hsn.uk.net
Higher
Mathematics
Integration
Contents
Integration
1
2
3
4
5
6
7
8
9
10
Indefinite Integrals
Preparing to Integrate
Differential Equations
Definite Integrals
Geometric Interpretation of Integration
Areas between Curves
Integrating along the y-axis
Integrating sinx and cosx
A Special Integral
Integrating sin(ax + b) and cos(ax + b)
1
RC
RC
A
RC
A
A
RC
RC
RC
RC
CfE Edition
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1
4
5
7
8
13
18
19
20
23
Higher Mathematics
Integration
Integration
1
Indefinite Integrals
RC
In integration, our aim is to “undo” the process of differentiation. Later we
will see that integration is a useful tool for evaluating areas and solving a
special type of equation.
We have already seen how to differentiate polynomials, so we will now look
at how to undo this process. The basic technique is:
x n +1
+c
n +1
n
dx
∫x =
n ≠ −1, c is the constant of integration.
Stated simply: raise the power (n) by one (giving n + 1 ), divide by the new
power ( n + 1 ), and add the constant of integration (c ) .
EXAMPLES
1. Find ∫ x 2 dx .
x3
+ c = 13 x 3 + c .
3
2
∫ x dx =
2. Find ∫ x −3 dx .
∫x
−3
1
x −2
dx = + c =− 2 + c .
−2
2x
5
3. Find ∫ x 4 dx .
5
4
∫ x dx =


9
x4
9
4 x 4 + c.
+
c
=
9
9
4
We use the symbol
∫
for integration.
The ∫ must be used with “dx” in the examples above, to indicate that we
are integrating with respect to x.

The constant of integration is included to represent any constant term in
the original expression, since this would have been zeroed by differentiation.

Integrals with a constant of integration are called indefinite integrals.
hsn.uk.net
Page 1
CfE Edition
Higher Mathematics
Integration
Checking the answer
Since integration and differentiation are reverse processes, if we differentiate
our answer we should get back to what we started with.
For example, if we differentiate our answer to Example 1 above, we do get
back to the expression we started with.
1 x3 + c
3
differentiate
x2
integrate
Integrating terms with coefficients
The above technique can be extended to:
ax n +1
dx a ∫ x =
dx
+c
∫ ax =
n +1
n
n
n ≠ −1, a is a constant.
Stated simply: raise the power (n) by one (giving n + 1 ), divide by the new
power ( n + 1 ), and add on c.
EXAMPLES
4. Find ∫ 6 x 3 dx .
6x 4
dx
+c
∫ 6x =
4
= 32 x 4 + c .
3
5. Find ∫ 4 x
− 32
dx .
−1
4x 2
x
dx
4
=
+c
∫
− 12
− 32
−1
=
−8 x 2 + c
8
=
−
+ c.
x
Note
It can be easy to confuse integration and differentiation, so remember:
dx
∫ x=
hsn.uk.net
1 x2 + c
2
∫ k dx=
Page 2
kx + c .
CfE Edition
Higher Mathematics
Integration
Other variables
Just as with differentiation, we can integrate with respect to any variable.
EXAMPLES
6. Find ∫ 2 p −5 dp .
2 p −4
+c
dp
∫2 p =
−4
1
=
− 4 + c.
2p
Note
dp tells us to integrate
with respect to p.
−5
7. Find
∫ p dx .
∫ p dx
Note
Since we are integrating
with respect to x, we
treat p as a constant.
= px + c .
Integrating several terms
The following rule is used to integrate an expression with several terms:
∫ ( f ( x ) + g ( x ) ) dx = ∫ f ( x ) dx + ∫ g ( x ) dx .
Stated simply: integrate each term separately.
EXAMPLES
(
)
1
 3 x 2 − 2 x 2 dx .
8. Find ⌠
⌡
(
⌠
 3x − 2 x
⌡
2
1
2
)
3
3x 3 2 x 2
− 3 +c
dx =
3
2
3
4x 2
=x −
+c
3
=−
x 3 34 x 3 + c .
3
(
)
−5
9. Find ⌠
 4 x 8 + 3 x + 7 dx .
⌡
(
⌠
 4x
⌡
− 58
)
3
+ 3 x + 7 dx =
3x 2
3 + 2 + 7x + c
8
4x 8
3
=83 × 4 x 8 + 23 x 2 + 7 x + c
3
8
= 32
+ 32 x 2 + 7 x + c .
x
3
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Page 3
CfE Edition
Higher Mathematics
2
Integration
Preparing to Integrate
RC
As with differentiation, it is important that before integrating, all brackets are
multiplied out, and there are no fractions with an x term in the denominator
(bottom line), for example:
1
= x −3
3
x
1
−1
=x 2
x
3
= 3x −2
2
x
1
= 14 x −5
5
4x
5
5 x − 23 .
=
2
4
43 x
EXAMPLES
⌠ dx
for x ≠ 0 .
⌡ x2
1. Find 
⌠ dx
⌠ 1
 2 is just a short way of writing  2 dx , so:
⌡x
⌡x
⌠ dx ⌠ 1
=
dx
 2 =
⌡ x2
⌡x
∫x
−2
dx
x −1
=
+c
−1
1
=
− + c.
x
⌠ dx
2. Find 
⌡
x
for x > 0 .
⌠ dx ⌠ 1
=
1 dx
=

