Math AI HL

394 10 Differentiation

Be the Examiner 10.2

Differentiate f(x) = 5 .
e2x
Which is the correct solution? Identify the errors in the incorrect solutions.

Solution 1 Solution 2 Solution 3

Let u = 5 and v = e2x f(x) = 5 = 5e−2 x f(x) = 5 = 5e−2 x
v 2eThen u′ = 0 and ′ = 2x e2x e2x
v eLet u = 5 and = −2x f′(x) = −10e−2x
f′(x) e2x × 0−5× 2e2x v 2eThen u′ = 0 and ′ = − −2x
= (e2 x )2

− 10e2x 10 f 5 2e e 0′(x) = (− −2x) + −2x ×
e e= 2 x )2 = − 2x = −10e−2x
(

Exercise 10C
yxFor d
questions 1 to 4, use the method demonstrated in Worked Example 10.10 to find d for the following graphs.

1 a y = x sin x 1 2 a y = x2 cos x
2 b y = x−1 cos x
b y = x sin x

3 a y x − 1 ex 4 a y x 2 ln x
2 3
= =

b y = x3ex b y = x4 ln x

For questions 5 to 9, use the method demonstrated in Worked Example 10.11 to find f ′(x) for the following functions.
xx2(2(3xx+−1)421)23
5 a f(x) = 6 a f(x) = x2 sin 3x 7 a f(x) = x cos(4x − 1)
b f(x) = b f(x) = 3 sin 2x b f(x) = x−2 cos5x
x4
1
f e8 a x2 4x+5 9 a f(x) = x−1 ln(2x − 3)
(x) =

f eb (x) = x−2 1−x b f(x) = x3 ln(5 − x)

For questions 10 to 12, use the method demonstrated in Worked Example 10.12 to find the derivative of the following

graphs. 11 a y= x−2 12 a y = ex
x +1 b y = (3xe+x 1)2
2 (2x − 1)3

y x x10 a = + 2
3x 4−x
b y = x−4 b y = (x − 3)2

13 Find the equation of the tangent to the curve y = sin x at x = π .
x
2

14 Find the equation of the normal to the curve y = x cos2x at x = π .
15
fIf (x) = xe2x, find f'(3) . 4

16 Differentiate xex ln x.
e17 Differentiate xsin x .
y e 11x18 Find the point on the curve = 3x − where the tangent has gradient 13.
y x19 Find the coordinates of the points on the curve = e−x where the gradient is zero.
y x ln x20 Find the coordinates of the points on the curve =
where the gradient is 2.
f x k x x21 Find and simplify an expression for the derivative of ( ) = + 2 , where k is a positive constant.
f 2 f x ax b22 If (x) = x + x, show that ′( ) = + , where a and b are integers to be found.
f 1 2 2 x x a x b23
Show that (x) = x2 is an increasing + if > or if < , where a and b are constants to be determined.

+x function

10D Related rates of change 395

24 The population of rabbits on an island, P thousands, n years after they were introduced is modelled by
P 1 43e= + −n
Pna d
Find an expression for the population growth rate, d , and explain why this predicts that the population is
always growing.

b Find

i the initial population

ii the initial population growth rate.

c What is the value of P as n gets very large?

d Hence sketch the population as a function of time.

0 e25 a Show that if x > , then xx = xln x .
y xb Hence find the equation of the tangent to the curve = x at x = 1.

10D Related rates of change

There are many situations where we are given one rate and want to link it to another
rate. For example, we might know the rate at which a barrel is leaking, and want to
know the rate at which the height of water is decreasing.

One of the most common tricks to do this is to use the chain rule to link the two rates.
ddxt dy
If we know and want to know dt we use

dy = dy × dx
dt dx dt
dy
So, the problem becomes finding dx . This can be done using some other link between
y and x, such as a given equation.

WORKED EXAMPLE 10.14

If y = x 2, find dy when x = 4 and dx = 12 .
dt dt

Use the chain rule to dy dy dx
dy dt dx dt
relate dt and dx = ×
dt

Use the fact that 2 x dx
dy dt
y = x2 to find dx = ×

Substitute in values = 8 × 12 = 96
dx
for x and dt

Frequently, the link between the variables comes from a geometric context.

396 10 Differentiation

WORKED EXAMPLE 10.15

A rectangle has width x and height y. When x = 2 cm and y = 4 cm, these values are changing

according to dx = 3 cm s−1 and dy = −1 cm s−1.
dt dt
What is the rate of change of the area at this time?

Define a variable Let the area be Acm2.
for the area A = xy

Differentiate both sides dA = d (xy)
with respect to t dt
dt
This requires the
product rule = x dy + y dx
dt dt
Substitute in the
given values = 2 × −1 + 4 × 3
= 10

So, the area is increasing at a rate of 10 cm s−1.

CONCEPTS – QUANTITY AND CHANGE

Worked Example 10.15 illustrates that the rate of change can depend on its current
size. For example, if the radius of a circle increases at a constant rate, then the area
will increase faster as the circle gets larger; when a balloon is inflated at a constant
rate (so that the rate of increase of volume is constant) the rate of change of the
radius will decrease with the size of the balloon. The chain rule is an important tool for
quantifying the relationship between the different rates.

Exercise 10D

For questions 1 to 4, use the method demonstrated in Worked Example 10.14 to find the required rate of change.

1 a If y = x3, find dy when x = 1 and dx = −1.
dt dt
dy
b If y = x2 + x, find dt when x = 0 and dx = 4.
a If A = dt
e2z , find the rate of increase of dz 6.
2 A when z = 0 and dt =

p dq 7b If = e q, find the rate of increase of p when q = 1 and = .
a dt3 a If = 1, find the rate of increase of a when b = 2 and b is increasing at a rate of 3 per second.

a bb1b If = 2 , find the rate of increase of a when b = 1 and b is increasing at a rate of 2 per second.
y ln x4 a If = , find the rate of increase of y when x = 2 and x is decreasing at a rate of 4 per hour.
y ln x 1b If = ( + ), find the rate of increase of y when x = 1 and x is decreasing at a rate of 3 per hour.

5 Given that A = x2 + y2, find dA when x = 3, y = 4, dx = 1, dy = −1.
dt dt dt

6 Given that B = x3 + y3, find dB when x = 1, y = 2, dx = 1, dy = −2.
dt dt dt

7 Given that C = x , find dC when x = 3, y = 4, dx = 3, dy = −1.
y dt dt dt

10E The second derivative 397

8 The sides of a square are increasing at a rate of 2 cm s–1. Find the rate of increase of the area when the area
is 25 cm2.

9 Circular mould is spreading on a leaf. When the radius is 3 mm, the rate of increase is 1.2 mm per day.
What is the rate of increase of the area?

10 The volume of a spherical balloon is increasing at a rate of 200 cm3 per second. Find the rate of increase of the

radius when the volume is 100 cm3.

11 An x by y rectangle is expanding, with dx = 4 cm s−1 and dy = −2cm s−1. When x = 3 cm and y = 4cm, find
dt dt
a the rate of increase of the rectangle’s area

b the rate of increase of the length of the diagonal.

12 An inverted cone is being filled with water at a constant rate of 5 cm3 s–1. The surface of the water is always

horizontal as it is being filled. The largest diameter of the cone is 10 cm and its height is 30 cm. If the volume of

water in the cone is V at time t , and h is the height of the water above the vertex of the cone,
πh3
a show that V = 108

18b find the rate that the height is increasing when h = cm.
cm cm13 A circular stain of radius r and area A 2 is increasing in size. At a certain time, the rate of increase of the
86.5cm sradius is 1.8 cm s–1 and the rate of increase of the area is 2 −1. Find the radius of the stain at this point.

14 A sportsman throws a ball. When it is 2 m above the sportsman and 4 m away horizontally, it is moving purely

horizontally with a speed of 3 m s–1. Find the rate at which the ball is moving away from the sportsman.

5gcm15 The density of a reactive substance is given by its mass divided by its volume. When the density is −3,
2 gs 1cm sthe mass is decreasing at a rate of −1 and the volume is decreasing at a rate of 3 −1. Determine, with
justification, whether the density is increasing or decreasing.

