© Boardworks Ltd 2006 1 of 57 © Boardworks Ltd 2006 1 of 57 A-Level Maths: Core 4 for Edexcel C4.5...

© Boardworks Ltd 20061 of 57 © Boardworks Ltd 20061 of 57

A-Level Maths: Core 4for Edexcel

C4.5 Integration 1

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Integrals of standard functions

Reversing the chain rule

Integration by substitution

Integration by parts

Volumes of revolution

Examination-style question



Review of integration

So far, we have only looked at functions that can be integrated using:

1

= + for all 11

nn kx

kx dx c nn

For example:

Integrate with respect to x.42

110 + + 6 3x x

x

42

110 + + 6 3 =x x dx

x

124 210 + + 6 3x dx x dx x dx dx

5 31= 2 + 4 3 +x x x c

x

510=

5

x 1

+1

x

32

32

6+

x3x + c


The integral of

Adding 1 to the power and then dividing would lead to the meaningless expression,

The only function of the form xn that cannot be integrated by this method is x–1 = .1

x

0

0

x

This does not mean that cannot be integrated.1x

Remember that 1

if = ln then =dy

y xdx x

Therefore 1

= ln + (where > 0)dx x c xx

1x


The integral of

We can only find the log of a positive number and so this is only true for x > 0.

We can get around this by taking x to be negative.

However, does exist for x < 0 (but not x = 0). So how do we integrate it for all possible values of x?

1x

If x < 0 then –x > 0 so:

1 1 = = ln( )+ (where < 0)dx dx x c x

x x

1 = ln + dx x c

x

We can combine the integrals of for both x > 0 and x < 0 by using the modulus sign to give:

1x

1x


The integral of

Find 2

3dx

x

This is just the integral of multiplied by a constant. 1x

2=

3dx

x2 1

=3

dxx

2ln +

3x c

Find 22

54x x dx

x

22

54 =x x dx

x 3 20

4x dxx

4= 20ln +x x c

1x


Definite integrals involving

It is particularly important to remember the modulus sign when evaluating definite integrals of functions involving . 1

x

Find the area under the curve y = – between x = –3, x = –1 and the x-axis, writing your answer in the form ln a.

1x

0 x

y

–1–3

1y

x

The area is given by .1

3

1

x

1 1

33

1= ln x

x

= ln 1 ln 3

= ln1+ ln3 Remember that ln 1 = 0

1x

= ln3 units squared


Definite integrals involving

We should note that definite integrals of the form can only1b

a xbe evaluated if x = 0 does not lie in the interval [a, b].

1x


1

= + ( 1)1

nn x

x dx c nn


By reversing the process of differentiation we can derive the integrals of some standard functions.

1 = ln + dx x c

x

= + x xe dx e c

cos = sin + x dx x c

sin = cos +x dx x c

2sec = tan + x dx x c

These integrals should be memorized.



Also, if any function is multiplied by a constant k then its integral will also be multiplied by the constant k.

Find 4sin + 7 xx e dx 4sin + 7 = 4sin + 7x xx e dx x dx e dx

= 4 sin + 7 xx dx e dx = 4( cos )+ 7 xx e

= 7 4cosxe x

In practice most of these steps can be left out.


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A very helpful technique is to recognize that a function that we are trying to integrate is of a form given by the differentiation of a composite function. This is sometimes called integration by recognition.

Let f +1= ( ( ))ny x

By the chain rule: f f= ( +1)( ( )) '( )ndyn x x

dx

So f f f +1( +1)( ( )) '( ) = ( ( )) +n nn x x dx x cIt follows that for n ≠ 1

f f f +11( ( )) '( ) = ( ( )) +

( +1)n nx x dx x c

n



Suppose we want to integrate (2x + 7)5 with respect to x.

If the integral is multiplied by a constant k:

Consider the derivative of y = (2x + 7)6.