⌡ x ⌡ x2
∫x
− 12
dx
1
=
x2
1
2
+c
= 2 x + c.
⌠ 7
3. Find 
⌡ 2p
2
dp where p ≠ 0 .
⌠ 7
dp = ∫ 72 p −2 dp

2
⌡ 2p
−1
7 p +c
=×
2 −1
7
=
−
+ c.
2p
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CfE Edition
Higher Mathematics
Integration
⌠ 3 x − 5x
4. Find 
dx .
4
⌡
5
⌠ 3 x − 5x
3 x 5 − 5 x dx
=
dx ⌠

4
4
⌡
4
⌡
5
(
)
3x 6
5x 2
−
+c
4×6 4×2
3 x6 − 5 x2 + c
= 24
8
=
= 18 x 6 − 58 x 2 + c .
3
Differential Equations
A
A differential equation is an equation
dy
= x2 .
involving derivatives, e.g.
dx
1 x3 + c
3
A solution of a differential equation is an
expression for the original function; in this
case=
y 13 x 3 + c is a solution.
differentiate
x2
integrate
In general, we obtain solutions using integration:
⌠ dy
dx
⌡ dx
y =
or
f ( x ) = ∫ f ′ ( x ) dx .
This will result in a general solution since we can choose the value of c.
y
O
x
The general solution corresponds to a “family”
of curves, each with a different value for c.
The graph to the left illustrates some of the
curves=
y 13 x 3 + c with different values of c.
If we have additional information about the function (such as a point its graph
passes through), we can find the value of c and obtain a particular solution.
hsn.uk.net
Page 5
CfE Edition
Higher Mathematics
Integration
EXAMPLES
1. The graph of y = f ( x ) passes through the point ( 3, − 4 ) .
dy
If = x 2 − 5 , express y in terms of x.
dx
⌠ dy
y =
dx
⌡ dx
= ∫ ( x 2 − 5 ) dx
= 13 x 3 − 5 x + c .
We know that when x = 3 , y = −4 so we can find c:
y = 13 x 3 − 5 x + c
=
−4 13 ( 3 )3 − 5 ( 3 ) + c
−4 = 9 − 15 + c
c =2
So y = 13 x 3 − 5 x + 2 .
2. The function f , defined on a suitable domain, is such that
1
f ′ ( x ) = x 2 + 2 + 23 .
x
Given that f (1) = 4 , find a formula for f ( x ) in terms of x.
f ( x ) = ∫ f ′ ( x ) dx
1
⌠
=   x 2 + 2 + 23  dx
x