16 A ladder of length 3 m is sliding down a vertical wall. The foot of the ladder is on horizontal ground. When the
point of contact with the wall is 2 m above the horizontal, that point is moving down at a rate of 0.1 m s–1. At what

speed is the foot of the ladder moving away from the wall, assuming that the ladder always stays in contact with

both the wall and the ground?

10E The second derivative

 Finding the second derivative

Differentiating the derivative f′(x) gives the second derivative f′′(x), which measures

the rate of change of the gradient.
xy xyIf the
notation d is used for the derivative, then d2 is used for the second derivative.
d d2

WORKED EXAMPLE 10.16

If f(x) = x3 + 3x, find f′′(2).

First find f ′(x) f′(x) = 3x2 + 3
Then differentiate again f 6′′(x) = x
f′′(2) = 12
Now substitute x = 2

398 10 Differentiation

In Section 9B  Concave-up and concave-down
of the
Mathematics: You already know that you can classify different parts of the graph of a function as
applications and being increasing or decreasing depending on whether the gradient (the derivative) is
interpretation SL positive or negative.
book, you saw that a
function is increasing There is another classification based on the gradient of the gradient (the second
derivative).
where f′(x) > 0 and
If the second derivative is positive, it means that the gradient is increasing. We say that
decreasing where the curve is concave-up. Depending upon whether the gradient is positive or negative it
can look like:
f′(x) < 0.
yy

xx

If the second derivative is negative, it means that the gradient is decreasing. This is
called concave-down. Depending upon whether the gradient is positive or negative it
can look like:

yy

xx

Tip

A curve does not have to have a turning point to be described as concave-up or concave-down.
Any part of the graphs above can be described as concave-up or concave-down. One useful
trick is to think where you would be naturally put your compass (or your elbow) when sketching
small sections of the curve. If it is above the curve it is concave-up. If it is below the curve it is
concave-down.

10E The second derivative 399

KEY POINT 10.10

A function f(x) is

f 0 concave-up where ′′(x) >
f 0 concave-down where ′′(x) < .

WORKED EXAMPLE 10.17

Using one or more of the terms ‘increasing’, ‘decreasing’, ‘concave-up’, ‘concave-down’,
describe these sections of the graph.

a A to B b C to E c F to G

A G

CD

B
E

F a Decreasing, concave-up
b Concave-down
The curve is increasing c Increasing, concave-up
C to D but decreasing
D to E. It is concave-
down from C to E

You are the Researcher

Just like the first derivative is related to the graphical concept of gradient of a
graph, the second derivative is related to the graphical concept of curvature.
You might like to investigate how to find a formula for the radius of curvature
of a graph. This is a core problem in an area of mathematics called differential
geometry.

400 10 Differentiation

WORKED EXAMPLE 10.18

f 3 2 1Find the values of x for which the function (x) = x3 + x2 + x + is concave-down.

Tip Find the second derivative f(x) = 3x2 + 6x + 2
f′′(x) = 6x + 6
You can check your f 0For concave-down ′′(x) < f 0′′(x) <
answer by graphing the 6x + 6 < 0
function on your GDC. Solve the inequality
x < −1

 Classifying local maximum and minimum points

You have already found that at a maximum or minimum point the derivative is zero.
This fact can be used to find local maxima and minima. However, you can now use the
concavity of the graph to decide whether the stationary point is a maximum or a minimum.

If the graph is concave-up at the stationary point, then it must be a minimum:

If the graph is concave-down at the stationary point, then it must be a maximum:

We can rephrase this in terms of derivatives.

KEY POINT 10.11

f 0Given ′(a) = , if
f 0, x a ′′(a) < then there is a local maximum at =
f 0 x a ′′(a) > , then there is a local minimum at = .

10E The second derivative 401

WORKED EXAMPLE 10.19

y x 3xFind and classify the stationary points on the curve = 3 − 2 .

Differentiate the curve dy = 3x2 − 6x
to find the gradient dx

At stationary points, 3x2 − 6x = 0
the gradient is zero
3x(x − 2) = 0
This equation is best
solved by factorising x = 0 or x = 2

We need to find When x = 0, y = 0
the y-coordinates When x = 2, y = −4
corresponding to the
d2 y = 6x − 6
x-coordinates dx2

We find the second When x = 0, d2 y = −6 < 0
derivative to classify dx2
the stationary points So, (0,0) is a local maximum.

Substitute in the x value d2 y
of each stationary point dx
2
to determine whether
the second derivative is

positive or negative

When x = 2, = 6 > 0

So, (2,−4) is a local minimum.

Tip

You can find stationary points using your calculator, but it is useful to practise this method in
case a question uses parameters.

In Key Point 10.11, it turns out that if f′′(a) = 0, the situation is more complicated.

y xIt might be a minimum (for example, = 4 at x = 0) or a maximum, or it might be a
y xpoint of inflection. This is a point where the concavity changes, for example = 3

at x = 0.

y

y = x3

ffꞌꞌꞌ((00)) 0 x
0
=
=

402 10 Differentiation

Points of inflection do not have to have zero gradient, they occur whenever the
concavity changes:

y

y = x3 + x

ffꞌꞌꞌ((00)) ≠ 0 x
0
=

This means that at any point of inflection, d2 y = 0.
dx2
Another equivalent interpretation of points of inflection is where the gradient reaches

a local maximum or minimum.

Exercise 10E

For questions 1 to 4, use the method demonstrated in Worked Example 10.16 to evaluate the second derivative of the

following functions when x = 2.

1 a f(x) = 5x2 + 2x + 1 2 a f(x) = 5x3 + x2 + 3x + 7
b f(x) = 6x2 + 3x + 5 b f(x) = x3 − 2x2 + 4x + 11
y e3 a = −2x 4 a y = ln x
b y = 2e3x b y = −4 ln(x) + 2

5 Use the method demonstrated in Worked Example 10.17 to give one or more of the terms ‘increasing’,
‘decreasing’, ‘concave-up’, ‘concave-down’ to describe the following sections of each graph.

aB bF

E G
D
A CF A C
E
B
DG
i A to B
ii C to F
iii F to G

i A to B
ii C to E
iii E to G

10E The second derivative 403

For questions 6 and 7, use the method demonstrated in Worked Example 10.18 to find the values of x for which the

function has the given description.

f 4 5 26 a (x) = x3 − x2 + x − is concave-down 7 a f(x) = 18x2 + 2x − 1 is concave-up
f 9 4 1b (x) = x3 + x2 − x + is concave-up f 2 7b (x) = x − − x2 is concave-up

For questions 8 to 12, use the method demonstrated in Worked Example 10.19 to find and classify the stationary points

on the given graph. 9 a y = x4 − 18x2 − 3 10 a y = 2 x 1 − 3x + 1
2
8 a y = 2x3 − 9x2 + 12x + 6
1
b y = x3 − 12x + 4 b y = 2x4 − x +1 b y = x − 2 + 4x − 2

11 a y = ex − 5x 12 a y = 3ln x − 2x

b y = x − ex b y = x 1 − ln x
2 2

13 If f(x) = x3 + kx2 + 3x + 1 and f′′(1) = 10 , find the value of k.
14 If f(x) = x3 + ax2 + bx + 1, with f 1′′(− ) = −4 and f′(1) = 4 , find a and b.

15 If y = xex , find d2 y .
dx2
16 Sketch a graph where, for all x:

f(x) < 0 , f′(x) > 0 , f′′(x) < 0
f 0 f 0 f 0 f 017 Sketch a graph where, for all positive x, ′(x) > and ′′(x) > and, for all negative x, ′(x) > and ′′(x) < .
y x kx 4x 718 a The graph of = 3 − 2 + + is concave-up for all x > 1 and this is the only section which is concave-up.
Find the value of k.

b What can be said about the graph at x = 1?

19 If f(x) = xn and f′′(1) = 12, find the possible values of n.
y x bx c20 The curve = 2 + + has a minimum value at (1, 2). Find the values of b and c.
y x bx c21 The curve = 4 + + has a minimum value at (1, −2). Find the values of b and c.
y x x22 Find and classify the stationary point on the curve = − 2 .
y x bx c23 Show that the graph of = 2 + + is always concave-up.

