hsn.uk.net
Higher
Mathematics
UNIT 2 OUTCOME 2
Integration
Contents
Integration
1
2
3
4
5
6
7
Indefinite Integrals
Preparing to Integrate
Differential Equations
Definite Integrals
Geometric Interpretation of Integration
Areas between Curves
Integrating along the y-axis
HSN22200
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89
89
92
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106
Higher Mathematics
Unit 2 – Integration
OUTCOME 2
Integration
1
Indefinite Integrals
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:
n
∫ x dx =
x n+1
+c
n +1
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 .
2
∫ x dx =
x3
+ c = 13 x 3 + c .
3
2. Find ∫ x −3 dx .
∫x
−3
x −2
1
dx =
+ c = − 2 + c.
−2
2x
5
3. Find ∫ x 4 dx .
9
∫x
5
4
x4
9
dx = 9 + c = 94 x 4 + c .
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.
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HSN22000
Higher Mathematics
Unit 2 – 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
∫ ax dx = a ∫ x dx = n + 1 + c
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
∫ 6 x dx = 4 + c
= 32 x 4 + c .
3
5. Find ∫ 4 x
∫ 4x
− 32
− 32
dx .
−1
4x 2
dx = 1 + c
−2
−1
= −8 x 2 + c
8
=−
+ c.
x
Note
It can be easy to confuse integration and differentiation, so remember:
∫ x dx = 12 x
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2
∫ k dx = kx + c .
+c
Page 90
HSN22000
Higher Mathematics
Unit 2 – 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
−4
1
= − 4 + c.
2p
−5
∫ 2 p dp =
7. Find
Note
dp tells us to integrate
with respect to p.
∫ 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 ⌠
⌡
3
(
1
⌠
 3x 2 − 2x 2
⌡
)
3x 3 2x 2
dx =
− 3 +c
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 .
⌡
3
8
2
4x
3x
−5
⌠
+ 7x + c
 4 x 8 + 3 x + 7 dx = 3 +
⌡
2
8
3
= 83 × 4 x 8 + 23 x 2 + 7 x + c
3
8
= 32
x
+ 32 x 2 + 7 x + c .
3
(
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HSN22000
Higher Mathematics
2
Unit 2 – Integration
Preparing to Integrate
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
3
= 3x −2
2
x
1
−1
=x 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
−2
 2 =  2 dx = ∫ x dx
⌡x
⌡x
x −1
=
+c
−1
1
= − + c.
x
⌠ dx
2. Find 
⌡
for x > 0 .
x
⌠ dx ⌠ 1
− 12
=
dx
=
x
1


∫ dx
⌡ x ⌡ x2
1
x2
= 1 +c
2
= 2 x + c.
⌠ 7
3. Find 
⌡ 2p
2
dp where p ≠ 0 .
⌠ 7
dp = ∫ 72 p −2 dp

2
⌡ 2p
−1
p
= 72 ×
+c
−1
=−
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7
+ c.
2p
Page 92
HSN22000
Higher Mathematics
Unit 2 – Integration
5
⌠ 3 x − 5x
4. Find 
dx .
4
⌡
5
⌠ 3x − 5x
3 x 5 − 5 x dx
dx = ⌠

4
4
⌡
4
⌡
(
)
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 differential equation is an equation
dy
involving derivatives, e.g.
= x2 .
dx
1 x3 + c
3
differentiate
x2
integrate
A solution of a differential equation is an
expression for the original function; in this
case y = 13 x 3 + c is a solution.
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, the
constant of integration.
y
The general solution corresponds to a “family” of
curves, each with a different value for c.
O
x
The graph to the left illustrates some of the
curves y = 13 x 3 + c for particular 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.
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HSN22000
Higher Mathematics
Unit 2 – 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 − 5x + 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
= 13 x 3 −
1 2
+ x + c.
x 3
We know that f (1) = 4 , so we can find c:
1
f ( x ) = 13 x 3 − + 23 x + c
x
4 = 13 (1)3 − 11 + 23 (1) + c
4 = 13 − 1 + 23 + c
c = 4.
So f ( x ) = 13 x 3 −
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1 2
+ x + 4.
x 3
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HSN22000
Higher Mathematics
4
Unit 2 – Integration
Definite Integrals
If F ( x ) is an integral of f ( x ) , then we define:
b
b
∫a f ( x ) dx = [ F ( x )]a = 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 
∫1 5x dx =  3 

1
3
2
 5 ( 3 )3   5 (1)3 
=
−

 3   3 
= 5 × 32 − 53
= 45 − 35 = 43 31 .
2. Find
2
∫0 ( x
3
+ 3 x 2 ) dx .
2
 x 4 3x 3 
∫0 ( x + 3 x ) dx =  4 + 3 

0
2
3
2
2
 x4

=  + x3 
4
0
 24
  04

=  + 2 3  −  + 03 
 4
 4

= 16
+8−0
4
= 4 + 8 = 12.
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HSN22000
Higher Mathematics
Unit 2 – Integration
4
⌠ 4
3. Find  3 dx .
⌡−1 x
4
4
⌠ 4
−3
 3 dx = ∫−1 4 x dx
⌡−1 x
4
 4 x −2 
=