⌡
∫(x
=
2
)
+ x −2 + 23 dx
= 13 x 3 − x −1 + 23 x + c
1
= 13 x 3 − + 23 x + c .
x
We know that f (1) = 4 , so we can find c:
1
f ( x )= 13 x 3 − + 32 x + c
x
=
4 13 (1)3 − 11 + 23 (1) + c
4 = 13 − 1 + 23 + c
c = 4.
1
So f ( x )= 13 x 3 − + 32 x + 4 .
x
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CfE Edition
Higher Mathematics
4
Integration
Definite Integrals
RC
If F ( x ) is an integral of f ( x ) , then we define:
b
) dx [ F ( =
x )]a
∫a f ( x=
b
F (b ) − F ( a )
where a and b are called the limits of the integral.
Stated simply:
Work out the integral as normal, leaving out the constant of integration.
 Evaluate the integral for x = b (the upper limit value).
 Evaluate the integral for x = a (the lower limit value).
 Subtract the lower limit value from the upper limit value.

Since there is no constant of integration and we are calculating a numerical
value, this is called a definite integral.
EXAMPLES
1. Find
3
∫1 5x
2
dx .
3
 5x 3 
5
x
dx
=
∫1
 3 

1
3
2
 5 ( 3 )3   5 (1)3 
= 
− 3 
3

 

=5 × 32 − 53
= 45 − 53 = 43 13 .
2. Find
∫0 ( x
2
+ 3 x 2 ) dx .
3
2
 x 4 3x 3 
∫0 ( x + 3 x ) dx =
4 + 3 

0
2
3
2
2
 x4

=  + x3 
4
0
 24
 04

3
=  + 2  −  + 03 
 4
  4

= 16
+8−0
4
= 4 + 8 = 12.
hsn.uk.net
Page 7
CfE Edition
Higher Mathematics
Integration
4
⌠ 4
3. Find  3 dx .
⌡−1 x
4
4
⌠ 4
dx = ∫−1 4 x −3 dx

3
⌡−1 x
4
 4 x −2 
=

 −2  −1
4
2
= − 2 
 x  −1

2
2 
=  − 2  −  −

 4   ( −1)2 
2 +2=
=− 16
1 78 .
5
Geometric Interpretation of Integration
A
We will now consider the meaning of integration in the context of areas.
Consider
2
 4 x − 13 x 3 
∫0 ( 4 − x ) dx =

0
2
2
(
)
= 8 − 83 − 0
= 5 13 .
On the graph of y= 4 − x 2 :
y
y= 4 − x 2
The shaded area is given by
2
∫0 ( 4 − x ) dx .
2
Therefore the shaded area is 5 13 square units.
–2
O
2 x
In general, the area enclosed by the graph y = f ( x ) and the x-axis, between
x = a and x = b , is given by
b
∫a f ( x ) dx .
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CfE Edition
Higher Mathematics
Integration
EXAMPLE
1. The graph of =
y x 2 − 4 x is shown below. Calculate the shaded area.
y
=
y x 2 − 4x
4 5 x
O
5
 x 3 4x 2 
∫4 ( x − 4 x ) dx =
3 − 2 

4
5
2
 53
  43
2
=  − 2 ( 5 )2  −  − 2 ( 4 ) 
3
  3

64
= 125
3 − 50 − 3 + 32
= 61
3 − 18
= 2 13 .
So the shaded area is 2 13 square units.
Areas below the x-axis
Care needs to be taken if part or all of the area lies below the x-axis. For
example if we look at the graph of =
y x2 − 4 :
y
The shaded area is given by
=
y x 2 − 4x
4
4
 x 3 4x 2 
2
∫1 ( x − 4 x ) dx =
3 − 2 

1
O
4
 43
2
=  − 2 ( 4 )  − 13 − 2
 3

(
x
1
)
1
= 64
3 − 32 − 3 + 2
63 − 30 =
=
−9.
21 − 30 =
3
In this case, the negative indicates that the area is below the x-axis, as can be
seen from the diagram. The area is therefore 9 square units.
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Page 9
CfE Edition
Higher Mathematics
Integration
Areas above and below the x-axis
Consider the graph from the example above, with a different shaded area:
y
=
y x 2 − 4x
From the examples above, the total shaded area is:
Area 1 + Area 2 = 2 13 + 9 = 11 13 square units.
Area 1
Using the method from above, we might try to
calculate the shaded area as follows:
4 5 x
5
5
 x 3 4x 2 
2
Area 2
∫1 ( x − 4 x ) dx =
3 − 2 

