a quadratic. It follows that for a quadratic

you only need two strips.


SELF ASSESSMENT EXERCISE No.1

1. Find the area under the graph of the following functions using integration, the mid-ordinate

rule and the trapezoidal rule.

y = 2x3 between the limits x = 0 and x = 5

y = ex between the limits of x = 1 and x = 5.

y = sin x between the limits x = 0 and x = 180o.

2. Estimate the value of the definite integral 5

1

4dxxI by Simpson's rule using four strips. What

is the error in the estimate? (625.3 and by integration it is 624.8 giving an error of 0.53 units)

In Engineering the area under the graph represents real things.

For example the area under a force – distance graph represents the work done or energy used and

the area under a pressure – volume graph also represents work done during the compression or

expansion of a gas.


WORKED EXAMPLE No. 3

The pressure (p) and volume (V) during a gas expansion is related by the law

p = 0.2V-1.2

. Determine the work done when the volume is expanded from 10 x 10-6

m3 to 100 x

10-6

m3.

Use calculus and the trapezoidal rule to find the answer.

SOLUTION

INTEGRATION

Joules 3.69 10-6.309-1W

10x1010 x 100-1V1W

0.2-

V0.2

11.2-

V0.2 dV0.2VW

0.2Vpbut pdV Work Area

0.260.26-10x10

1x100.2

V

V

0.2-V

V

11.2-V

V

1.2

1.2V

V

6

6

2

1

2

1

2

1

2

1

TRAPEZOIDAL RULE w = 10 x 10-6

m3

Joules 3.872 (28.0922x1010 x 212.6210 x 5 W

....2.8993.7895.3528.70610 x 210 x 1.26210 x 22

10 x 10W


wW

436-

4456-


SELF ASSESSMENT EXERCISE No. 2

1. The electric current charging a capacitor is related to time by the following law.

I = 10(1 – e-t/2

) Amps

Calculate the charge Q (the area under the graph) between the limits t = 0 and t = 6 s. Use

calculus and the trapezoidal rule. (Graph and ordinates are calculated for you)

(Answer around 41 Coulombs)

2. Find the area (with units) under the following function between the limits x = 0 and x = 10 m

using integration, the mid-ordinate rule and the trapezoidal rule.

y = 4 + 20x – x2 m

(Answers around 706.7 m2)

3. Find the area under the following function between the limits t = 0 and t = 1 s using integration,

the mid-ordinate rule and the trapezoidal rule with steps of 0.1s.

v = 2t + e2t

m/s

(Answers around 4.2 m)

4. Find the area (with units) under the following function between the limits = 0 and = 1.4 radian using integration, the mid-ordinate rule and the trapezoidal rule with steps of 0.2 radian

T = 3 cos (Nm)

(Answers around 2.96 Joules)

5. Find the area (with units) under the following function between the limits V = 1 and V = 5 m3

using integration, the mid-ordinate rule and the trapezoidal rule with steps of 1.

p = 2 ln V N/m2

(Answers around 8.09 Nm or Joules)


2. INTEGRATING POLYNOMIALS

The rule for integrating a single polynomial is: x2

x1

1nx2

x1

n

1n

xaax

Note that no power shown against a variable (e.g. x), means x1 and that this integrates as x

2/2.

Anything raised to the power of zero is 1 so a number on its own integrates e.g. 2 could be written

as 2x0 and this integrates to 2x

1/1 = 2x.

If the polynomial is a sum each term integrates separately.


Evaluate

4

2

3 3x)dx(2xF(x)

1022104F(x)

6824128F(x)

2

23

4

22

2

43

4

42F(x)

2

x3

4

x23x)dx(2xF(x)

2424

4

2

244

2

3


1. Integrate the following expressions.

i. y = 3x5 –x

2/2

ii. y = x3 –x/2

2. Evaluate the following.

i.

4

0

2 3x)dx(2x (Answer 18.67)

ii.

5

1

23 )dx3x(4x (Answer 500)


2. INTEGRATING TRIGONOMETRIC EXPRESSIONS

The standard integrals are given in the table and used in the following examples.