24 If y = sin3x + 2cos3x , show that d2 y = −9 y .
dx2
d2 y
25 If y = ln(a + x) , find dx2 . Hence show that the graph of y = ln(a + x) is always concave-down.

y x ln x26 a Find the point on the graph = 2 where the second derivative equals 1.

b Given that the graph has a point of inflection, find its x-coordinate.

y x 5x 4x 227 a Find the interval in which the gradient of the graph = 3 − 2 + − is decreasing.

b Hence find the point of inflection on this graph.

f ln28 x x
The function f is defined by (x) = x for > 0.

a Find f ′(x) .

y xb Find the coordinates of the stationary point on the curve = f ( ) .

c Show that f′′ (x) = 2ln x − 3 .

x3

d Hence classify the stationary point.

y 1x p x 029 Find and classify the stationary points on the curve = + 2 with p > , giving your answer in terms of p.

404 10 Differentiation

y e 0 2π30 Find and classify the stationary points on the curve = sin x for < x < .

31 Find and classify the stationary points on the curve y = sin x − x , 0 < x < 2π.
2
y x32 Find and classify the stationary points on the curve = 2 .
1 x+

33 A business produces x thousand of a particular product each month, where x > 0 . The profit P (in thousands of

dollars) is given by P = 16x 3 − x3 − 25.
2

a Find the number of items the business should produce each month to maximise its profit, showing that this is
indeed a maximum.

b Find this maximum monthly profit.

a x34 A cuboid has a square base of side length x cm. The other side of the cuboid is of length − cm. Find the
maximum volume of the cuboid in terms of a, justifying that it is a maximum.

y x35 Find and classify the stationary point on the curve = e px, p ≠ 0, giving your answer in terms of p.
V t 3t 5t36 The volume of water, V litres, in a barrel filled by rain is modelled by = 3 − 2 + , where t is the time in

hours.

Find the time at which the barrel is filling most slowly, and state that minimum rate.

2

P 1 t t37 The population of fish in a lake, P thousands, is modelled by = + 2 , where t is the time in years since

the fish were introduced. At what population is the fish population growing most quickly?

38 Find the range of the function f(x) = ex − ax in terms of a.
y e 6e 4x 839 Find the exact coordinates of the stationary points on the curve = 2x − x + + , classifying them.
h x e e40 The height, h metres of a point on a hanging chain at a given x-coordinate is modelled by ( ) = x + −2x .

Find the exact value of the minimum height of the chain.

y x ax bx 741 The curve = 3 + 2 + + is concave-up for all x > 2 and this is the only section which is concave-up.

a Find the value of a.

b Find a condition on b for the curve to be strictly increasing.

y ax bx42 The curve = 3 − 2 (where a and b are positive) is increasing for x < 0 and x > 4.

a Evaluate b .
a

b Prove that the curve is concave-up for x > 2.

43 If y x′′( ) = 10 , y′(0) = 10 and y(0) = 2, find y (x) .
y x44 The graph below shows the graph of = f ′( ) .

y

x

Checklist 405

On a copy of the diagram
a mark all points corresponding to a local minimum of f(x) with a P
b mark all points corresponding to a local maximum of f(x) with a Q
c mark all points corresponding to a point of inflection of f(x) with a R.
d Are any of the points of inflection stationary points of inflection? Explain your answer.

Checklist

 You should be able to differentiate functions such as x , sin x and ln x:

x xIf f( ) = n, where n ∈ , then f′(x) = nxn−1.
If f(x) = sin x, then f′(x) = cos x.
If f(x) = cos x, then f′(x) = − sin x.
f e f eIf (x) = x, then ′(x) = x.
1
If f(x) = ln x, then f′(x) = .
x
 You should know how to use the chain rule to differentiate composite functions:

If y = f(u), where u = g(x), then

dy dy du
dx = du × dx

 You should know how to differentiate products and quotients:

If y = u(x)v(x), then

dy = u dv + v du
dx dx dx

If y = u(x) , then
v(x)

dy v du u dv
dx dx − dx
=
v2

 You should know how to convert related rates of change.

 You should know how to interpret the second derivative of a function. A function f (x) is

f 0concave-up where ′′(x) >
f 0concave-down where ′′(x) < .

 You should know how to find local minimum and maximum points on a graph.

f 0Given ′(a) = , if
f 0 x a– ′′(a) < , then there is a local maximum at =
f 0 x a– ′′(a) > , then there is a local minimum at = .

406 10 Differentiation

 Mixed Practice
1 The curve y = x2 + bx + c has a minimum value at (2, 3). Find the values of b and c.
e2 Find the derivative of x2.
d2 y 1
3 Find dx2 if y = x 2 .
+

4 Differentiate sin x .
x

5 Find f ′(x) if f(x) = x ln x.

y x 16 Find the gradient of the tangent to the curve = − at x = 5.

y ln x7 Find the equation of the tangent to the curve = at x = 1.

y e8 Find the equation of the normal to the curve = 2x at x = 0.

0.6 s9 The side of a cube is increasing at a rate of cm −1. Find the rate of increase of the volume of the
cube when the side length is 12 cm.
7
10 Given that y = 3sin 2πx , find the rate of change of y when x = 12 .

11 A cuboid has a square base of side a cm and height h cm. The volume of the cuboid is 1000 cm3.
4000
a Show that the surface area of the cuboid is given by S = 2a2 + a .
b Find the value of a for which
dS = 0.
da
c Show that this value of a gives the minimum value of the surface area and find this minimum

value.

12 Find the equation of the tangent to the curve y = x cos x at x = 0.

13 If y = x2e2x, find d2 y .
dx2

14 The two shorter sides of a right-angled triangle are x cm and 6 − x cm.
a Find the maximum area of the triangle.
b Find the minimum perimeter of the triangle.

15 If f(x) = xekx and f′′(0) = 10, find f′(1).

x kx 8x 216 a The graph of 3 − 2 + + is concave-up for all x > 4 and this is the only section which is
concave-up. Find the value of k.
b Hence find the coordinates of the point of inflection on the graph.

17 Differentiate xex sin x .

f18 If (x) = x x , show that f′(x) = ax + b , where a, b and c are constants to be found.
2(4 + x)c
4+

f e19 a Find the set of values for which (x) = x −x is concave-down.
b Hence find the point of inflection on the curve y = f(x).

20 Find the coordinates of the stationary point on the curve y = e−x sin x for 0 < x < π.

Mixed Practice 407

21 Consider z = xey . When x = 2 and y = 1, dx = −1 and dy = 0.5, find dz .
dt dt dt

22 The volume of water, V million cubic metres, in a lake t hours after a storm is modelled by

V = 2te−t + 5.
a What is the initial volume of the lake?

b What is the maximum volume of the lake?

c When is the lake emptying fastest?

23 A tranquilizer is injected into a muscle from which it enters the bloodstream. The concentration
C t 2t ,t 0C in mg l, of tranquilizer in the bloodstream can be modelled by the function
() = +2
3 twhere t is the number of minutes after the injection.

Find the maximum concentration of tranquilizer in the bloodstream.

Mathematics HL November 2014 Paper 1 Q5

20 km h24 Two cyclists are at the same road intersection. One cyclist travels north at −1; the other travels
west at 15km h−1.
Use calculus to show that the rate at which the distance between the two cyclists changes is

independent of time.

Mathematics HL November 2014 Paper 2 Q4

25 Let f(x) = –2x2 x – 2     x  4, x  1 , x         
2
+5x

y

4

3

2

1 A (1, 1) 3 4 x
−2 −1 1 2
−1

−2

−3

−4

The graph of f has a local minimum at A (1, 1) and a local maximum at B.

a Use the quotient rule to show that f′(x) = 2x2 – 2 2)2.
(–2x2 +5x –
b Hence find the coordinates of B.

c Given that the line y = k does not meet the graph of f, find the possible values of k.

Mathematics SL May 2012 Paper 1 TZ2 Q10

408 10 Differentiation

y ln x kx26 Prove that the equation of the tangent to the curve = + at x = 1 always passes through the

point (0, −1) independent of the value of k.

27 Let f(x) = (1 100 x ) . Part of the graph of f is shown below.
50e–0.2
+

y

50 x

a Write down f(0).

b Solve f(x) = 95.

c Find the range of f.
1000e–0.2x
d Show that f′(x) = (1 + 50e–0.2x)2 .

e Find the maximum rate of change of f.