1
O
1
 53

=  − 2 ( 5 )2  − 13 − 2
3

(
)
1
= 125
3 − 50 − 3 + 2
124 − 48 =
=
−6 32 .
3
Clearly this shaded area is not 6 23 square units since we already found it to
be 11 13 square units. This problem arises because Area 1 is above the x-axis,
while Area 2 is below.
To find the true area, we needed to evaluate two integrals:
∫1 ( x
4
2
− 4 x ) dx
and
∫4 ( x
5
2
− 4 x ) dx .
We then found the total shaded area by adding the two areas together.
We must take care to do this whenever the area is split up in this way.
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CfE Edition
Higher Mathematics
Integration
EXAMPLES
2. Calculate the shaded area shown in the diagram below.
y
y =3 − 2 x − x 2
2
x
O 1
–3
To calculate the area from x = −3 to x = 1 :
1
2x 2 x 3 

∫−3 ( 3 − 2 x − x ) dx = 3 x − 2 − 3  −3
1
2
1
= 3 x − x 2 − 13 x 3 
−3
(3 (1) − (1) − 13 (1) ) − (3 ( −3 ) − ( −3 )
(3 − 1 − 13 ) − ( −9 − 9 + 9 )
2
=
=
3
2
− 13 ( −3 )3
)
= 3 − 1 − 13 + 9
= 10 23
So the area is 10 23 square units.
We have already integrated the equation of the curve, so we can just
substitute in new limits to work out the area from x = 1 to x = 2 :
2
2
2
3
∫1 ( 3 − 2 x − x ) dx = 3 x − x − 13 x 1
2
(3 ( 2 ) − ( 2 ) − 13 ( 2 ) ) − (3 (1) − (1)
( 6 − 4 − 83 ) − (3 − 1 − 13 )
2
=
=
3
= 2 − 83 − 2 + 13
2
− 13 (1)3
)
Remember
=
−2 13 . So the area is 2 13 square units. The negative sign just
So the total shaded area is 10 23 + 2 13 =
13 square units .
hsn.uk.net
Page 11
indicates that the area
lies below the axis.
CfE Edition
Higher Mathematics
Integration
3. Calculate the shaded area shown in the diagram below.
y
y = x 2 + 2 x − 24
O
x
5
2
First, we need to calculate the root between x = 2 and x = 5 :
x 2 + 2 x − 24 =
0
0
( x − 4)( x + 6) =
x = 4 or x = −6.
So the root is x = 4
To calculate the area from x = 2 to x = 4 :
4
 x 3 2x 2