Evaluate π

0

dxsin x A

SOLUTION

From the list of standard integrals we see sin x = - cos x so:

2 (-1) - (-1)- 0 cos- - πcos xcosAπ0


Evaluate

π/2

0

2 dx x sinA

SOLUTION

From the list of standard integrals we see sin2 x = ½ x – ¼ sin2x so:

785.0000-0.785 A

4

sin0

2

0

4

sinπ

4

π

4

sin2x

2

x dxx sinA

π/2

0

π/2

0

2


Solve the following integrals. All angles are in radian.

1.

0.2

0.1

1dVV220W (Answer 152.49)

2.

π

0

dθ ) θ sin(2A (Answer 4)

3.

1.5

0.5

dθ ) θ ( cosA (Answer 0.518)

4. 1

0

2 (x)dxsinA (Answer 0.273)


3. INTEGRATING LOGARITHMIC AND EXPONENTIAL EXPRESSIONS

The standard integrals are given in the table and used in the following examples.


Evaluate 3

0

2x dx eA

SOLUTION

From the list of standard integrals we see eax

=eax

/a so:

26.80.5-27.3 2

e -

2

e

2

eA

0 x 242

0

2x


Evaluate 4

0

dx ln(x)I

SOLUTION

Using the standard integral 1.545014ln41ln(x)xI4

0


Evaluate

3

1

dxx

24ln(x)I

SOLUTION

Using the standard integrals

2.98641.01388I

042.1971.1832ln(1)1ln(1)42ln(3)1ln(3)12I

2ln(x)1ln(x)4xdxx

2 - ln(x) 4I

3

1

3

0


1.

10

0

x dxeln(x)I (-22012)

2.

2

0

dxln(x)sin(x)I (2.03)

3.

5

0

2x dx)ln(2eI x (11000)

4. 2

0

2 dxln(x) - (x)sinI (1.803)


4. AVERAGE VALUES OF FUNCTIONS

The mean value of a function can be defined accurately as the area under the graph divided by the

base length of the graph. Consider a simple sine function such as used to describe a sinusoidal

voltage. The function is v = V sin(θ) where θ is the angle in radians and V the amplitude.

The plot is simply as shown.

Let’s find the mean value of V over the range θ = 0 to θ = π.

π

0dθ ) Vsin(θArea

2V1)1V(cos(0)) πcos(V) θ Vcos(Areaπ

0

Base length = π

Mean = 2V/π

Repeat for the range 0 to 2π

01)V(1cos(0)) πcos(2V) θ Vcos(Area2π

0

Mean = 0 as expected for a full cycle.


Find the mean value of the function y = 2x2 over the range x = 0 to x = 4

SOLUTION

42.6673

2xdx2xA

4

0

34

0

2

Mean = 42.667/4 = 10.67

5. ROOT MEAN SQUARE VALUES (rms)

This is mainly used in electrical engineering as a way of expressing alternating current and voltage

as a meaningful quantity. Basically r.m.s. values are the equivalent D.C. values that will give

correct power dissipation into a resistive load.

The mean values of sinusoidal voltages and currents are

zero and cannot be used to calculate electric power. The

power formula is P = I2 R or V

2/R or VI so if we used the

mean values of I2

or V2

then the power formula will

work. To get these we first find the mean value of the

function for I2

or V2. For a sinusoidal voltage this would

be done as follows.

v = V sin(θ) v2 = V sin

2(θ)

Over one cycle the plot is as shown. Note that all values

of v2 are positive. The area under the graph for one complete cycle from θ = 0 to θ =2π is :

2π

0

22 dθ ) θ (sinVArea

2

0sin(0) 0) cos(

2

1V

2

2π) πsin(2 ) π2 cos(

2

1V

2

θθ)sin(θ) cos(

2

1VArea 22

2π

0

2

22 πV02

2π0VArea

The mean value is πV

2/2π = 0.5V

2 remember V is the amplitude.


If we did the same for current we would find the mean of the i2 is 0.5I

2.

We could use this value to find the mean power using the P = I2 R or V

2/R

If we take the square root of these values we have the r.m.s. value.

