Chapter 4 Integration

Calculus (Integration)

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

Mathematics 1Level 4

The following presentation is an introduction to the Algebraic Methods – part one for level 4 Mathematics. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1st year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Integration Terminology Integration of xn Indefinite Integrals Definite Integrals Integrals of Sin x, Cos x, ex and 1/x

Integrals of ex and 1/x

Area under curves Volumes of Solids of Revolution Worked Example 1 Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see:•KA Stroud & DJ Booth, Engineering Mathematics, 8th Editon, Palgrave 2008.•http://www.mathcentre.ac.uk/•Derive 6

Calculus Integration

IntegrationWhere differentiation measured the rate of change of a function, integration measures the area under the function between limits.

It is the reverse function of differentiation. and therefore if we differentiate a function then integrate it we will get back to where we started.


TerminologyA definite integral is typically explained by the area problem as follows. Suppose we attempt to find the area of the region, S, under the curve y = (x) bounded from lower limit a to upper limit b, where is a continuous function.

Sx

y

(x)

To estimate S, we divide the range[a, b] into n subintervals, that is, [x0, x1], [x1, x2], [x2, x3], . . . , [xn−1, xn], Each of width Δx = (b − a)/n (so xi = a + iΔx).

a b


Terminology

x

y

(x)

ba

xi-1 xi

(xi*)

The area Si of the strip between xi−1 and xi can be approximated as the rectangular area with width Δx and height f(xi*), where xi* is a sample point in [xi-1, xi]. So the total area under the curve is approximated:

which is called a Riemann Sum.

xxxxxxxx n

n

i i *)(...*)(*)()*( 211


TerminologyThe estimation is better as the strips gets thinner, and the exact area under the graph of with the limit can be identified as:

xxArean

i i )*(lim

1n∞

As long as is continuous, the value of the limit is independent of the sample points xi* used.That limit is represented by

b

a

dxx )( and is called definite integralof from a to b

Note that in intervals where (x) is negative the graph ofy = (x) lies below the x-axis and the definite integral takes a negative value. In general a definite integral gives the net area between the graph of y = (x) and the x-axis.

Integration of xn

It has been stated that integration is the reverse process of differentiation.So if y = x5 dy/dx = 5x4

Therefore if we start with 5x4 and we integrate it we will get to x5

But note y = x5 + 6 dy/dx = 5x4

y = x5 - 12 dy/dx = 5x4

y = x5 + constant dy/dx = 5x4

It is therefore important to note that the integration of 5x4 gives us x5 + C where C is the

constant of integration

In general Cnax

dxaxn

n

1

1

ExamplesC

xdxx 8

871.

2.

3.

4.

Cxxxxx

dxxxxx 62

7

3

3

45673

2345234

Cxx

xxx

62

7

45

23

45

Cxx

dxxxdxxx 35

45)5(

34232

Cx

Cx

dxxdxx

4

45

5 41

41


Examples cont.C

xC

xdxxdxx 3

2

23

323

21

5.

6.

7.

8.

dxxxxdx

xx

x

x

xdx

x

xx 432444

2

4

2

656565

CxCx

dxxdxdx 11

10

Cx

Cx

xdxx

2

15

325

55 3 23

2

31

3

Cxxx

Cxxx

32

321 2251

36

25

1


Indefinite IntegralsYou will note that the integrals we have been working out have no limits – numbers at the top and bottom of the integral sign. These do not generate numerical results and are referred to as indefinite integrals.

All indefinite integrals must have a constant of integration.


Definite IntegralsDefinite integrals will have limits – numbers at the top and bottom of the integral sign. The one on the top is the upper limit and the one on the bottom is the lower limit. Definite integrals will have a value associated with them – to determine this value we integrate then substitute in the upper limit into the new equation and then subtract from this the value when the lower limit is substituted. Due to the subtraction the constant of integration disappears and therefore, for definite integrals it is not normally included.


ExampleDetermine

Note where the limits are, the square brackets and no

constant

3

1

2 63 dxxx

3

1

23

62

33

x

xx

3

28

6

14

2

1416

2

13

3

136

2

33

3

3 2323


Integrals of Sin x, Cos x, ex and 1/xSin xIf we have y = cos ax then dy/dx = -a sin ax.We can therefore say that ∫-asin ax dx = cos axNote as well that if y = cos ax +5 then dy/dx = -a sin ax.

So ∫sin ax dx = -1/a cos ax + C C = constant of integration

Note – the number we divide by is the d/dx of the sin partLikewise Cos xIf we have y = sin ax then dy/dx = a cos ax.We can therefore say that ∫acos ax dx = sin axNote as well that if y = sin ax +5 then dy/dx = a cos ax.

