A-Level Maths:
Core 4
for Edexcel
C4.5 Integration 1
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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
ò dx = n + 1 + c (n ¹ 1)
ò sin x dx = - cos x + c
1
ò x dx = ln x + c
ò cos x dx = sin x + c
x
x
e
dx
=
e
+c
ò
2
sec
x dx = tan x + c
ò
These integrals should be memorized.
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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
ò dx = n + 1 + c for all n ¹ 1
For example:
1
Integrate 10 x + 2 + 6 x - 3 with respect to x.
x
4
1
1
æ
ö
4
4
-2
2
ò çè10 x + x2 + 6 x - 3 ÷ø dx = ò 10 x dx + ò x dx + ò 6 x dx - ò 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
ò 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
ò x dx = ò - 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
ò x dx = ln x + c
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The integral of
1
x
2
Find ò dx
3x
This is just the integral of 1x multiplied by a constant.
2 1
2
2
dx
=
ln x + c
dx
=
ò
ò 3x
3 x
3
æ 2 5ö
4x
Find ò ç x - 2 ÷ dx
x ø
è
20
æ 2 5ö
3
4
x
x
dx
=
ò çè x2 ÷ø ò 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 ò-3 - .
x
-1 1
-1
ò-3 - x = éë- ln x ùû -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 òa
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
Also, if any function is multiplied by a constant k then its
integral will also be multiplied by the constant k.
(
)
Find ò 4 sin x + 7e x dx
ò ( 4 sin x + 7e ) dx = ò 4 sin x dx + ò 7e dx
x
x
= 4 ò sin x dx + 7 ò 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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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 ò (5 x + 2) dx.
Let u = 5x + 2 so that
3
3
u
(5
x
+
2)
dx
=
ò dx
ò
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 Þ
=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 Þ du = 5dx Þ dx = du
dx
5
1
So if u = 5x + 2 and dx = du:
5
3
3 1
(
5
x
+
2
)
dx
=
ò
ò 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
ò
20
3
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Reversing the chain rule for logarithmic functions
Use a suitable substitution to find ò 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
=
ò
ò xu 4 x du
x’s cancel out.
1
= ò 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
ò
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 .
ò
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
ò x(1 - 2 x) dx = ò 2 (1 - u )u (- 2 )du
1
= - ò (1 - u )u 4 du
4
1
= - ò (u 4 - u 5 )du
4
4
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Integration by substitution
5
6
æ
ö
1
1
u
u
4
5
- ò (u - u )du = - ç - ÷ + c
4
4è 5
6ø
1 æ 6u 5 - 5u 6 ö
=- ç
÷+c
4è
30
ø
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
ò
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 Þ 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
The area is given by ò
dx . Rewrite this in terms of u:
4
8-x
2
7
2 2(8 - u )
2x
ò4 8 - x dx = ò1 u (-2u )du
2
= ò -4(8 - u 2 ) du
1
2
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= 4 ò (u 2 - 8) du
1
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Definite integration by substitution
2
éu
ù
4 ò (u - 8) du = 4 ê - 8u ú
1
ë3
û1
2
2
3
1
æ8
ö
= 4 ç - 16 - + 8 ÷
3
è3
ø
æ7
ö
= 4ç - 8÷
è3
ø
= -22 32
Therefore, the required area is 22 32 units squared.
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