∫2 ( x + 2 x − 24 ) dx =  3 + 2 − 24 x  2
4
2
=  13 x 3 + x 2 − 24 x 
4
2
( 13 ( 4 ) + ( 4 ) − 24 ( 4 )) − ( 13 ( 2 )
( 643 + 16 − 96 ) − ( 38 + 4 − 48)
3
=
=
2
3
+ ( 2 )2 − 24 ( 2 )
)
= 56
3 − 36
=
−17 13
So the area is 17 13 square units.
To calculate the area from x = 4 to x = 5 :
5
2
3
2
∫4 ( x + 2 x − 24 ) dx =  13 x + x − 24 x  4
5
=
=
( 13 (5) + (5) − 24 (5) ) − ( 13 ( 4 ) + ( 4 )
( 1253 + 25 − 120) − ( 643 + 16 − 96 )
3
2
3
2
− 24 ( 4 )
= 61
3 − 15
= 5 13
So the area is 5 13 square units.
So the total shaded area is 17 13 + 5 13 =
22 23 square units.
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Page 12
CfE Edition
)
Higher Mathematics
6
Integration
Areas between Curves
A
The area between two curves between x = a and x = b is calculated as:
b
∫ ( upper curve − lower curve ) dx
a
square units.
So for the shaded area shown below:
y
y = f (x)
a
The area is
O
y = g (x) b
b
∫ ( f ( x ) − g ( x ) ) dx square units.
a
x
When dealing with areas between curves, areas above and below the x-axis do
not need to be calculated separately.
However, care must be taken with more complicated curves, as these may give
rise to more than one closed area. These areas must be evaluated separately.
For example:
y
y = g (x)
c
O a
In this case we apply
b
x
y = f (x)
b
∫ ( upper curve − lower curve ) dx
a
to each area.
So the shaded area is given by:
b
c
∫a ( g ( x ) − f ( x ) ) dx + ∫b ( f ( x ) − g ( x ) ) dx .
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Page 13
CfE Edition
Higher Mathematics
Integration
EXAMPLES
1. Calculate the shaded area enclosed by the curves with equations
y= 6 − 3 x 2 and y =−3 − 2 x 2 .
y
y= 6 − 3 x 2
x
O
y =−3 − 2 x 2
To work out the points of intersection, equate the curves:
6 − 3 x 2 =−3 − 2 x 2
6 + 3 − 3x 2 + 2 x 2 =
0
9 − x2 =
0
( 3 + x )( 3 − x ) =
0
x=
−3 or x =
3.
Set up the integral and simplify:
3
∫ ( upper curve − lower curve ) dx
= ∫ ( ( 6 − 3 x ) − ( −3 − 2 x ) ) dx
= ∫ ( 6 − 3 x + 3 + 2 x ) dx
= ∫ ( 9 − x ) dx .
−3
3
2
2
−3
3
2
−3
3
2
2
−3
Carry out integration:
3
x3 

∫−3 ( 9 − x ) dx =
9 x − 3 

 −3
3
2

( 3 )3  
( −3 )3 
=  9 (3) −
 −  9 ( −3 ) − 3 
3

 

( 27 − 273 ) − ( −27 + 273 )
=
= 27 − 9 + 27 − 9
= 36.
Therefore the shaded area is 36 square units.
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Page 14
CfE Edition
Higher Mathematics
Integration
2. Two functions are defined for x ∈  by f ( x ) = x 3 − 7 x 2 + 8 x + 16 and
g ( x=
) 4 x + 4 . The graphs of y = f ( x ) and y = g ( x ) are shown
below.
y
y = f (x)
y = g (x)
–1 O
2
6
x
Calculate the shaded area.
Since the shaded area is in two parts, we apply
Area from x = −1 to x = 2 :
b
∫ ( upper − lower ) dx twice.
a
2
∫ ( upper − lower ) dx
= ∫ ( x − 7 x + 8 x + 16 − ( 4 x + 4 ) ) dx
= ∫ ( x − 7 x + 4 x + 12 ) dx
−1
2
−1
2
3
2
3
2
Note
The curve is at the top
of this area.
−1
2
 x 4 7x 3 4x 2

= −
+
+ 12 x 
3
2
4
 −1
4
3

 2 4 7 × 23
  ( −1) 7 ( −1)
2
2
=  −
+ 2 × 2 + 12 × 2  − 
−
+ 2 ( −1) + 12 ( −1) 

3
3
 4
  4

=
( 4 − 563 + 8 + 24 ) − ( 14 − 73 + 2 − 12 )
= 99
4
= 24 34 .
So the area is 24 34 square units.
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Page 15
CfE Edition
Higher Mathematics
Integration
Area from x = 2 to x = 6 :
6
∫ ( upper − lower ) dx
= ∫ ( 4 x − 4 − ( x − 7 x + 8 x + 16 ) ) dx
= ∫ ( − x + 7 x − 4 x − 12 ) dx
2
6
2
6
3
3
Note
The straight line is at the
top of this area.
2
2
2
6
 x 4 7x 3 4x 2

