Transcript C4.5 Integration 1

A-Level Maths:
Core 4
for Edexcel
C4.5 Integration 1
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Integrals of standard functions
Contents
Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
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Review of integration
So far, we have only looked at functions that can be integrated
using:
n 1
kx
n
kx
 dx = n  1 + c for all n  1
For example:
1
Integrate 10 x + 2 + 6 x  3 with respect to x.
x
4
1
1


4
4
2
2
 10 x + x2 + 6 x  3  dx =  10 x dx +  x dx +  6 x dx   3 dx
3
1
5
6x2
x
10 x
+ 3  3x + c
+
=
1
5
2
1
5
= 2 x  + 4 x3  3 x + c
x
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The integral of
1
x
The only function of the form xn that cannot be integrated by
this method is x–1 = 1x .
Adding 1 to the power and then dividing would lead to the
meaningless expression,
x0
0
This does not mean that 1x cannot be integrated.
Remember that
Therefore
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dy 1
if y = ln x then
=
dx x
1
 x dx = ln x + c (where x > 0)
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The integral of
1
x
We can only find the log of a positive number and so this is
only true for x > 0.
However, 1x does exist for x < 0 (but not x = 0). So how do we
integrate it for all possible values of x?
We can get around this by taking x to be negative.
If x < 0 then –x > 0 so:
1
1
 x dx =   x dx = ln( x) + c (where x < 0)
We can combine the integrals of 1x for both x > 0 and x < 0 by
using the modulus sign to give:
1
 x dx = ln x + c
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The integral of
1
x
2
dx
Find 
3x
This is just the integral of 1x multiplied by a constant.
2 1
2
2
dx
=
ln x + c
dx
=

 3x
3 x
3
 2 5
4x
Find   x  2  dx
x 

20
 2 5 
3
4
x
x

dx
=
  x2 
 4x  x dx
= x 4  20ln x + c
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Definite integrals involving 1x
It is particularly important to remember the modulus sign when
evaluating definite integrals of functions involving 1x .
Find the area under the curve y = – 1x between x = –3,
x = –1 and the x-axis, writing your answer in the form ln a.
y
–3
1
The area is given by 3  .
x
1 1
1
3  x =  ln x  3
1
–1 0
x
y
1
x
=  ln 1   ln 3
=  ln1+ ln3
Remember
that ln 1 = 0
= ln3 units squared
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Definite integrals involving 1x
1
We should note that definite integrals of the form a
can only
x
be evaluated if x = 0 does not lie in the interval [a, b].
b
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Integrals of standard functions
By reversing the process of differentiation we can derive the
integrals of some standard functions.
n 1
x
n
x
 dx = n  1 + c (n  1)
 sin x dx =  cos x + c
1
 x dx = ln x + c
 cos x dx = sin x + c
x
x
e
dx
=
e
+c

2
sec
x dx = tan x + c

These integrals should be memorized.
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Integrals of standard functions
Also, if any function is multiplied by a constant k then its
integral will also be multiplied by the constant k.
Find   4 sin x + 7e x  dx
  4 sin x + 7e  dx =  4 sin x dx +  7e dx
x
x
= 4  sin x dx + 7  e x dx
= 4(  cos x ) + 7e x
= 7e x  4cos x
In practice most of these steps can be left out.
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Reversing the chain rule
Contents
Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
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Reversing the chain rule
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.
y = (f ( x ))n+1
Let
By the chain rule:
dy
= ( n +1)(f ( x ))n f '( x )
dx
n
n+1
(
n
+1)(
f
(
x
))
f
'(
x
)
dx
=
(
f
(
x
))
+c

So
It follows that for n ≠ 1
1
n+1
(
f
(
x
))
f
'(
x
)
dx
=
(
f
(
x
))
+c

( n +1)
n
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Reversing the chain rule
If the integral is multiplied by a constant k:
k
n+1
k
(
f
(
x
))
f
'(
x
)
dx
=
(
f
(
x
))
+c