Mathematics SL May 2013 Paper 2 TZ1 Q9

28 The function f(x) = x − a is defined for x ≠ b.
x − b
a b f 0a Show that if > , then ′(x) > at all values of x in the domain.
p q p qb Does this mean that if > , then f( ) > f( )? Justify your answer.

y 4 x29 a State the largest possible domain for the function = − 2 .

y 4 xb Prove that, for any point on the graph of = − 2 , the normal passes through the origin.

Mixed Practice 409

y kx30 The graph = n has the property that, at every point, the product of the gradient and the y value
equals 1.

a Find the value of n.
b Find the possible values of k.

31 A bird is flying at a constant speed at a constant height of 35 m in a straight line that will take it
directly over an observer at ground level. At a given instant, the observer notes that the angle θis
1
1.2 radians and is increasing at 60 radians per second. Find the speed, in metres per second, at which

the distance between the bird and the observer is decreasing.

Bird

35 m



Observer

11 Integration

ESSENTIAL UNDERSTANDINGS

 Calculus describes rates of change between two variables and the accumulation of limiting areas.
 Understanding these rates of change and accumulations allows us to model, interpret and analyse

real-world problems and situations.
 Calculus helps us to understand the behaviour of functions and allows us to interpret features of

their graphs.

In this chapter you will learn…

 how to integrate functions such as x , sin x and ex
 how to integrate by inspection
 how to find definite integrals

 how to link definite integrals to areas between a curve and the x-axis, and also between two curves

 how to use integration to find volumes of certain solids.

CONCEPTS

The following concepts will be addressed in this chapter:
 The area under a function on a graph has a meaning and has applications in space

and time.

In Chapter 12, you will learn how to apply integration to one of the most important
topics in physics – the study of motion.

 Figure 11.1 What is being accumulated in each photograph?

Integration 411

PRIOR KNOWLEDGE

Before starting this chapter, you should already be able to complete the following:

1 Find y if dy = x2 + 2 and y = 4 when x = 0.
dx
y x2 Use technology to evaluate the area between the curve = 4 and the x-axis between x = 1 and x = 3.
dy
3 If y = e2x + sin x + 2, find dx .

f 34 If (x) = x2 + , find f ′(x).

In the Mathematics: applications and interpretation SL book, you saw that the
operation which reverses differentiation is called integration, but that surprisingly this
has applications in evaluating areas and accumulations. Now that you know how to
differentiate more functions, it is natural to extend your knowledge of integration to
reverse these new derivatives. We can then apply these new integrals to more complex
situations, including finding areas and volumes of various shapes.

Starter Activity

Look at the photos in Figure 11.1 and discuss the following questions. What rate is being shown? What is being
accumulated?

Now look at this problem: Which of the following can you evaluate? a+1

If you know that a ∫ f(x − 1) dx

a ∫f(x − 1) dx 1

∫f(x) dx = A 0 a
2
0 a
∫f(2x) dx
∫f(2x) dx
0
0

412 11 Integration

11A Further indefinite integration

 Extending the list of integrals

In Chapter 10, you saw that the derivative of xn with respect to x is nxn−1 for any
n ∈
. This means that xn+1 differentiates to (n + 1) xn so n 1 1 x n+1 differentiates to

+
xn as long as n ≠ −1 (to avoid dividing by zero). Since integration is the reverse of

differentiation, this means the following.

KEY POINT 11.1

∫xn dx = x n+1 + c for n ≠ −1
n+1

You saw a similar rule in Chapter 10 of the Mathematics: applications and
interpretation SL book, but we have now extended it to all rational values of n other
than −1. Remember that we always need a ‘+ c’ when integrating since this will

disappear when it is differentiated.

WORKED EXAMPLE 11.1 ∫ 3x 1 dx
2
Find ∫ 3 x dx.
= 3∫ 1 dx
Rewrite the integral to make x2
it explicitly include xn
= 3x(21+1) + c
The constant factor can be 1
taken out of the integral 2 + 1

Apply Key Point 11.1 = 3x 3 + c
3 2
Simplify

2

= 2x 3 + c
2

You also found, in Chapter 10, that sin x differentiates to cos x when x is measured in
radians. This can be reversed to turn this statement into an integral.

KEY POINT 11.2

∫ cos x dx = sin x + c

Since cos x differentiates to −sin x, it follows that −cos x differentiates to sin x. Reversing
this result gives an integral.

11A Further indefinite integration 413

Tip KEY POINT 11.3

Because the rules ∫ sin x dx = −cosx + c
of differentiating
and integrating Since tan x differentiates to 1 x, this also can be reversed.
trigonometric functions cos2
are so similar they are
often confused. Make KEY POINT 11.4
sure you remember
which way round they 1 dx tan x c
work. cos2

∫ x = +

You also saw, in Chapter 10, that ex differentiates to ex . This statement can be reversed.

KEY POINT 11.5

∫ ex dx = ex + c

∫You 1 ln x. 1 dx = x + c.
know that differentiates to This suggests that ln However,
x x
there is a small complication that you need to consider. This integral can be linked to

the area between the curve and the x-axis and this area exists for negative values of x

even though ln x is not defined in this region:

y

y = 1x
B
−2 −1 12 x
A

This fills y xWe can use the symmetry of the graph of = 1 to see that that area A and area B are

in the hole equal. This leads to the following interpretation.

in Key

Point 11.1 –

xintegrating n where

n = −1.

KEY POINT 11.6

∫ 1 dx = ln x + c
x

414 11 Integration

WORKED EXAMPLE 11.2

If dy = 3ex + cos x + 2, find a general expression for y.
dx

We reverse differentiation y = ∫ 3ex + cos x + 2 dx
using integration = 3∫ ex dx + ∫ cos x dx + ∫ 2 dx
= 3ex − sin x + 2x + c
We can split up the sum
into separate integrals and

take constant factors out

The integrals can be
evaluated using the

key points above

 Integrating f(ax + b)

From the chain rule, you know that when differentiating f(ax + b) you get af′(ax + b).

In words, this says that you can differentiate the function as normal and then multiply

by a. When integrating, we can reverse this logic to say that when integrating f(ax + b)
we can integrate as normal and then divide by a.

KEY POINT 11.7

If ∫ f(x)dx = F(x) , then ∫ f(ax + b) dx = 1 F(ax + b).
a

WORKED EXAMPLE 11.3 ∫ sin (5x + 2) dx = − 1 cos(5x + 2) + c
5
Find ∫ sin(5x + 2) dx.

We can identify f(x) = sin x.
This integrates to − cos x

Key Point Tip
11.7 is a
special In Worked Example 11.3, you might have wondered whether you needed to divide the constant
case of Key Point by 5 too. However, c is just another constant so it actually does not matter.
11.8 (slightly
rearranged) with 5

g (x) = ax + b. You  Integration by inspection

will meet a further The chain rule says that the derivative of a composite function, f(g (x)) is g′(x) f′(g(x)).
generalization,
called integration This is a very common type of expression to look out for – a composite function
by substitution, in multiplied by the derivative of the inner function – because it can easily be integrated.
Section 11D.
KEY POINT 11.8

∫ g′(x)f′(g(x)) dx = f(g(x)) + c

11A Further indefinite integration 415

Tip WORKED EXAMPLE 11.4 ∫ 2xe(x2 ) dx = e(x2 ) + c

You can check Find ∫ 2x e(x2 ) dx .
your answer by
differentiating the f e g′(x) = x and (x) = x2. The
result. composite function is
multiplied by 2x which
is g′(x) so we can apply
Key Point 11.8 directly

Sometimes the expression multiplying the composite function is not exactly the
derivative of the inner function, but is within a constant factor of it. In this situation,
we can introduce the required factor in the integral, but to keep the value unchanged
we need to divide by the same factor too.

WORKED EXAMPLE 11.5

Find ∫ x 4 dx.

3x2 −

We can identify f′(x) = 1 ∫ x 4 dx = 1 ∫ 6x 4 dx
= 3x2 − 4x; 6 3x2 −
and g(x) 3x2 −

g 6therefore ′(x) = x. To get
this, we need to introduce

an extra factor of 6, which
1
must be balanced by 6

The integral is now of 1 ln 3x2 4 c
6
the required form to = − +

apply Key Point 11.8. We

can use the fact that the
1 is ln x +c
integral of x

Tip

Worked Example 11.5 makes use of the very common integral

∫ f′ (x) dx = ln f(x) +c
f(x)

This integral occurs so often you might like to learn this result.