=−
 4 + 3 − 2 − 12 x 

2
4
3
2
 6
  2 4 7 × 23 4 × 2 2

7×6 4×6
= − +
−
− 12 × 6  −  − +
−
− 12 × 2 
3
2
3
2
 4
  4

(
= ( −324 + 504 − 72 − 72 ) − −4 + 56
3 − 8 − 24
)
= 160
3
= 53 13 .
So the area is 53 13 square units.
1 square units.
So the total shaded area is 24 34 + 53 13 =
78 12
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Page 16
CfE Edition
Higher Mathematics
Integration
3. A trough is 2 metres long. A cross-section of the trough is shown below.
y
y = x 2 − 4x + 5
y =2
x
O
The cross-section is part of the parabola with equation y = x 2 − 4 x + 5 .
Find the volume of the trough.
To work out the points of intersection, equate the curve and the line:
2
x 2 − 4x + 5 =
0
x 2 − 4x + 3 =
( x − 1)( x − 3=
) 0 so =
x 1 or =
x 3.
Set up the integral and integrate:
2
∫1 ( upper − lower ) dx = ∫1 ( 2 − ( x − 4 x + 5) ) dx
3
3
=
∫ ( −x
3
1
2
+ 4 x − 3 ) dx
3
 x 3 4x 2

=−
 3 + 2 − 3x 

1
 ( 3 )3
  (1)3

2
+ 2 (3) − 3 (3)  −  −
+ 2 (1)2 − 3 (1) 
= −
 3
  3

(
= ( −9 + 18 − 9 ) − − 13 + 2 − 3
)
= 0 + 13 − 2 + 3
= 34
= 1 13 .
Therefore the shaded area is 1 13 square units.
=
Volume cross-sectional area × length
= 34 × 2
= 83= 2 32 .
Therefore the volume of the trough is 2 23 cubic units.
hsn.uk.net
Page 17
CfE Edition
Higher Mathematics
7
Integration
Integrating along the y-axis
RC
For some problems, it may be easier to find a shaded area by integrating with
respect to y rather than x.
EXAMPLE
The curve with equation y = 19 x 2 is shown in the diagram below.
y
y = 19 x 2
y = 16
y=4
x
O
Calculate the shaded area which lies between y = 4 and y = 16 .
We have y = 19 x 2
9 y = x2
x
x = ± 9y
x = ±3 y .
x =3 y
The shaded area in the diagram to the
right is given by:
∫
16
4
16
1
3 y dy = ∫ 3 y 2 dy
O
4
y=4
y = 16
y
16
3 y 2 
= 3 
 2  4
3
16
3
= 2 y 
4
3
= 2 16 − 2 4
3
= 2 × 64 − 2 × 8
= 112.
Since this is half of the required area, the total shaded area is 224 square units.
hsn.uk.net
Page 18
CfE Edition
Higher Mathematics
8
Integration
Integrating sinx and cosx
RC
We know the derivatives of sin x and cos x , so it follows that the integrals are:
dx
∫ cos x=
− cos x + c .
∫ sin x dx =
sin x + c ,
Again, these results only hold if x is measured in radians.
EXAMPLES
∫ (5sin x + 2cos x ) dx .
−5cos x + 2sin x + c .
∫ (5sin x + 2cos x ) dx =
1. Find
2. Find
π
4
∫ ( 4cos x + 2sin x ) dx .
0
π
4
π
[ 4sin x − 2cos x ]04
∫ 4cos x + 2sin x dx =
0
( )
( )
=  4sin π4 − 2cos π4  − [ 4sin0 − 2cos0]