( n +1)
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.
Suppose we want to integrate (2x + 7)5 with respect to x.
Consider the derivative of y = (2x + 7)6.
dy
= 6(2 x + 7)5 × 2 = 12(2x + 7)5
Using the chain rule:
dx
5
6
12(2
x
+
7)
dx
=
(2
x
+
7)
+c
So


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5
(2
x
+
7)
dx =

1
(2 x + 7)6 + c
12
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Reversing the chain rule
In general, you can integrate any linear function raised to a
power using the formula:
n
 (ax  b) dx =
1
( ax  b)n+1 + c
a( n +1)
With practice, integrals of this type can be written down
directly. For example:
1
9
(
x

5)
dx
=
(
x

5)
+c

9
8
2
(4
x
+
7)
dx =

1
(4 x + 7)3 + c
12
5
1
10
10
5(3

2
x
)
dx
=
(3

2
x
)
+
c
=
(3

2
x
)
+c

20
4
9
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Reversing the chain rule
Let’s look at some more integrals of functions of the form
k(f(x))n f ’(x).
Integrate y = x(3x2 + 4)3 with respect to x.
Notice that the derivative of 3x2 + 4 is 6x.
Now consider the derivative of y = (3x2 + 4)4.
Using the chain rule:
dy
= 4(3 x 2 + 4)3 × 6 x = 24x(3x2 + 4)3
dx
2
3
2
4
24
x
(3
x
+
4)
dx
=
(3
x
+
4)
+c

So
1
  x(3 x + 4) dx =
(3 x 2 + 4)4 + c
24
2
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Reversing the chain rule
Find  7 x 2 (2 x3  9)2 dx.
Notice that the derivative of 2x3 – 9 is 6x2.
Now consider the derivative of y = (2x3 – 9)3.
Using the chain rule:
dy
= 3(2 x3  9)2 × 6 x 2 = 18x2(2x3 – 9)2
dx
2
3
2
3
3
18
x
(2
x

9)
dx
=
(2
x

9)
+c

So

2
3
2
x
(2
x

9)
dx =

1
(2 x3  9)3 + c
18
7
  7 x (2 x  9) dx = (2 x3  9)3 + c
18
2
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Reversing the chain rule
Find

x2
x 1
3
dx .
Start by writing
x2 is the derivative
of (x3 – 1).
x2
x 1
3
 21
as x ( x  1) .
2
3
 21 plus 1 is 21 .
1
2
Now consider the derivative of y = ( x  1) .
dy 1 3
3 2 3
 21
 21
2
= ( x  1) × 3 x = x ( x  1)
Using the chain rule:
dx 2
2
1
3 2 3
 21
3
2
So
 2 x ( x  1) dx = ( x  1) + c
1
2 3
 21
2
3
2
  x ( x  1) dx = ( x  1) + c
3
3

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
2 x3  1
dx =
+c
3
3
x 1
x2
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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:
If y = e
f ( x)
dy
then
= f '( x )ef ( x )
dx
We can reverse this to integrate functions of the form k f ’(x)ef(x).
For example:
5 x 3
5 x 3
5
e
dx
=
e
+c

A numerical adjustment is usually necessary.
1 5 x 3
e
+c
5
2 5 x 3
5 x 3
2
e
dx
=
e
+c

5
5 x 3
e
 dx =
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Reversing the chain rule for exponential functions
In general,
Find
e
ax b
1 ax b
dx = e
+c
a
6
 e3 x dx .
6
3 x
dx
=
6
e
 e3 x
 dx
6 3 x
=
e +c
3
= 2e3 x + c
2
=  3x + c
e
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Reversing the chain rule for exponential functions
With practice, this method can be extended to cases where the
exponent is not linear. For example:
Find
 xe
2 x2
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).
 xe
2 x2
1
2 x2
dx =  4 xe dx
4
1 2 x2
= e +c
4
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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:
If y = ln f ( x) then
dy f '( x)
=
dx f ( x)
kf '( x)
We can reverse this to integrate functions of the form
f ( x)
For example:
1
Remove a factor of 5
1
1
5
 5 x + 4 dx = 5  5 x + 4 dx to write the function
f '( x )
in the form
1
= ln 5 x + 4 + c
5
In general,
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f ( x)
1
1
 ax + b dx = a ln ax + b + c
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.
Reversing the chain rule for logarithmic functions
5
dx.
Find 
2  6x
This is now of
the form f '( x ) .
5
5
6
dx
=