Using the chain rule: 5= 6(2 + 7) ×2dy

xdx

= 12(2x + 7)5

So5 612(2 + 7) = (2 + 7) +x dx x c5 61

(2 + 7) = (2 + 7) +12

x dx x c

f f f +1( ( )) '( ) = ( ( )) +( +1)

n nkk x x dx x c

n

Don’t try to learn this formula, just try to recognize that the function you are integrating is of the form k(f(x))n f ’(x) and compare it to the derivative of (f(x))n + 1.



In general, you can integrate any linear function raised to a power using the formula:

+11( ) = ( ) +

( +1)

n nax b dx ax b ca n

With practice, integrals of this type can be written down directly. For example:

8( 5) =x dx 91( 5) +

9x c

2(4 + 7) =x dx 31(4 + 7) +

12x c

95(3 2 ) =x dx 105(3 2 ) + =

20x c

101

(3 2 ) +4

x c



Integrate y = x(3x2 + 4)3 with respect to x.

Notice that the derivative of 3x2 + 4 is 6x.

Using the chain rule: 2 3= 4(3 + 4) ×6dy

x xdx

= 24x(3x2 + 4)3

So2 3 2 424 (3 + 4) = (3 + 4) +x x dx x c2 3 2 41

(3 + 4) = (3 + 4) +24

x x dx x c

Let’s look at some more integrals of functions of the form k(f(x))n f ’(x).

Now consider the derivative of y = (3x2 + 4)4.



Now consider the derivative of y = (2x3 – 9)3.

Using the chain rule: 3 2 2= 3(2 9) ×6dy

x xdx

= 18x2(2x3 – 9)2

So2 3 2 3 318 (2 9) = (2 9) +x x dx x c 2 3 2 3 31(2 9) = (2 9) +

18x x dx x c

Find . 2 3 27 (2 9)x x dx

Notice that the derivative of 2x3 – 9 is 6x2.

2 3 2 3 377 (2 9) = (2 9) +

18x x dx x c



So1 12 22 3 33

( 1) = ( 1) +2

x x dx x c 1 12 22 3 32

( 1) = ( 1) +3

x x dx x c

Find .

2

3 1

xdx

x

Start by writing as 2

3 1

x

x

122 3( 1) .x x

Now consider the derivative of y =123( 1) .x

x2 is the derivative of (x3 – 1).

plus 1 is12 1

2 .

Using the chain rule:123 21

= ( 1) ×3 =2

dyx x

dx

122 33

( 1)2

x x

2 3

3

2 1= +

31

x xdx c

x


Reversing the chain rule for exponential functions

When we applied the chain rule to functions of the form ef(x) we obtained the following generalization:

We can reverse this to integrate functions of the form k f ’(x)ef(x). For example:

f ff( ) ( )If = then = '( )x xdyy e x e

dx

5 35 =xe dx 5 3 +xe c

A numerical adjustment is usually necessary.

5 3 =xe dx 5 31+

5xe c

5 32 =xe dx 5 32+

5xe c



In general, 1= +ax b ax be dx e c

a

Find . 3

6x

dxe

36= +

3xe c

33

6= 6 x

xdx e dx

e

3= 2 +xe c

3

2= +

xc

e



With practice, this method can be extended to cases where the exponent is not linear. For example:

Find .

22xxe dx

Notice that the derivative of 2x2 is 4x and so the function we are integrating is of the form k f ’(x)ef(x).

221= +

4xe c

22 =xxe dx221

44

xxe dx


Reversing the chain rule for logarithmic functions

When we applied the chain rule to functions of the form ln f(x) we obtained the following generalization:

ff

f

'( )If = ln ( ) then =

( )

dy xy x

dx x

We can reverse this to integrate functions of the form For example:

f

f

'( )

( )

k x

x

1=

5 + 4dx

x1

= ln 5 + 4 +5

x c

In general, 1 1= ln + +

+dx ax b c

ax b a

1 5

5 5 + 4dx

xRemove a factor of to write the function in the form .

15

ff'( )( )

xx



Find . 5

2 6dx

x5

=2 6

dxx

5 6

6 2 6dx

x

5

= ln 2 6 +6

x c

Evaluate , writing your answer in the form a ln b. 2

1

4

2 7dx

x

First of all, note that the graph of y = has a discontinuity

when 2x – 7 = 0, that is when x = 3.5.

