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14 Integration 1

Jakob Bernoulli was a Swiss

mathematician born in Basel,

Switzerland, on 27 December 1654.

Along with his brother, Johann, he is

considered to be one of the most

important researchers of calculus after

Newton and Leibniz. Jakob studied

theology at university, but during this

time he was studying mathematics and

astronomy on the side, much against

Jakob Bernoulli

the wishes of his parents. After

graduating in theology he travelled

around Europe and worked with a number of the great mathematicians of the time.

On return to Basel, it would have been natural for him to take an appointment in the

church, but he followed his first love of mathematics and theoretical physics and took

a job at the university. He was appointed professor of mathematics in 1687 and, along

with his brother, Johann, started studying Leibniz’s work on calculus. At this time

Leibniz’s theories were very new, and hence the work done by the two brothers was at

the cutting edge. Jakob worked on a variety of mathematical ideas, but in 1690 he

first used the term “integral” with the meaning it has today. Jakob held the chair of

mathematics at the university in Basel until his death in 1705. Jakob had always been

fascinated by the properties of the logarithmic spiral, and this was engraved on his

tombstone along with the words “Eadem Mutata Resurgo” which translates as “I shall

arise the same though changed.”

14.1 Undoing differentiation

In Chapters 8–10 we studied differential calculus and saw that by using the techniques

of differentiation the gradient of a function or the rate of change of a quantity can be

found. If the rate of change is known and the original function needs to be found, it is

necessary to “undo” differentiation. Integration is this “undoing”, the reverse process

to differentiation. Integration is also known as anti-differentiation, and this is often the

best way of looking at it.

dy

2x, what is the original function y?

dx

This is asking what we started with in order to finish with a derived function of 2x.

Remembering that we differentiate by multiplying by the power and then subtracting

one from the power, we must have started with x2.

dy

Similarly, if

4, then this must have started as 4x.

dx

If

Exercise 1

373

14 Integration 1

Exercise 1

Find the original function.

1

dy

5

dx

2

dy

10

dx

3

dy

2

dx

4

dy

4x

dx

5

dy

12x

dx

6

dy

3x2

dx

7

dy

4x3

dx

8

dy

5x4

dx

9

dy

9x2

dx

10

dy

4x2

dx

Looking at the answers to Exercise 1, we can form a rule for anti-differentiation.

If we describe the process of “undoing” differentiation for this type of function, we

could say “add 1 to the power and divide by that new power”.

In mathematical notation this is

This symbol means “the

integral of”.

冮x

n

dx

xn1

n1

This means “with

respect to x” – the

variable we are

concerned with.

14.2 Constant of integration

dy

4. Geometrically, this means that the gradient of

dx

the original function, y, is constant and equal to 4.

Again consider the situation of

So y 4x c (c is the y-intercept of the line from the general equation of a line

y mx c ).

Consider the lines y 4x 3, y 4x and y 4x 5.

dy

4 (as the gradient is 4 each time). Remember that, when

dx

differentiating, a constant “disappears” because the gradient of a horizontal line is zero

For each line

or alternatively the derivative of a constant is zero.

So

冮4 dx 4x c.

Unless more information is given (a point on the line), then the value of c remains

dy

4. In fact

unknown. So any of the lines below could be the original function with

dx

there are an infinite number of lines that could have been the original function.

y 4x 2

y 4x 1

y 4x

y 4x 2

y

2

1

0.5

2

374

x

14 Integration 1

This “c” is called the constant of integration. It must be included in the answer of any

integral, as we do not know what constant may have “disappeared” when the function

was differentiated.

Example

Find y if

dy

8x.

dx

冮

y 8x dx 4x2 c

Example

Find

冮10x 7dx.

冮10x 7dx 冮10x dx 冮7 dx

So 冮10x 7dx 5x 7x c Example

So

2

We established in

Chapter 8 that a function

could be differentiated

“term by term”. In a

similar way, this can be

integrated term by term.

Example

Integrate 8x2 x4.