42 TOOLKIT: Problem Solving
1
∞ Consider ∫ 2x dx. Using the method of Worked Example 11.5, we can make the top

3

π924

of the fraction the derivative of the bottom:

∫ 1 dx = 1 ∫ 2 dx = 1 ln 2x + c
2x 2 2x 2
to 1
Alternatively, we can use algebra take out a factor of 2 from the integral.

∫ 1 dx = 1 1 dx = 1 ln x + c
2x 2∫ x 2

Which of these two solutions is correct?

416 11 Integration

Be the Examiner 11.1

Find ∫ sin x cos x dx.

Which is the correct solution? Identify the errors in the incorrect solutions.

Solution 1 Solution 2 Solution 3

Since d (sin x) = cos x ∫ sin xcos x dx Since d (cos x) = − sin x
dx dx
1
∫ (sin x)1 cos x dx = 2 ∫ sin 2x dx −∫ − sin x(cos x)1 dx

= sin2 x + c = − 1 cos 2 x + c = − cos2 x + c
2 4 2

Exercise 11A

For questions 1 to 6, use the method demonstrated in Worked Example 11.1 to find the following integrals.

1 a ∫ x 2 dx 2 a ∫ 1 dx 3 a ∫ 10x23 dx
3 x−2

b ∫ 3 dx b ∫ 4 dx b ∫ 5x41 dx
x4 x−3

4 a 4 x − 2 dx 5 a 1 3 x dx 6 a 6 dx
3 2
∫ ∫ ∫ 4x

b ∫ 3x− 2 dx b ∫ 1 5 x dx b ∫ 7 dx
5 3
x

For questions 7 to 10, use the method demonstrated in Worked Example 11.2 to find an expression for y.

7 a dy = 3cos x 8 a dy = −2 sin x
dx dx

b dy = − cos x b dy = 1 sin x
dx dx 2

9 a dy = 5ex 10 a dy 2
dx dx = x

b dy = − 4 ex b dy 1
dx 3 dx = 2x

For questions 11 to 16, use the method demonstrated in Worked Example 11.3 to find the following integrals.

11 a ∫ 2x + 1 dx 12 a ∫ 3 1 5x dx 13 a ∫ cos(2 − 3x) dx

−

b ∫ 3 1 − 2x dx b ∫ 4 1 7 dx b ∫ cos(4x + 3) dx
2x −
⎛1 1
14 a ∫ sin ⎝2 x − 5⎞ dx 15 a ∫ e5x+2 dx 16 a ∫ 4x − 5 dx
⎠

b ∫ sin(5 − 2x) dx b ∫ e1−3x dx b ∫ 3 1 dx
− 2x

For questions 17 to 19, use the method demonstrated in Worked Example 11.4 to find the following integrals.

17 a ∫ 2x(x2 + 4)3 dx 18 a ∫ cos xsin3 x dx 19 a ∫ 3x2 + 2 dx
b ∫ 4x3 (x4 − 2)23 dx b ∫ − sin x cos2 x dx x3 + 2x
4x3 − 5
b ∫ x4 − 5x dx

11A Further indefinite integration 417

For questions 20 to 22, use the method demonstrated in Worked Example 11.5 to find the following integrals.

20 a ∫ x2 x3 + 4 dx 21 a ∫ 4 ex −x2 dx 22 a ∫ x2 5 dx
b ∫ x 3 1 − x2 dx b ∫ 9x2ex3 dx x3 +

b ∫ x3 3 dx
x4 −

23 Given that dy = 3 x 12and that y = when x = 9, find an expression for y in terms of x.
dx
24 Function f satisfies f′(x) = 2cos x − 3sin x and f(0) = 5. Find an expression for f (x).

25 Given that dy = 2ex − 5 , and that y = 0 when x = 1, find an expression for y in terms of x.
dx x

26 Find ∫ 4x2 + 3 dx .
2x
dV 1
27 Given that dt = t − 2 sin t, and that V = 2 when t = 0, find an expression for V in terms of t.

28 It is given that x (0) = 5 and that dx = 5 − 2et. Find an expression for x in terms of t.
dt
29 Find ∫ (5 − 2x)6 dx.

30 Find ∫ (3sin(2x) − 2cos(3x)) dx.

31 Find ∫ 2x(x2 + 1)5 dx .
dy
32 Given that dx = x sin(3x2) , find an expression for y in terms of x.

33 Find ∫ 3x dx.
x2 + 2

f 2 1 3 y x 1, 134
Given that ′(x) = x+ )2 , and that the graph of = f ( ) passes through the point (− ), find an expression for f (x).
(

35 The curve y =f (x) passes through the point (1, 1) and fhas gradient ′(x) = 2ln x . Find an expression for f (x) .
36 Given that f′ (x) value of f(−3). 3x
2 f(−1) 5,
= and that = find the exact
x

37 Find ∫ 4cos3 x sin x + kx2 + 1 dx, where k is a constant.

38 e e e∫Find 2x − −2x dx .

x

d39 Find ∫ x4 + x2 x for x > 0 .
40 a Express cos2x in terms of sin x.

b Hence find ∫ sin2 x dx .
41 Find ∫ tan x dx, showing all your working.

42 Find ∫ sin3 x dx.

43 Find ∫ 1 dx for x > 0.
x ln x

418 11 Integration

11B Definite integrals and the area

between a curve and the x-axis

In Chapter 10 of the Mathematics: applications and interpretation SL book, you saw

that we can represent the area under a curve y = f(x) between x = a and x = b (called
b

∫the limits of the integral) using f(x) dx and you evaluated this using technology.
a

In this section, you will see how this can be evaluated analytically.

It turns out the definite integral is found by evaluating the integral at the value of the
upper limit and subtracting the value of the integral at the lower limit. This is often
expressed using ‘square bracket’ notation.

KEY POINT 11.9

b

∫f′(x) dx = [f(x)]ba = f(b) − f(a)

a

It is not at all obvious that the process of reversing differentiation has anything to
do with the area under a curve. This is something called the ‘fundamental theorem
of calculus’ and it was formalized by English mathematician and physicist Isaac Newton
and German mathematician and philosopher Gottfried Leibniz in the late seventeenth
century. Although, today, most historians agree that the two made their discoveries about
calculus independently, at the time there was a great controversy over the question of who
developed these ideas first, and whether one had stolen them from the other.

Proof 11.1 b

∫Prove that the area under the curve y = f(x) is f(x) dx.
a

Consider a general representation of y = f(x). y

The increase in the area when b is changed

is approximately a rectangle of height f(b)

Area = A ∆ A f(b)

a b∆b x

The change in area, ΔA, is approximately
the red rectangle, so ΔA = f(b) Δb.

So ΔA = f(b). In the limit as Δb gets very
Δb
this expression becomes dA
small, db .



420 11 Integration

TOK Links

e−x2 is an example of a function which you might be told cannot be integrated. This really

means that the indefinite integral cannot be written in terms of other standard functions, but

who decides what is a ‘standard function’? The area under the curve does exist, so we could

(and indeed some mathematical communities do) define a new function that is effectively the

2 erfarea under this curve from 0 up to x (if you are interested it is π (x)). Does j ust giving the

answer a name increase your knowledge?

 Area between a curve and the x-axis

Definite integrals give the area between a curve and the x-axis when the function
between the limits is entirely above the x-axis. If the curve is below the x-axis, then

the integral will give a negative value. The modulus of this value is the area.

WORKED EXAMPLE 11.8

Find the area enclosed by x = −1, x = 1, y = 3x2 − 4 and the x-axis.

It is always useful to y
sketch the graph to
visualise the area

−1 1 x

We need to evaluate 1 = ⎡ − ⎤11
an appropriate
∫3x − = (1 − 4) − ( 1 4( ))
definite integral 3 3= ( ) ( )
−1
The area is the modulus = −6
Tip of this answer
Area = −6 6
If you are not told any
units for x and y, then
you do not give units
in the answer.