4

2 7x

This is outside the interval [–1, 2] and so the integral is valid.

This is now of the form .f

f'( )( )

xx



2 2

1 1

4 2= 2

2 7 2 7dx dx

x x

2

1= 2 ln 2 7x

= 2(ln 4 7 ln 2 7 )

= 2(ln3 ln9)

This can be written in the required form by using the rule that ln a – ln b = ln .

a

b2

1

4 3= 2ln

2 7 9dx

x


f'( )( )

xx

1= 2ln

3



Find . 2

37 + 2

xdx

x2

3=

7 + 2

xdx

x2

3

1 6

6 7 + 2

xdx

x

31= ln 7 + 2 +

6x c


f'( )( )

xx

Find . 2

2

4

+1

x

x

edx

e2

2

4=

+1

x

x

edx

e2

2

22

+1

x

x

edx

e

2= 2ln +1 +xe c


f'( )( )

xx


The integral of tan x

We can find the integral of tan x by writing it as and

recognizing that this fraction is of the form .

sin

cos

x

xf

f

'( )

( )

x

x

sintan =

cos

xx dx dx

x = ln cos +x c

It is ‘tidier’ to rewrite this without a minus sign at the front, using the fact that –ln a = ln a–1:

1ln cos + = ln cos +x c x c

= ln sec +x c

1= ln +

cosc

x

tan = ln sec +x dx x c


Reversing the chain rule for trigonometric functions

When we applied the chain rule to functions of the form sin f(x) and cos f(x) we obtained the following generalizations:

We can reverse these to integrate functions of the form f ’(x) cos f(x) and f ’(x) sin f(x). For example:

f f fIf = sin ( ) then = '( )cos ( )dy

y x x xdx

f f fIf = cos ( ) then = '( )sin ( )dy

y x x xdx

3cos3 =x dx sin3 +x c

22 sin =x x dx 2cos +x c


Reversing the chain rule for trigonometric functions

As with other examples a numerical adjustment is often necessary.

In general, when dealing with the cos and sin of linear functions:

sin(5 2) =x dx1

5sin(5 2)5

x dx This is now of the form –f ’(x) sin f(x).

1= cos(5 2)

5x dx

1cos( + ) = sin( + )+ax b dx ax b c

a

1sin( + ) = cos( + )+ax b dx ax b c

a


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With practice, the technique of integration by recognition can save a lot of time.

However, when it is too difficult to use integration by recognition we can use a more formal method of reversing the chain rule called integration by substitution.

To see how this method works consider the integral 3

(5 + 2) .x dxLet u = 5x + 2 so that

3(5 + 2) =x dx 3u dx

The problem now is that we can’t integrate a function in u with respect to x. We therefore need to write dx in terms of du.

= 5 + 2 = 5du

u xdx



Now change the variable back to x:

When we used the chain rule for differentiation we saw that we can treat informally as a fraction, so:

du

dx

= 5 = 5du

du dxdx

1

=5

dx du

So if u = 5x + 2 and dx : 1

=5

du

3( )5 + 2 =x dx 3 1

5u du

41= +

20u c

3 41(5 + 2) = (5 + 2) +

20x dx x c



Use a suitable substitution to find . 2 5(2 5)x x dx

Let u = 2x2 – 5 = 4du

xdx

1=

4

dx

du x1

=4

dx dux

Substituting u and dx into the original problem gives:

Notice that the x’s cancel out.

2 52 5( ) =x x dx 5 1

4ux du

x51

=4

u du61

= +24

u c



This integral could also have been found directly by recognition.

Now we need to change the variable back to x :

2 5 2 61(2 5) = (2 5) +

24x x dx x c

However, there are functions that can be integrated by use of a suitable substitution but not by recognition. For instance:

Use the substitution u = 1 – 2x to find . 4(1 2 )x x dx

If u = 1 – 2x then = 2du

dx

1=

2

dx

du 1

=2

dx du



Substituting these into the original problem gives:

We also have to substitute the x so that the whole integrand is in terms of u.