So ∫cos ax dx = 1/a sin ax + C C = constant of integration


Worked examples

dxxxfdxxxe

dxxddxxc

dxxbdxxa

3sin3)(])63cos(7sin5[)(

)76sin()()45cos()(

)32sin()(3cos)(

Cxa 3sin31

)( Cxb )32cos(21

)(

Cxc )45sin(51

)( Cxd )76cos(71

)(

Cxxe )63sin(37

cos5)(

Cxx

f )3cos(31

23

)(2


Worked examples

4

0

2sin

dxxxEvaluate

4

0

2

2cos21

2

x

x

02cos

21

20

42cos21

24 22

Note all angles in radians, cos 0 = 1, cos π/2 = cos 90 = 0

808.021

0032

2


Integrals of ex and 1/xRecall the results from differentiation.If y = eax then dy/dx = aeax

And y = ln x then dy/dx = 1/x

Therefore∫aeax dx = eax + C In General ∫eax dx = 1/a eax +

C Note – the number we divide by is the d/dx of the power part

Also

∫1/x dx = ln x + C

∫1/(ax + b) dx = 1/a ln (ax + b) + C

Note – the number we divide by is the d/dx of the reciprocal part


Worked ExamplesIntegrate the following with respect to x

xe5 dxe x5 Ce x 5

51

671x

dx

x 671

Cx )67ln(71

xe 35 dxe x35 Ce x 35

31

x751

dxx75

1Cx )75ln(

71

Evaluate the following:

4

0

5.04 dxe x

4

0

5.0

5.04 xe 02 88 ee 917.6

10 e


Worked ExamplesEvaluate the following:

3

0 3105

dxx

3

0

)310ln(3

15 x

)10ln(35

)910ln(35

)10ln(3

5)1ln(

3

5838.3


Area under curvesWe are aware that the process of integration is the determining of the area under a curve.

e.g. What is the area under the curve y = x2 + 2 between the vertical lines x = 1 and x = 3.This is simply:

3

1

2 )2( dxx

3

1

3

)23

x

x

12

31

3233 33

32

1231

215 Calculus Integration

Volumes of Solids of RevolutionTo find the volume of revolution:

Consider the curve y = (x) shown on the left. Suppose we want the volume of the solid formed by rotating the area ABCD under the curve, about the x-axis.

x

y

y = (x)

a b

y

dx

A B

CD

Consider the area ABCD as being made up of an infinite number of small strips (there are so many that each approximates to a small rectangle), of which a typical one is y in height and dx in thickness, as shown.


Volumes of Solids of RevolutionWhen rotated the rectangle of area = y dx becomes a cylinder of volume = π y2 dx.

[note volume of a cylinder = πr2h where r = y and h = dx]

Thus the volume generated in rotating the area ABCD about the x-axis will be the sum of all these small cylindrical volumes, formed by rotating the area of strips about the x-axis from x = a to x = b.

i.e. in calculus form

N.B. substitute x in terms of y before you integrate

b

a

dxyV 2

Worked Example 1

Find the volume of a solid of revolution formed by revolving the area enclosed by the curve y = 3x2 + 4, the x-axis and the lines at x = 1 and x = 2, through 360º about the x-axis.

First make a rough sketch of the curve and show the area to be rotated. Y = 3x2 + 4

a) It is a parabola b) x = 0, y = 4 c) x = 1 y = 7 d) x = 2 y = 16

Area to be rotated

y = 3x2 + 4

y2 = (3x2 + 4)2

y2 = 9x4 +24x2 +16

b

a

dxy 2VolumeRequired

Worked Example 1

2

1

24 16249V dxxx

2

1

35 163

2459

V

xxx

1161

324

159

21623

242

59

V 3535

1161

324

159

21683

2432

59

V

54

2553

15316854

1326453

57V

3units5.40154

127V


Worked Example 2

Sketch the curve y = x2 – 2x and find the volume of the solid formed when the area contained between the curve and the x-axis is rotated about the x-axis.

Sketch:a) It is a parabola b) it passes through x =0 y = 0

Where else does it cross the x-axis x2 - 2x = 0x(x – 2) = 0 either x = 0 or x = 2

b

a

dxy 2VolumeRequired

y = x2 – 2x

y2 = (x2 – 2x)2

y2 = x4 -4x3 +4x2

Worked Example 2

2

0

234 44V dxxxx

2

0

345

34

44

51

V

xxx

02

34

244

251

V 345

8

34

1644

3251

V

3units35.3151

132

101652

6V


Worked Example 3

What is the equation for the volume of a cone height h and radius r?

b

a

dxy 2VolumeRequired

x

y

h

r

We would need to rotate the area shown about the x-axis.What is the equation of the line? y = mx + c (m = slope c = y intersect)

Intersect = 0 slope = r/h equation is xhr

y

22

22 x

hr

y

h

dxxhr

0

22

2

V h

xhr

0

3

2

2

3V

3

2

2

3V h

hr hr 2

31

V

Examples

1. The curve y = 2x2 + 3 is rotated about the x-axis through 360º, between the limits x = 1 and x = 3. Find the volume of the of the solid of rotation formed.

[Ans = 991.5 units3]

2. Find the volume of the solid obtained by rotating about the x-axis the part of the curve y = 2x2 – 7x – 4 lying below the axis.

[Ans = 772.9 units3]


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© 2009 University of Wales Newport

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Chapter 4 Integration

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