11B Definite integrals and the area between a curve and the x-axis 421

Sometimes areas are partly above and partly below the x-axis. In that case, you have to
find each part separately.

WORKED EXAMPLE 11.9 y y = x2 − 2x

Find the shaded area on the curve of y = x2 − 2x.

Tip 4x

A common mistake in First find where it goes from It crosses the x-axis when
above the axis to below the axis x− =
problems like that in
Split the integral into two 0x x − =
Worked Example 11.9 parts. Consider first the
region below the x-axis So, it crosses when x = 0 and x = 2
is t4o think that the area
Then the second region
is ∫x2 − 2x dx , which 2 4 0x x ⎡x3 2
106 Combine the two areas = − x2⎤ = − −
3 ∫x 3 30
is . This is the ‘net
0 4 4
−3 3
area’ – how much more So, the first area is =

is above the x-axis than ( )4

below – and this is ∫x 4 4 20
x2⎤ 3=3
2
sometimes very useful, 16x x = ⎡x3 − = −
3 3⎣ 2
but it is not the answer
4 20
to the given question. Total area is 3 + 3 = 8

x a x b y f xWith a calculator, you can find the total area between = , = , the curve = ( )

and the x-axis using the formula in Key Point 11.10.

KEY POINT 11.10

b

Area = ∫a f(x) dx

Tip

Check that Key Point 11.10 gives the correct answer to Worked Example 11.9.

422 11 Integration

WORKED EXAMPLE 11.10

y x xFind the area between the curve = 3 − and the x-axis between x = −1 and x = 1.

Write the required area in terms of a 1
definite integral from Key Point 21.9
∫Area = x x x
Evaluate using technology 1

= 0.5 from the GDC

42 ProofTOOLKIT:

∞ Justify each of the following rules for definite integrals:

π3 ba

924 ∫ ∫f(x) dx = − f(x) dx

ab
bc c

∫ ∫ ∫f(x) dx + f(x) dx = f(x) dx

aba

bb b

∫ ∫ ∫f(x) dx + g(x) dx = f(x) + g(x) dx

aa a

Exercise 11B

questions 1 to 6, use the in Example 11.6 to f

41 x4
a x2 x
For

∫ ∫ ∫d d d1
method demonstrated1 Worked ind the given definite integrals.
x
2a −1
3 a x2 x

1 −2 −3
8 0 −2
1
∫b x3 dx ∫b x5 dx ∫b x3 dx

0 −1 −4

π ln 3 5 1

2 ∫5 a ex dx a∫6 x dx
0
∫4 a sin x dx 1
π
3
π
ln 7 3 1
2
∫b ex dx ∫b x dx
∫b cos x dx ln 2
0 2

11B Definite integrals and the area between a curve and the x-axis 423

For questions 7 and 8, use the method demonstrated in Worked Example 11.7 to find the given definite integrals.

3 8 a 3 1 dx
ln x
7 a ∫ sin(x2) dx ∫

0 2

69

b ∫ cos(x3) dx b ∫ ln(ln x) dx
14
For questions 9 and 10, use the method demonstrated in Worked Examples 11.8 and 11.9 to find the shaded areas.
9a y
b y y = x2 − 1

y = x2 − 2x

xx

y10 a by

y = x3 − 3x2

4x −2 y = x − x3 x

For questions 11 and 12, use the method demonstrated in Worked Example 11.10 to find the shaded areas.
by
11 a y

−1 2 x

y = x4 − 2x3 − x2 + 2x −2 2 x
y = −x4 − x3 + 4x2 + 4x

424 11 Integration

y12 a by
y = sin x − cosx
y = sin x + cosx

πx 43π x

5

∫13 Evaluate 3x + 1 dx.
y 3sin2x1
crosses the x-axis at x = 0 and x = π . Find the exact area enclosed by this
14 The curve with equation =
2
part of the curve and the x-axis.

y ln x15 Find the area enclosed by the graph of = , the x-axis and the lines x = 2 and x = 5.
a
3 dx 24.
Find the value of a such that x
∫16 =

1
y 9 x17 a Find the coordinates of the points where the curve = − 2 crosses the x-axis.

b Find the area enclosed by the curve and the x-axis. y

18 The graph of y = x3 − 2x2 − x + 2 is shown in the diagram.

12

∫ ∫a Evaluate y dx and y dx.
−1 1

b Hence find the total area of the shaded region.

−1 1 2 x

19 The graph of y = ln (10 − x2 ), shown in the diagram, crosses the x-axis y
at the points (−3, 0) and (3, 0). Find the shaded area. ln 10

−3 3 x

11B Definite integrals and the area between a curve and the x-axis 425

ln 3 πx

∫20 Find the exact value of 2e−3x dx.0 2

−3 is independent of k.

∫ 5 d21 Find the exact value of −9 x x . 2k π y

1 dx22 Show that the value of the integral 2

∫ xk
 23 The diagram shows the graphs of y = sin x and y = cos x for 0 x
.

a Write down the coordinates of the point of intersection of the two

graphs.

b Find the exact area of the shaded region.

 24 a On the same axes, sketch the graphs of y = cos(x2 ) and y = 2 − 2sin(x2 ) for 0x π .

b Find the area enclosed by the two graphs. 2

 25 Show that the area enclosed by the graph of y = sin 2x and the x-axis for 0 x π equals 2.

3

∫ 1 d26 Find the exact value of x x2 + x. y
0

y y e27 The curves shown in the diagram have equations
= ex and = −2x.

Find the area of the shaded region.

1

4 −2 1 x
1
28 1 dx.
2x +
∫ ∫29
Find the exact value of 3

Evaluate 3 x 1 5 dx .
2
−

30 Solve for a

a

∫x2 + x dx = 90

0

426 y 11 Integration
A
31 Solve for a B
x
2a

∫x + 1dx = 8

a

y e y32 The curves with equations = −x2 and = e−x are shown in the diagram.

a Find the coordinates of the points A and B.
b Find the area enclosed by the two curves.

55 y
a bc
33 Given that ∫f(x) dx = 10, evaluate ∫(3f(x) − 1) dx.

22

63

34 Given that ∫f(x) dx = 7 , evaluate ∫5f(2x) dx.

00

ln35 a Differentiate x x.

e

b Hence evaluate ∫ ln x dx .

1

y x36 The diagram shows the graph of = f ( ) .

dd d c

Given that ∫f(x) dx = 1, ∫f(x) dx = 5 and ∫ f(x) dx = 17, find ∫f(x) dx .
ba a a

d x

You are the Researcher

1

4 1 dUse technology to evaluate ∫ − x2 x. What do you notice about the answer?
0

Justify your answer (i) graphically, (ii) using the substitution x = sin t. This is an

example of an integral which can be done analytically but is not of the form from

Key Point 11.8. Can you explain why the substitution x = sin t works here? Can you

find any other integrals which need surprising substitutions?

11C Further geometric interpretation of integrals 427

11C Further geometric interpretation
of integrals

 Area of the region enclosed by a curve and the y-axis
b

f x dxYou already know that the area between a curve and the x-axis is given by ∫a ( ) .

The diagram below shows the area between a curve with equation y = f(x) and the

y-axis.

y

Ûy = f(x) x = g(y)

d

c
x

If you imagine swapping the x- and the y-axes, you can find the area by using integration,
as before. However, the equation needs to be for x in terms of y, that is, in the form
x = g(y), and the limits need to be the y-values.

KEY POINT 11.11

The area bounded by a curve x = g(y), the y-axis and the lines y = c and y = d is given by
d

∫c g(y) dy

428 11 Integration

WORKED EXAMPLE 11.11

  Find the area enclosed by the curve y = x, the y-axis and the lines y = 0 and y = 1.

A sketch is always helpful y
to make sure that you are

finding the correct area

1

x

Rearrange the equation to y =  x so x = e y
express x in terms of y
1
d
∫Area = ey dy
∫Area = g(y ) dy . The limits 0

for y are cgiven in the question = 1.72 (3 s.f.)

Evaluate the integral
using your GDC

 Volumes of revolution about the x-axis or y-axis

When a part of a curve is rotated 360° about the x-axis (or the y-axis), it forms a shape
known as a solid of revolution. The volume of this solid is a volume of revolution.