4( )1 2 =xx dx 4 1(

1(1 )

2)

2duuu

41= (1 )

4u u du

Also if u = 1 – 2x then1

= (1 )2

x u

4 51= ( )

4u u du



5 64 51 1

( ) = +4 4 5 6

u uu u du c

Changing the variable back to x gives:

4(1 2 ) =x x dx 51(1 2 ) (6 5(1 2 ))+

120x x c

51= (1 2 ) (10 +1)+

120x x c

5 61 6 5= +

4 30

u uc

51= (6 5 )+

120u u c

51= (2 1) (10 +1)+

120x x c


Definite integration by substitution

When a definite integral is found by substitution it is easiest to rewrite the limits of integration in terms of the substituted variable.

12

1= (8 ) ( 1)

2

dux

dx If u = then

12(8 )x

Using the chain rule for differentiation.

12

1=

2(8 )x

1

=2u

So = 2dx

udu

= 2dx u du

Use the substitution u = to find the area under the curve8 x

y = between x = 4 and x = 7. 2

8

x

x



Rewrite the limits in terms of u:

Now we need to find x in terms of u.

u2 = 8 – x

x = 8 – u2

= 8 = 24u when x = 3,

= 8 =17u when x = 1,

7

4

2=

8

x

xdx

2 2

1

2( 2

))

(8 u

uu du

2 2

1= 4(8 )u du

2 2

1= 4 ( 8)u du

The area is given by . Rewrite this in terms of u:7

4

2

8

xdx

x

12(8 )xIf u = then



232 2

11

4 ( 8) = 4 83

uu du u

8 1= 4 16 + 8

3 3

7= 4 8

3

23= 22

Therefore, the required area is units squared.2322


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Suppose we wish to integrate the product of two functions, such as x sin x, where one of the functions is not related to the derivative of the other.

An expression such as this can be integrated using the method of integration by parts.When we differentiate the product of two functions u and v we use the product rule:

( ) = +d dv du

uv u vdx dx dx

Integrating throughout with respect to x gives:

where u and v are functions of x.

= +dv du

uv u dx v dxdx dx



This can be rearranged to give:

It is important to choose u and so that

=dv du

u dx uv v dxdx dx

To integrate a product using this formula we let one part equal

u and the other equal . dv

dx

We find by differentiating the part we called u. du

dx

We find v by integrating the part we called .dv

dx

duv dx

dxdv

u dxdxis easier to integrate than .

dv

dx



So, to integrate x sin x with respect to x:

Let and=u x = sindv

xdx

So =1du

dxWe don’t need the “+ c” here.Now, using the formula : =

dv duu dx uv v dx

dx dx

sin = cos cosx x dx x x x dx = cos sin +x x x c

= sin cos +x x x c

differentiate integrate

= cosv xand



So and=1du

dx

2

=2

xev

Now, apply the formula : =dv du

u dx uv v dxdx dx

2 2

2 =2 2

x xx e e

xe dx x dx

Find .2xxe dx

2 2

= +2 4

x xxe ec

2

= (2 1)+4

xex c

2= xdve

dxLet and =u x



let and = lnu x = 8dv

xdx

so and1

=du

dx x2= 4v x

Now, using the formula, : =dv du

u dx uv v dxdx dx

2 2 1

8 ln = 4 ln 4 ×x x dx x x x dxx

Find .8 lnx x dxWe don’t know the integral of ln x so:

2= 4 ln 4x x x dx 2 2= 4 ln 2 +x x x c2= 2 (2ln 1)+x x c


The integral of ln x

We can also use integration by parts to find the integral of ln x.

We do this by writing ln x as (1 × ln x).