dx
 2  6x

6 2  6x
f ( x)
5
=  ln 2  6 x + c
6
4
dx, writing your answer in the form a ln b.
Evaluate 
1 2 x  7
4
First of all, note that the graph of y =
has a discontinuity
2x  7
when 2x – 7 = 0, that is when x = 3.5.
2
This is outside the interval [–1, 2] and so the integral is valid.
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Reversing the chain rule for logarithmic functions
2
4
2
dx
=
2
1 2 x  7
1 2 x  7 dx
2
This is now of
the form f '( x ) .
f ( x)
2
= 2 ln 2 x  7  1
= 2(ln 4  7  ln 2  7 )
= 2(ln3  ln9)
This can be written in the required form by using the rule that
a
ln a – ln b = ln .
b
2
4
3
dx
=
2ln
1 2 x  7
9
1
= 2ln
3
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Reversing the chain rule for logarithmic functions
x2
Find 
dx.
3
7 + 2x
x2
1
6 x2
 7 + 2 x3 dx = 6  7 + 2 x3 dx
This is now of
the form f '( x ) .
f ( x)
1
= ln 7 + 2 x3 + c
6
4e2 x
Find  2 x
dx.
e +1
4e2 x
2e2 x
 e2 x +1 dx = 2 e2 x +1 dx
This is now of
the form f '( x ) .
f ( x)
= 2ln e2 x +1 + c
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The integral of tan x
sin x
cos x
f '( x )
.
f ( x)
We can find the integral of tan x by writing it as
recognizing that this fraction is of the form
and
sin x
 tan x dx =  cos x dx
=  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:
1
 ln cos x + c = ln cos x + c
1
= ln
+c
cos x
= ln sec x + c
 tan x dx = ln sec x + c
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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:
dy
If y = sin f ( x ) then
= f '( x )cos f ( x)
dx
dy
If y = cos f ( x ) then
= f '( x)sin f ( x)
dx
We can reverse these to integrate functions of the form
f ’(x) cos f(x) and f ’(x) sin f(x). For example:
 3cos3 x dx = sin3 x + c
2
2
cos
x
+c

2
x
sin
x
dx
=

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Reversing the chain rule for trigonometric functions
As with other examples a numerical adjustment is often
necessary.
1
is now of the
 sin(5 x  2)dx =  5  5 sin(5 x  2)dx This
form –f ’(x) sin f(x).
1
=  cos(5 x  2) dx
5
In general, when dealing with the cos and sin of linear functions:
1
 cos(ax + b) dx = a sin(ax + b) + c
1
 sin(ax + b) dx =  a cos(ax + b) + c
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Integration by substitution
Contents
Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
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Integration by substitution
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.
3
To see how this method works consider the integral  (5 x + 2) dx.
Let u = 5x + 2 so that
3
3
u
(5
x
+
2)
dx
=
 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.
du
u = 5x + 2 
=5
dx
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Integration by substitution
When we used the chain rule for differentiation we saw that we
du
can treat
informally
as a fraction, so:
dx
du
1
= 5  du = 5dx  dx = du
dx
5
1
So if u = 5x + 2 and dx = du :
5
3
3 1
 (5 x + 2) dx =  u 5 du
1 4
=
u +c
20
Now change the variable back to x:
1
4
(5
x
+
2)
dx
=
(5
x
+
2)
+c

20
3
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Reversing the chain rule for logarithmic functions
Use a suitable substitution to find  x(2 x 2  5)5 dx.
du
= 4x
Let u =
5
dx
dx
1
=
du 4 x
1
dx =
du
4x
Substituting u and dx into the original problem gives:
2
5
5 1
Notice that the
x
(
2
x

5
)
dx
=

 xu 4 x du
x’s cancel out.
1
=  u 5 du
4
1 6
=
u +c
24
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2x2 –
Integration by substitution
Now we need to change the variable back to x :
1
2
6
x
(2
x

5)
dx
=
(2
x

5)
+c

24
2
5
This integral could also have been found directly by recognition.
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
If u = 1 – 2x then
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du
= 2
dx
dx
1
=
du
2
4
x
(1

2
x
)
dx .