KEY POINT 11.12

The volume of revolution formed when the part of the curve y = f(x), between x = a and x = b,
b

∫Vis rotated around the x-axis is given by = πy 2 dx.
a



430 11 Integration

You are the Researcher

There are also formulae to find the surface area of a solid formed by rotating a
region around an axis. Some particularly interesting examples arise if we allow
one end of the region to tend to infinity. For example, rotating the region formed

y xby the lines = 1, x = 1 and the x-axis results in a solid called Gabriel’s horn, or

Torricelli’s trumpet, which has a finite volume but infinite surface area!

When a curve is rotated around the y-axis, you can obtain the formula for the resulting
volume of revolution simply by swapping x and y.

KEY POINT 11.13

The volume of revolution formed when the part of the curve y = f(x), between y = c and y = d,
d

∫Vis rotated around the y-axis is given by = πx2 dy .
c

Notice that, to use this formula, you need to write x in terms of y and find the limits on
the y-axis.

WORKED EXAMPLE 11.13

Find the volume of revolution when the curve y = x2 − 1, 1 < x < 5, is rotated around the
y-axis. Give your answer in terms of π.

You need to express y = x2 − 1, so x2 = y + 1
x2 in terms of y
Limits:
Find the limits for y
when x = 1, y = 0
The volume is given when x = 5, y = 24

24 24

∫by πx2 dy ∫V = π(y + 1) dy
0 0
= 312π
Evaluate the integral
on your GDC

You are the Researcher

An alternative formula for the volume of revolution when y = f(x), a < x < b, is
b
∫ 2πxy dxrotated around the y-axis is given by
. Can you j ustify this and find any

applications? a

11C Further geometric interpretation of integrals 431

Exercise 11C

For questions 1 to 4, use the technique demonstrated in Worked Example 11.11 to find the area between the given curve,

the y-axis and the lines y = c and y = d.

1a y y =x by
3

y = x3

8

1

1 x
x

2a y b y
y = e2x y = e3x

5

4

2

3a y y = x 1  x 1 x
by x
y = x22

6

2 4

1
x

432 yb 11 Integration
y = tan x
4a y
4 y = 3sin x

2
2

1

x

x

For questions 5 to 8, use the technique demonstrated in Worked Example 11.12 to find the volume of revolution formed
when the given part of the curve is rotated 360° about the x-axis. Give your answers to three significant figures.

5 a y = 3x2 between x = 0 and x = 3 6 a y = x2 + 3 between x = 1 and x = 2

b y = 2x3 between x = 0 and x = 2 b y = x2 − 1 between x = 2 and x = 4
2
7 a y = e2x between x = 0 and x = ln 2 8 a y = x +1 between x = 1 and x = 3
b y = e3x between x = 0 and x = ln 2
y = 3 between x = 0 and x = 2
b x 2
+

For questions 9 to 12, use the technique demonstrated in Worked Example 11.13 to find the volume of revolution formed

when the given part of the curve y = g(x), for x < a < b, is rotated 360° about the y-axis.

9 a g(x) = 4x2 + 1, a = 0, b = 2 10 a g(x) =  x + 1, a = 1, b = 3
x2 − 1,
b g(x) = a = 1, b = 4 b g(x) = ln (2x − 1), a = 1, b = 5
a 3 a b
b g(x) = cos x, a = 0, b = π g(x) 1 5, a 6, b 8
11 = 0, = π2 12 a = x = =
g(x) 4 −
1
= tan x, b g(x) = x 2, a = 3, b = 8
−

13 The shaded region in the diagram is bounded by the curve y

y = 1x , the y-axis and the lines y = 1 and y = 5.

a Find the area of the shaded region.

b Find the volume of the solid generated when the 5
shaded region is rotated about the y-axis.

1 y = 1x

x

11C Further geometric interpretation of integrals 433

y14 The part of the curve with equation = 1 between x = 1 and x = a is rotated 360° about the x-axis. The volume of
23π xthe resulting solid is . Find the value of a.
15 The part of the parabola y = x2 between x = 0 and x = a is rotated about the y-axis. The volume of the resulting

solid is 8π. Find the value of a. y

y x16 The diagram shows the curve with equation = + 3.

a Write down the x-coordinates of point A. y = x + 3

b The region bounded by the curve, the x-axis and the

line x = 3 is rotated completely about the x-axis.

Find the volume of the resulting solid.

17 The diagram shows the region bounded by the y-axis and the A x
3
curve with equation x = 4 − y2. y
2 x = 4  y2
Find
x
a the area of the region
y =  x sin x
b the volume of the solid generated when the region is
rotated about the y-axis.

18 The diagram shows the region bounded by the curve 

y = x sin x and the x-axis. y

Find

a the area of the region

b the volume of the solid generated when the region is

rotated through 2π radians about the x-axis.

0 x

434 11 Integration

y x 919 The curve in the diagram has equation = 3 + and y = x3+ 9 y
9, 0intersects the coordinate axes at the points A(− 3 ) and B

B(0, 3). Region R is bounded by the curve, the x-axis and
the y-axis.

x y 9a Show that 2 = 3 ( 2 − )2 .

b Find the area of R.

c Find the volume of revolution generated when R is
rotated fully about the

i x-axis

ii y-axis.

x
A

y x20 The part of the curve = 1, between x = 1 and x = 3, is rotated about the y-axis. Find the volume of the
resulting solid.

y x21 The diagram shows the part of the curve = − 4 between y

x = 4 and x = 20.

The shaded region bounded by the curve, the line x = 20 and the x-axis y =x  4

is rotated 360° about the y-axis. Find the resulting volume

of revolution. Give your answer to the nearest integer.

x
4 20

22 The part of the curve y = sin x between x = 0 and x = π is rotated around the x-axis. Find the exact value of the

volume generated.

y x23 a Sketch the curve with equation = .

b The part of the curve between x = 0 and x = 9 is rotated about the x-axis. Find the volume of the resulting solid.

c Find the volume of the solid generated when the same part of the curve is rotated about the y-axis.

24 The diagram shows the graph of y = (2 − x)ex. The region y

R is bounded by the curve and the coordinate axes.

a Find the coordinates of the points A and B. y = (2 x)ex
b Find the area of R.

c Find the volume of the solid generated when R is rotated A

360° about the x-axis.

R
Bx

11C Further geometric interpretation of integrals 435
y
25 yThe region between the curve = x2 + 1 and the line y = 3,
shown in the diagram, is rotated fully about the y-axis. y=3
Find the volume of the resulting solid, giving your answer
    

y = x2 + 1 x

26 The diagram shows the region bounded by the curve with y
x=2
equation x = 3 − (y − 1)2 and the line x = 2.

The region is rotated around the y-axis. Find the volume of

the resulting solid, giving your answer as a multiple of π.

x = 3  (y 1)2

27 The diagram shows the region bounded by the graph of y x
ln a y = ln x
y = ln x, the coordinates axes and the line y = ln a.

a Find, in terms of a, the area of the shaded region.

a

∫b Hence show that ln x dx = a(ln a − 1) + 1.
1

x

28 a Find the equation of the line passing through the points (r, 0) and (0, h), giving your answer in the

form ax + by = c.

b By considering the solid formed when a part of this line is rotated about the y-axis, show that the volume
1
of a cone with radius r and height h is given by 3 πr2h.

436 11 Integration

29 a Write down the equation of a circle with centre at the origin and radius r.

43 πb By considering a suitable solid of revolution, prove that the volume of a sphere with radius r is r3.
y x30 a Find the coordinates of the intersection points of the curves y = x2 and = .

b Find the volume of revolution generated when the region between the two curves is rotated about the x-axis.

31 The region shown in the diagram is bounded by the x-axis y

and the curves y = sin x and y = cos x.

Find the exact volume of the solid generated when the region y = sinx y = cosx
is rotated around the x-axis.

x

32 a Find the points of intersection of the curve y = x2 and the line y = 2x.

b The region bounded by the line and the curve is rotated fully about the y-axis. Find the volume of the
resulting solid.

33 The diagram shows the graph of y = x3 and two regions, A and B.
y

y = x3

B

A x
pq

Show that the ratio (area of A) : (area of B) is independent of p and q, and find the value of this ratio.