V rms = √0.5V2 = 0.707 V or V

2

1

I rms = √0.5I2 = 0.707 I or I

2

1

If we use the r.m.s. value we could calculate the power with the formula P = Vrms Irms

For all sinusoidal quantities the r.m.s. are always given by2

Amplitude

Sometimes we have alternating quantities other than sinusoidal so the same basic process should be

used. The mid ordinate rule, trapezoidal or Simpson’s rule could be used to find the mean of a

function.

SELF ASSESSMENT EXERCISE No.6

An alternating voltage has a saw tooth form as shown. Calculate the r.m.s. value.

The following is not stated as a requirement in the syllabus but students would do well to have a go

at it.


6. INTEGRATION BY SUBSTITUTION

A complex equation may be simplified with a substitution but it takes experience to recognise these

cases.


Evaluate

dx48x

7xI

2

SOLUTION

A suitable substitution is 48xz 2 Differentiate to get 48x

8x

dx

dz

2

48x

dxx

8

dz

2

Substitute back into the original equation C8

7zdz

8

7I

Substitute for z C8

48x7I

2


The voltage Vc across a capacitor when it discharges through a resistance is given by

dt

dVTVV C

CS where T is a time constant and Vs is the voltage at t = 0.

Find the equation relating Vc with time t.

SOLUTION

Let VS – VC = x Differentiate and since VS is a constant we find -dVC = dx

The equation becomes dt

dxTx x Rearranging

x

dx

T

dt

Integrating ln xx

dxdt

T

1t

0

The limits become clear when we substitute x = VS – VC

CV

0CS

t

0

VVln x

dxdt

T

1 SCS VlnVVln

T

t

SS

CS

V

V1ln

V

VVln

T

t

Take antilogs and S

CT

t

V

V1e

Rearrange

T

t

SC e-1 VV



Find dx 2x)x(3I 4

SOLUTION

Substitute z = 3 – 2x and x = (3 – z)/2

Differentiate to get dz = -2dx and substitute dx = -dz/2

Substitute to get rid of all the x terms

dx 2x)x(3I 4

dz

4

z-3z- dz

4

zz3

2

dz )(z

2

z-3 5444

C24

z

20

3zC

4

6

z

5

3z

I65

65

Now substitute back

C20

2x-33

24

2x-3I

56


Find dθθcos θsinI 3

SOLUTION

Substitute sin(θ) = x and noting that

dθ

dx

dθ

θdsinθcos hence cos(θ) dθ = dx

dxxdθθcos θsinI 33

C4

θsinC

4

xI

44


Solve the following integrals.

1.

5x

xdxI C5x10x5x

3

22

1

2

3

2.

3

2

x6

dxxI Cx6ln

3

1 3

3.

2x1

dxI Cxsin 1

4. dθθcos θsinI 2 C3

θsin3

5.

3x

2xdxI 2x – 6 ln(x + 3) + C


7. INTEGRATION BY PARTS

The rule is without explanation vduuvudv and best shown with an example.


Find dx ex I x

SOLUTION

Let x = u and let ex = dv du = dx and xx eev

vduuvudv C1xe Cexedxexe xxxxx


Find dx e xI 2x2

SOLUTION

Let x2 = u and let e

2x = dv du = 2xdx and

2

eev

2x2x

vduuvudv dx2x 2

e

2

ex 2x2x2

dxx e2

ex 2x2x2

We must repeat the process for the integration of dxx e2x

Let u = x and 2xedv du = dx 2

eev

2x2x

4

e

2

xedx

2

e

2

xedxx e

2x2x2x2x2x

Now put the two parts together including the constant of integration.

C12x2x4

eI

C4

1

2

x

2

xeC

4

e

2

xe

2

exCdxx e

2

exI

22x

22x

2x2x2x22x

2x2


Solve the following integrals.

1. dx ln(x) xI 3 C1)ln(416

xI

4

x

2. dx sin(2x) xI 2 C4

cos(2x)sin(2x)

2

xcos(2x)

2

xI

2

3. dx cos(x)eI x Ccos(x)sin(x)2

eI

x

4. dx sin(3x)eI -2x Csin(3x)23cos(3x)13

e-I

-2x

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