=  4 × 1 − 2 × 1  − [ −2 ]
Note
2
2 

It is good practice to
= 4 − 2 +2
rationalise the
(
) (
2
2
= 2 × 2 +2
2
2
(
=
)
)
denominator.
2 + 2.
4
3. Find the value of ∫ 12 sin x dx .
0
4
1 sin x dx =  − 1 cos x  4
∫2
 2
0
0
=
− 12 cos ( 4 ) + 12 cos ( 0 )
= 1 ( 0.654 + 1)
2
= 0.827 (to 3 d.p.)
hsn.uk.net
Page 19
Remember
We must use radians
when integrating of
differentiating
trigonometric functions.
CfE Edition
Higher Mathematics
9
Integration
A Special Integral
RC
n
The method for integrating an expression of the form ( ax + b ) is:
n
∫ ( ax + b )
n +1
( ax + b )
=
dx
a ( n + 1)
+c ,
where a ≠ 0 and n ≠ −1 .
Stated simply: raise the power (n ) by one, divide by the new power and also
divide by the derivative of the bracket ( a ( n + 1)) , add c.
EXAMPLES
1. Find
∫ ( x + 4)
∫ ( x + 4)
7
7
dx =
dx .
( x + 4 )8
+c
8 ×1
8
( x + 4)
=
+ c.
8
2. Find ∫ ( 2 x + 3 )2 dx .
∫ ( 2x + 3)
2
dx=
( 2 x + 3 )3
+c
3×2
3
(2x + 3)
=
+ c.
6
1
dx where x ≠ − 95 .
3
⌡ 5x + 9
⌠
3. Find 
⌠

⌡
⌠
1
1
=
dx

1 dx

3
5x + 9
+
x
5
9
)3
⌡(
=
−
∫ (5x + 9 ) 3 dx
1
2
=
=
(5x + 9 ) 3
2 ×5
3
3
5x + 9
10
3
+c
2
+c
2
3 3 5x + 9 + c .
= 10
hsn.uk.net
Page 20
CfE Edition
Higher Mathematics
4. Evaluate
3
∫
0
3
∫
0
Integration
3 x + 4 dx where x ≥ − 34 .
3
1
3 x + 4 dx = ∫ ( 3 x + 4 ) 2 dx
0
3
 (3x + 4 )2 
= 3

 2 × 3  0
3
2
=


2
= 


=
3
2 13
2 4
−
9
9
= 29
3
( 3 x + 4 ) 
9

0
3
( 3 ( 3 ) + 4 )   2
−
 
9
 
3
(
3
13 − 8
(3 (0) + 4 )
3
9




3
)
(or 8.638 to 3 d.p.).
Note
Changing powers back
to roots here makes it
easier to evaluate the
two brackets.
Remember
To evaluate 43 , it is
easier to work out 4
first.
Warning
Make sure you don’t confuse differentiation and integration – this could lose
you a lot of marks in the exam.
Remember the following rules for differentiation and integrating expressions
n
of the form ( ax + b ) :
d  ax + b )n  = an ( ax + b ) n −1 ,

dx (
n
∫ ( ax + b )
=
dx
( ax + b )n+1
a ( n + 1)
+c .
These rules will not be given in the exam.
hsn.uk.net
Page 21
CfE Edition
Higher Mathematics
Integration
Using Differentiation to Integrate
Recall that integration is the process of undoing differentiation. So if we
differentiate f ( x ) to get g ( x ) then we know that ∫ g ( x =
) dx f ( x ) + c .
EXAMPLES
5
with respect to x.
( 3 x − 1)4
1
⌠
(b) Hence, or otherwise, find 
dx .
5
⌡ ( 3 x − 1)
5. (a) Differentiate y =
y
(a)=
5
−4
=
5 ( 3 x − 1)
4
( 3 x − 1)
dy
−5
= 5 × 3 × ( −4 )( 3 x − 1)
dx
60
= −
5.
−
3
x
1
)
(
60
5
dx =
+ c . So:
5
( 3 x − 1)4
⌡ ( 3 x − 1)
⌠
(b) From part (a) we know  −
⌠
1
5
5 dx = 4 + c
( 3 x − 1)
⌡ ( 3 x − 1)
−60
Note
We could also have used
the special integral to
obtain this answer.