1
dx =  du
2
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Integration by substitution
Also if u = 1 – 2x then
1
x = (1  u )
2
We also have to substitute
the x so that the whole
integrand is in terms of u.
Substituting these into the original problem gives:
1
1
4
 x(1  2 x) dx =  2 (1  u )u ( 2 )du
1
=   (1  u )u 4 du
4
1
=   (u 4  u 5 )du
4
4
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Integration by substitution
5
6


1
1
u
u
4
5
  (u  u )du =     + c
4
4 5 6 
1  6u 5  5u 6 
= 
+c
4
30

1 5
=
u (6  5u ) + c
120
Changing the variable back to x gives:
1
4
5
x
(1

2
x
)
dx
=

(1

2
x
)
(6  5(1  2 x )) + c

120
1
=
(1  2 x )5 (10 x +1) + c
120
1
=
(2 x  1)5 (10 x +1) + c
120
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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.
Use the substitution u = 8  x to find the area under the curve
2x
y=
between x = 4 and x = 7.
8x
1
2
If u = (8  x) then
So
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du 1
 21
= (8  x ) ( 1)
dx 2
1
=
1
2
2(8  x )
1
=
2u
Using the chain rule
for differentiation.
dx
= 2u  dx = 2u du
du
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Definite integration by substitution
Now we need to find x in terms of u.
1
2
If u = (8  x) then
u2 = 8 – x
x = 8 – u2
Rewrite the limits in terms of u:
when x = 3, u = 8  4 = 2
when x = 1, u = 8  7 = 1
7
2x
dx . Rewrite this in terms of u:
The area is given by 
4
8x
2
7
2 2(8  u )
2x
4 8  x dx = 1 u (2u )du
2
=  4(8  u 2 ) du
1
2
= 4  (u 2  8) du
1
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Definite integration by substitution
2
u

4  (u  8) du = 4   8u 
1
3
1
2
3
2
1
8

= 4   16  + 8 
3
3

7

= 4  8
3

= 22 32
Therefore, the required area is 22 32 units squared.
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Integration by parts
Contents
Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
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Integration by parts
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 + v
dx
dx
dx
where u and v are functions of x.
Integrating throughout with respect to x gives:
uv =  u
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dv
du
dx +  v dx
dx
dx
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Integration by parts
This can be rearranged to give:
dv
du
 u dx dx = uv   v dx dx
To integrate a product using this formula we let one part equal
dv
u and the other equal
.
dx
du
We find
by differentiating the part we called u.
dx
dv
We find v by integrating the part we called
.
dx
dv
It is important to choose u and
so that
dx
du
dv
 v dx dx is easier to integrate than  u dx dx .
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Integration by parts
So, to integrate x sin x with respect to x:
dv
= sin x
Let
and
u=x
dx
differentiate
du
=1
v =  cos x
So
and
dx
dv
du
Now, using the formula  u dx = uv   v dx :
dx
dx
integrate
We don’t need
the “+ c” here.
 x sin x dx =  x cos x    cos x dx
=  x cos x   sin x + c
= sin x  x cos x + c
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Integration by parts
Find
2x
xe
 dx .
dv
= e2 x
u=x
Let
and
dx
du
e2 x
=1
So
and
v=
dx
2
dv
du
Now, apply the formula  u dx = uv   v dx :
dx
dx
e2 x
e2 x
2x
 xe dx = x 2   2 dx
xe2 x e2 x
=