 −2  −1
4
2
= − 2 
 x  −1

2
2 
=  − 2  −  −

 4   ( −1)2 
2 + 2 =17 .
= − 16
8
5
Geometric Interpretation of Integration
We will now consider the meaning of integration in the context of areas.
Consider
2
2
2
3
∫0 ( 4 − x ) dx =  4 x − 13 x  0
(
)
= 8 − 83 − 0
= 5 13 .
On the graph of y = 4 − x 2 :
y
y = 4 − x2
The shaded area is given by
2
2
∫0 ( 4 − x ) dx .
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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HSN22000
Higher Mathematics
Unit 2 – 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
=  − 2 (5)  −  − 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 = x 2 − 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
3 − 30 = 21 − 30 = −9.
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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HSN22000
Higher Mathematics
Unit 2 – 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
Area 1
From the working above, the total shaded area is:
Area 1 + Area 2 = 2 13 + 9 = 11 13 square units.
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
2
= 124
3 − 48 = −6 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:
4
∫1 ( x
2
− 4 x ) dx
and
5
∫4 ( x
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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HSN22000
Higher Mathematics
Unit 2 – Integration
EXAMPLES
2. Calculate the shaded area shown in the diagram below.
y
y = 3 − 2x − x 2
2
–3
x
O 1
To calculate the area from x = −3 to x = 1 :
1
2x 2 x3 

∫−3 ( 3 − 2 x − x ) dx = 3 x − 2 − 3  −3
1
2
1
= 3 x − x 2 − 13 x 3 
−3
(
) (
= ( 3 − 1 − 13 ) − ( −9 − 9 + 9 )
= 3 (1) − (1)2 − 13 (1)3 − 3 ( −3 ) − ( −3 )2 − 31 ( −3 )3
)
= 3 − 1 − 13 + 9
= 10 32
So the area is 10 32 square units.
We have already carried out the integration, so we can just substitute in
new limits to work out the area from x = 1 to x = 2 :
2
2
2
2
3
∫1 ( 3 − 2 x − x ) dx = 3 x − x − 13 x 1
(
) (
= ( 6 − 4 − 83 ) − ( 3 − 1 − 13 )
= 3 ( 2 ) − ( 2 )2 − 13 ( 2 )3 − 3 (1) − (1)2 − 13 (1)3
)
= 2 − 83 − 2 + 13
= −2 13 .
Remember
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 .
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Page 99
indicates that the area
lies below the axis.
HSN22000
Higher Mathematics
Unit 2 – 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
( x − 4)( x + 6) = 0
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
(
) (
8
= ( 64
3 + 16 − 96 ) − ( 3 + 4 − 48 )
3
2
= 13 ( 4 ) + ( 4 ) − 24 ( 4 ) − 13 ( 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
5
2
3
2
∫4 ( x + 2 x − 24 ) dx =  13 x + x − 24 x  4
(
) (
64
= ( 125
3 + 25 − 120 ) − ( 3 + 16 − 96 )
3
2
= 13 ( 5 )3 + ( 5 )2 − 24 ( 5 ) − 13 ( 4 ) + ( 4 ) − 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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Higher Mathematics
6
Unit 2 – Integration
Areas between Curves
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
x
b
y = f (x)
In this case we apply
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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HSN22000
Higher Mathematics
Unit 2 – Integration
EXAMPLES
1. Calculate the shaded area enclosed by the curves with equations
y = 6 − 3x 2 and y = −3 − 2 x 2 .
y
y = 6 − 3x 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 + 2x 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
= 27 − 27
3 − −27 + 3
)
= 27 − 9 + 27 − 9
= 36.
Therefore the shaded area is 36 square units.
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HSN22000
Higher Mathematics
Unit 2 – Integration
2. Two functions are defined for x ∈ ℝ by f ( x ) = x 3 − 7 x 2 + 8x + 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
b
∫ ( upper − lower ) dx
a
twice.
Area from x = −1 to x = 2 :
2
∫ ( upper − lower ) dx
= ∫ ( x − 7 x + 8 x + 16 − ( 4 x + 4 ) ) dx
= ∫ ( x − 7 x + 4 x + 12 ) dx
−1
2
3
2
3
2
−1
2
Note
The curve is at the top
of this area.
−1
2
 x 4 7x3 4x 2

= −
+
+ 12 x 
3
2
4
 −1

 2 4 7 × 23
  ( −1)4 7 ( −1)3
2
= −
+ 2 × 2 + 12 × 2  − 
−
+ 2 ( −1)2 + 12 ( −1) 
3
3
 4
  4

(
) (
1 7
= 4 − 56
3 + 8 + 24 − 4 + 3 + 2 − 12
)
= 99
4
= 24 34 .
So the first area is 24 34 square units.
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HSN22000
Higher Mathematics
Unit 2 – 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
3
Note
The straight line is at the
top of this area.
2
2
6
3
2
2
6
 x 4 7x 3 4x 2

= − +
−
− 12 x 
3
2
 4
2
 6 4 7 × 63 4 × 6 2
  2 4 7 × 23 4 × 2 2

= − +
−
− 12 × 6  −  − +
−
− 12 × 2 
3
2
3
2
 4
  4

(
= ( −324 + 504 − 72 − 72 ) − −4 + 56
3 − 8 − 24
)
= 160
3
= 53 31 .
So the second 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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HSN22000
Higher Mathematics
Unit 2 – 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:
x 2 − 4x + 5 = 2
x 2 − 4x + 3 = 0
( x − 1)( x − 3 ) = 0 so x = 1 or x = 3.
Set up the integral and integrate:
3
3
2
∫1 ( upper − lower ) dx = ∫1 ( 2 − ( x − 4 x + 5) ) dx
3
= ∫ ( − x 2 + 4 x − 3 ) dx
1
3
 x3 4x2

= − +
− 3x 
2
 3
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
= 38 = 2 32 .
Therefore the volume of the trough is 2 23 cubic units.
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HSN22000
Higher Mathematics
7
Unit 2 – Integration
Integrating along the y-axis
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.
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