⌠
1
5
1
=
−
+
dx
c



5
60  3 x − 1 4
(
)
⌡ ( 3 x − 1)


1
=
−
4 + c1
12 ( 3 x − 1)
6. (a) Differentiate y =
(x
1
3
− 1)
⌠
x2
(b) Hence, find 
 x3 −1
⌡
)
(
(a) =
y
1
=
5
( x 3 − 1)
(x
3
6
− 1)
5
where c1 is some constant.
with respect to x.
dx .
−5
−6
dy
=
−5 ( x 3 − 1) × 3 x 2
dx
15 x 2
= −
6 .
3
( x − 1)
hsn.uk.net
Page 22
CfE Edition
Higher Mathematics
Integration
⌠
15 x 2
− 3
(b) From part (a) we know 

x −1
⌡
(
⌠
−15
(
x2
)
 x3 −1
⌡
⌠
x2

 x3 −1
⌡
(
)
6
1
dx =3
5 + c . So:
( x − 1)
1
dx = 5 + c
( x 3 − 1)
6

)
Note
In this case, the special
integral cannot be used.


5 +c

 ( x − 1)

1
=
−
where c 2 is some constant.
5 + c2
3
15 ( x − 1)
1
1
dx =
− 15
 3
6
10 Integrating sin(ax + b) and cos(ax + b)
RC
Since we know the derivatives of sin ( ax + b ) and cos ( ax + b ) , it follows that
the integrals are:
1 sin ( ax + b ) + c ,
a
dx
∫ cos ( ax + b )=
− 1a cos ( ax + b ) + c .
∫ sin ( ax + b ) dx =
These are given in the exam.
EXAMPLES
1. Find ∫ sin ( 4 x + 1) dx .
− 14 cos ( 4 x + 1) + c .
∫ sin ( 4 x + 1) dx =
(
)
2. Find ∫ cos 32 x + π5 dx .
) dx
∫ cos ( 2 x + π5=
3
(
)
3
π
2
3 sin 2 x + 5 + c .
1
3. Find the value of ∫ cos ( 2 x − 5 ) dx .
0
1
1
dx  12 sin ( 2 x − 5 ) 
∫ cos ( 2 x − 5) =
0
0
= 12 sin ( −3 ) − 12 sin ( −5 ) .
1 ( −0.141 − 0.959 )
=
2
= −0.55 (to 2 d.p.).
hsn.uk.net
Page 23
Remember
We must use radians
when integrating or
differentiating
trigonometric functions.
CfE Edition
Higher Mathematics
Integration
(
)
4. Find the area enclosed be the graph=
of y sin 3 x + π6 , the x-axis, and
the lines x = 0 and x = π6 .
y
(
=
y sin 3 x + π6
O
π
6
(
⌠ sin 3 x + π
6
⌡
0
)
)
x
π
6
(
)
π
 − 1 cos 3 x + π  6
dx =
6 0
 3
(( ) )
(
)
=  − 13 cos 3 π6 + π6  −  − 13 cos 3 ( 0 ) + π6 


 
=  − 13 cos ( 90 + 30 ) ° +  13 cos ( 30 ) °
( )
=  − 13 × ( − 12 )  +  13 × 23 


 
= 16 + 63
1+ 3
.
6
1+ 3
square units.
So the area is
6
=
5. Find ∫ 2cos ( 12 x − 3 ) dx .
) dx
∫ 2cos ( 12 x − 3=
=
(
)
4sin ( 12 x − 3 ) + c
2 sin 1 x − 3 + c
1
2
2
6. Find ∫ 5cos ( 2 x ) + sin ( x − 3 ) dx .
∫ 5cos ( 2 x ) + sin ( x −
hsn.uk.net
3 )=
dx 52 sin ( 2 x ) − cos ( x − 3 ) + c
Page 24
CfE Edition
Higher Mathematics
Integration
1
with respect to x.
cos x
tan x
(b) Hence find ⌠
dx .

⌡ cos x
1
−1
d ( cos x )−1 = −1( cos x )−2 × − sin x
(a)
= ( cos x ) , and dx
cos x
sin x
.
=
cos 2 x
7. (a) Differentiate
sin x
tan x
sin x
cos x
(b) = =
.
cos x cos x cos 2 x
sin x
1
From part (a) we know ⌠
dx
+c .
 2=
⌡ cos x
cos x
tan x
1
Therefore ⌠
=
dx
+c.

⌡ cos x
cos x
hsn.uk.net
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CfE Edition