+c
2
4
e2 x
=
(2 x  1) + c
4
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Integration by parts
Find  8 x ln x dx.
We don’t know the integral of ln x so:
let
u = ln x
du 1
=
dx x
and
dv
= 8x
dx
v = 4 x2
dv
du
u
dx
=
uv

v
dx :
Now, using the formula, 

dx
dx
1
2
2
 8 x ln x dx = 4 x ln x   4 x × x dx
= 4 x 2 ln x   4 x dx
so
and
= 4 x2 ln x  2x2 + c
= 2 x 2 (2ln x  1) + c
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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).
u = ln x
Let
du 1
=
dx x
So
and
dv
=1
dx
and
v=x
Now we can integrate by parts:
1
 1×ln x dx = x ln x   x x dx
= x ln x   1dx
= x ln x  x + c
 ln x dx = x ln x  x + c
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Integration by parts
To evaluate a definite integral using integration by parts we use:
b du
dv
b
a u dx dx = uva  a v dx dx
b

Evaluate  2 2 x cos x dx.
0
u = 2x
Let
du
=2
dx
So


2
0
and
v = sin x


2 x cos x dx = 2 x sin x 0   2 2sin x dx
sin π/2 = 1
π/
2
cos
=0
cos 0 = 1
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and
dv
= cos x
dx
2
0
= (
sin 
2

 0)   2cos x 02
=   ( 2cos 2 + 2cos0)
= 2
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Volumes of revolution
Contents
Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
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Volumes of revolution
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.
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Volumes 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
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Volumes of revolution
The total volume of the solid is given by the sum of the volume
of the slices.
x =b
V    y 2 x
x=a
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.
x =b
V = lim   y 2 x
 x 0
x=a
This limit is represented by the following integral:
b
V =   y 2 dx
a
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Volumes of revolution
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:
b
V =   y 2 dx
a
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:
b
V =   x 2 dy
a
Volumes of revolution are usually given as multiples of π.
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Volumes of revolution
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
V =   y 2 dx
0
2
=   (2 x  x 2 )2 dx
0
2
=   (4 x 2  4 x 3  x 4 )dx
0
2
 4x
4x
x 
= 

 
4
5 0
 3
3
4
5
32  16
 32
=
cubic units
= 
 16 +

15
5 
 3
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Volumes of revolution
Find the volume of the solid formed by rotating the area
between the curve y = 1x , the y-axis, y = 1, and y = 2
360° about the y-axis.
2
V =   x 2 dy
1
2
=
2
1
 1
 y  dy
 
Rearranging
1
y = x gives x = 1y .
2
=   y 2 dy
1
2
 1
=   
 x 1
 1  
=    +1 = cubic units
 2  2
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Examination-style question
Contents
Integrals of standard functions
Reversing the chain rule
Integration by substitution
Integration by parts
Volumes of revolution
Examination-style question
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Examination-style question
a) Show that cos2x = 21 (1 + cos 2x)
b) The diagram below shows the finite region R which is
bounded by the curve y = 1  x 2 the x-axis, the y-axis and
the line x = 21 .
y
R
0
1
2
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.
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Examination-style question
a) RHS = 21 (1 + cos 2x)
Using the double angle
identity cos 2A ≡ 2cos2A – 1.
= 21 (1 + (2cos2x – 1))
= 21 (2cos2x)
= cos2x = LHS
b) The area of R is given by

1
2
0
1  x 2 dx
Using the substitution x = sin u we have:
dx
= cos u  dx = cos u du
du
Find u when x is 0 and ½:
when x = 0, u = 0
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and

when x = 21 , u = 6
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Examination-style question
Now make the substitution:

1
2
0

1  x dx =  6 1  sin2 u cos u du
2
0

=  6 cos2 u du
Using the identity
cos2A = 1 – sin2A.
0

=  6 21 (1+ cos2u ) du
0
Using the identity
cos2A = 21 (1 + cos 2A).

=
1 u + 1 sin2u  6
2
2
0

3
=
+
12 8
Therefore, the area of region R is 0.478 units squared (to 3 s.f.).
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Examination-style question
c) The required volume of revolution is given by

1
2
0
1
2
 y dx =   (1  x 2 ) dx
2
0
1
2

x 
=  x  
3 0

3
 1 81 
=   
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.).
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