Let and = lnu x =1dv

dx

So 1

=du

dx x

Now we can integrate by parts:1

1×ln = lnx dx x x x dxx

= ln 1x x dx = ln +x x x c

ln = ln +x dx x x x c

=v xand



To evaluate a definite integral using integration by parts we use:

Let and = 2u x = cosdv

xdx

So = 2du

dx

=b bb

aa a

dv duu dx uv v dx

dx dx

Evaluate .2

02 cosx x dx

2 22

00 02 cos = 2 sin 2sinx x dx x x x dx

2

02= ( sin 0) 2cos x

2= ( 2cos + 2cos0) sin π/2 = 1

cos π/2 = 0cos 0 = 1 = 2

= sinv xand


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Consider the area bounded by the curve y = f(x), the x-axis and x = a and x = b.

If this area is rotated 360° about the x-axis a three-dimensional shape called a solid of revolution is formed.

The volume of this solid is called its volume of revolution.



We can calculate the volume of revolution by dividing the volume of revolution into thin slices of width δx.

The volume of each slice is approximately cylindrical, of radius y and height δx, and is therefore approximately equal to

πy2δx



The total volume of the solid is given by the sum of the volume of the slices. =

2

=

x b

x a

V y x

The smaller δx is, the closer this approximate area is to the actual area.

We can find the actual area by considering the limit of this sum as δx tends to 0. =

2

0=

= limx b

xx a

V y x

This limit is represented by the following integral:

2=b

aV y dx



So in general, the volume of revolution V of the solid generated by rotating the curve y = f(x) between x = a and x = b about the x-axis is:

Similarly, the volume of revolution V of the solid generated by rotating the curve x = f(y) between y = a and y = b about the y-axis is:

Volumes of revolution are usually given as multiples of π.

2=b

aV y dx

2=b

aV x dy



Find the volume of the solid formed by rotating the area between the curve y = x(2 – x), the x-axis, x = 0, and x = 2 360° about the x-axis.

2 2

0=V y dx

2 2 2

0= (2 )x x dx

2 2 3 4

0= (4 4 )x x x dx

23 4 5

0

4 4=

3 4 5

x x x

32 32= 16 +

3 5

16cubic u s

15= nit



2 2

1=V x dy

22

1

1= dy

y

2 2

1= y dy

2

1

1=

x

1= +1

2

cubic u= nits

2

Find the volume of the solid formed by rotating the area between the curve y = , the y-axis, y = 1, and y = 2 360° about the y-axis.

1x

Rearranging y = gives x = .

1x

1y


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a) Show that cos2x = (1 + cos 2x)

b) The diagram below shows the finite region R which is bounded by the curve y the x-axis, the y-axis and the line x = .

Using the substitution x = sin u, or otherwise, find the area of the region R to 3 significant figures.

c) Find the volume of the solid formed when R is rotated through 360° about the x-axis to 3 significant figures.

12

12

2= 1 x

12

x

y

0

R



a) RHS = (1 + cos 2x)12

= (1 + (2cos2x – 1))12

= (2cos2x)12

= cos2x = LHS

Using the double angle identity cos 2A ≡ 2cos2A – 1.

b) The area of R is given by12 2

01 x dx

Using the substitution x = sin u we have:

Find u when x is 0 and ½:

= cos = cosdx

u dx u dudu

when x = 0, u = 0 and when x = , u =12 6



Now make the substitution:12 2

01 =x dx

6 2

0= cos u du

6

0

12= (1+ cos2 )u du

Using the identity cos2A = (1 + cos 2A).1

2

6

0

1 12 2= + sin2u u

6 2

01 sin cosu u du

3= +

12 8

Therefore, the area of region R is 0.478 units squared (to 3 s.f.).

Using the identity cos2A = 1 – sin2A.



181

=2 3

11=

24

Therefore, the volume formed when R is rotated about the x-axis is 1.44 units cubed (to 3 s.f.).

c) The required volume of revolution is given by 12 2

0=y dx

123

0

=3

xx

12 2

0(1 )x dx

© Boardworks Ltd 2006 1 of 57 © Boardworks Ltd 2006 1 of 57 A-Level Maths: Core 4 for Edexcel C4.5...

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Transcript of © Boardworks Ltd 2006 1 of 57 © Boardworks Ltd 2006 1 of 57 A-Level Maths: Core 4 for Edexcel C4.5...