34 a The graph of y = ln x is translated 2 units to the right. Write down the equation of the resulting curve.

b Hence find the exact volume generated when the region bounded by x = 1, y = 1 and the curve y = ln x,

 for 1 x e, is rotated around the line x = −2.
 35 a Sketch the graph of y = cos x for −π x π.

The curve from part a is rotated through 2π radians about the line y = −1.

b Show that the volume of the resulting solid is given by
π
∫1π −π (cos 2 x + 4 cos x + 3) dx

2

c Find the exact value of the volume.

11D Integration by substitution 437

11D Integration by substitution

Note that this is an optional section. All the questions in this section can be done
using Key Point 11.8, but some people find this method easier.

Instead of using Key Point 11.8, you can use a substitution to find integrals.

This involves using a substitution u = f(x) to replace all instances of x and dx in the

integral, turning it into one which is easier to evaluate.

Generally, if trying to integrate ∫ kg′(x) f′(g(x)) dx , the substitution u = g (x) will work.

WORKED EXAMPLE 11.14

Evaluate ∫ x sin(x2) dx using the substitution u = x2.

Use the given substitution If u = x2 , then du = 2 x
dx
du So, du = 2x dx
to find dx and rearrange

it to find dx dx = du
2x

Replace all instances ∫ x sin (x 2 ) dx = ∫ x sin (u) du
of x2 and dx 2x

The factor of x which = ∫ sin u du
was causing problems 2
can now be cancelled,
leaving something which = − 1 cos u + c
2
can be integrated
1
Perform the integration 2
and then replace u

using the substitution

= − cos (x 2 ) + c

TOK Links
uxTechnically,
d is not a fraction, so splitting it up as shown in Worked Example 11.14 is
d

frowned upon in formal mathematics (although it will work for all the integrals you will meet

in this course). Is analogy a valid way of arguing in mathematics?

438 11 Integration

Exercise 11D

For questions 1 to 5, use the method demonstrated in Worked Example 11.14 to evaluate the given integral using the

given substitution. 1 ex
− 5x + ex
1 a ∫e2x dx, u = 2x 2 a ∫ 2 dx, u = 2 − 5x 3 a ∫ 1 dx, u = 1 + ex
b ∫ cos(5x) dx, u = 5x
b ∫ e1+4x dx, u = 1 + 4x b ∫ 1 x2 dx, u = 1 + x3
+ x3
5 a ∫ sin x x dx, u = x
4 a ∫ x x2 + 2 dx, u = x2 + 2

b ∫ cos x sin x dx, u = sin x b ∫ cos (ln x) dx , u = ln x
x

For questions 6 to 10, use a substitution to evaluate the given integral.
1
6 a ∫ sin(4x) dx 7 a ∫ 3 + 4x dx e d∫8 a x2 3(x ) x

b ∫ e3x dx b ∫ 2 1 x dx b ∫ x4 cos x5 dx

−

9 a ∫ 1 x dx 10 a ∫ sin x cos x dx

+ 3x2

b ∫ 1 cos x x dx b ∫ tan x dx
+ sin
u x11 Use the substitution = + 1 to integrate

∫ x3 + x2 dx

for x > 0 .

12 Use the substitution x = sinu to evaluate
1
∫ 1− x2 dx

Checklist 439

Checklist

sin You should be able to integrate functions such as x , x and ex:

∫ xn dx = x n+1 + c for n ≠ −1 ∫ cos x dx = sin x + c ∫ sin x dx = −cos x + c
n+1

∫ ex dx = ex + c ∫ 1 dx = ln x +c
x
 You should be able to integrate by inspection or substitution:
1
If ∫ f(x) dx = F (x) , then ∫ f(ax + b) dx = a F (ax + b) .

∫ g′(x) f′(g(x)) dx = f(g(x)) + c

 You should be able to find definite integrals:

b

∫g′(x) dx = [g(x)]ba = g(b) − g(a)

a

 You should be able to link definite integrals to areas between a curve and the x-axis:

b

Area = ∫ f(x) dx

a

x y You should be able to find the area bounded by a curve = g( ), the y-axis and the lines y = c and y = d:

d

∫g(y) dy

c

 You should be able to find volumes of revolution:

y xThe volume of revolution formed when the part of the curve = f ( ) , between x = a and x = c, is rotated around

b

the x-axis is given by V = ∫πy 2 dx.

a

y xThe volume of revolution formed when the part of the curve = f ( ) , between y = c and y = d, is rotated around

d

the y-axis is given by V = ∫πx2 dy .

c

440 11 Integration

 Mixed Practicea

1 Evaluate ∫3x2 − 4 dx.
0
dy
2 Given that dx = 1 and that y = 3 when x = 4 , find y in terms of x. π
3 4x graph x-axis and the lines x = 0 6
4 Find the area between the of y = cos x − sin x , the and x .
5 3x − 2 = ln x 2e is rotated 360° around the =
4x2 between x = 1 and x =
Find ∫ dx.

The part of the graph of y x-axis.

Find the volume generated.
y cos x π6 The part of the graph =
between x = 0 and x = is rotated about the x-axis. Find the
2resulting volume of revolution.

y x7 Find the area enclosed by the line = , the y axis and the lines y = 1 and y = 2.
12 52 . The graph of f passes through (4, 0).
8 f(x) 2x − dx, x
Let = ∫ 5 for

fFind (x).

Mathematics SL May 2013 Paper 1 TZ1 Q6

9 Find ∫ 3x2 − 2 dx.
10 x
x 3cos x sin2 , 2(π ). of the curve.
A curve has gradient x and passes through the point Find the equation
π
11 a Sketch the graph of y = cos x for 0 x  2 .

b The graph intersects the y-axis at the point A and the x-axis at the point B. Write down the

coordinates of A and B.

c Find the equation of the straight line which passes through the points A and B.

d Calculate the exact area enclosed by the line and the curve between intersections A and B.

 012 Function f is defined on the domain x 2 by f(x) = x2 − kx . On the graph of y = f(x), the area
below the x-axis equals the area above the x -axis. Find the value of k.

y x 1 y x13 The diagram shows the graphs of = 2 + and = 7 − . y

a Find the coordinates of the points A, B and C.
b Find the area of the shaded region.

B

A Cx

Mixed Practice 441

14 The diagram shows the region R bounded by the y
a
curve y = ln(5x + 1), the y-axis and the line y = a .
R
a Write down the exact value of a.
b Find the area of R.
c The region R is rotated fully about the y-axis.

Find the volume of the resulting solid.

x
2

15 The part of the curve y = ln(x2) between x = 1 and x = e2 is rotated 360° around the y-axis. Find the
exact value of the resulting volume of revolution.

16 The diagram shows the region bounded by the y
2
graphs of y = 2 − x4 and y = x .

The region is rotated about the y-axis to form a
solid of revolution. Find the volume of the solid.

   

x

17 The diagram shows an ellipse with equation 4x2 + 9 y 2 = 36. y
x
Show that the volume generated when the ellipse is rotated
around the x-axis is not the same as the volume generated
when the ellipse is rotated around the y-axis.

442 11 Integration

18 a Find an expression for the volume when the curve y = 1 , for 0 < x < a < π , is rotated 2π
radians around the x-axis. cos x 2

b Given that this volume equals π, find the value of a.

f19 The following diagram shows the graph of (x) = x , for y

0   4 1x , and the line x = 4. x2 +

Let R be the region enclosed by the graph of f, the x-axis and the
line x = 4 .

Find the exact area of R.

R
4x

Mathematics SL November 2014 Paper 1 Q6

20 Let f(x) = cos x, for 0  x  2π. The following y
diagram shows the graph of f. 1
π 3π
There are x-intercepts at x = 2 , 2 .

The shaded region R is enclosed by the graph of f,
3π
the line x = b, where b > 2 , and the x-axis. π Rx
32π 2π
1 3The area of R is ⎛ − ⎞. Find the value of b. 2π
2⎝ ⎠
−1

21 A large vase can be modelled by a solid of Mathematics SL May 2015 Paper 1 TZ1 Q7
revolution formed when the cubic curve, shown
in the diagram, is rotated about the x-axis. y
The units of length are centimetres.
a Find the equation of the curve in the form (20.5, 25)

y = ax3 + bx2 + cx + d.

b Hence find the volume of the vase in litres.

(55, 18)

(50, 17.5)

(0, 10)

x