3

冮8x

2

x4 dx 8x1

3

4 74

x c

7

The coefficient of a term has no effect on the process of integration, and so a constant

can be “taken out” of the integral. This is demonstrated in the next example.

Example

Find the solution of

dy

2 1

x 3.

dx

9

So

冮9 x

2

y 冮x

9

y

2

13

dx

13

dx

This is asking to find y

by integrating.

2 # 3 23

x c

9 2

1 2

y x3 c

3

y

As with the expressions differentiated in Chapters 8–10, it is sometimes necessary to

simplify the function prior to integrating.

Example

375

14 Integration 1

Example

x2 x5

冮 32x dx.

Find

x2 x5

冮 32x dx

The integral sign and the dx

remain until the integration is

performed.

冮

1 12 2

x 1x x5 2 dx

3

冮

1 32

9

x x2 dx

3

The c can remain outside all

brackets as it is an arbitrary

constant and so, when

multiplied by another constant,

it is still a constant.

1 2 52

2 112

B x

x Rc

3 5

11

2 112

2 52

x

x c

15

33

Example

冮p dp.

4

Find

Here p is the variable and so

the integral is with respect to p.

2

冮p dp

4

2

冮

4p2 dp

4p1 c

4

c

p

Exercise 2

Integrate these expressions.

1 2x 1

2 x2

3 x3

4 x4

5 6x2 5

6 8x3 4x 3

7 5x2 4

8 x2

10 x3

1

11 7 4x3

2

9 x2

Find these integrals.

12

冮x

15

冮4x

12

3

dx

13

冮2x

4x 9 dx

16

冮1 2x 6x

19

dy

1

2x

dx

2x

6

5x4 dx

2

x3 dx

14

冮5x 4x

17

冮x dx

20

dy

3

8 1

dx

x3

3

2

3

6

3

Find the solution of these.

18

376

dy

2

5

dx

x

dx

14 Integration 1

dy

8x12x2 32

dx

dy

x2 5

24

dx

x5

21

dy

1x 92 12x 32

dx

dy

4x3 7x

25

dx

2x

22

dy

13x 42 2

dx

3

dy

7x4 6x4

26

1

dx

2x4

23

Find y by integrating with respect to the relevant variable.

27

dy

12

3

dp

p

28

dy

5

8k4

dk

30

1t 32 2

dy

dt

t4

31

dy

2t 4t3

dt

3t

29

dy

1

z3 ¢z2 2 ≤

dz

z

14.3 Initial conditions

In all of the integrals met so far, it was necessary to include the constant of integration,

c. However, if more information is given (often known as initial conditions), then the

value of c can be found.

dy

Consider again the example of

4. If the line passes through the point (1, 3) then

dx

c can be evaluated and the specific line found.

So

dy

4

dx

冮

1 y 4 dx

1 y 4x c

Since we know that when x 1, y 3 these values can be substituted into the

equation of the line.

So

341c

1 c 1

The equation of the line is y 4x 1.

When the value of c is unknown, this is called the general solution.

If the value of c can be found, this is known as the particular solution.

Example

Given that the curve passes through (1, 3) and

of the curve.

dy

2x 1, find the equation

dx

dy

2x 1

dx

冮

1 y 2x 1 dx

1 y x2 x c

Using (1, 3)

1 3 12 1 c

1c3

So the equation of the curve is y x2 x 3.

377

14 Integration 1

Example

Find P given that

dP

1

and P 7 when t 100

dt

2t

1

dP

dt

2t

1P

冮 2t dt

1

冮

1 P t2 dt

1

1

1 P 2t 2 c

Since P 7 when t 100, 7 2# 2100 c

1 c 13

1

2

Hence P 2t 13

Exercise 3

Given the gradient of each curve, and a point on that curve, find the equation

of the curve.

1

dy

6, 12, 82

dx

2

dy

4x, 11, 5 2

dx

3

dy

8x 3, 12, 42

dx

4

dy

2x 5, 14, 42

dx

5

dy

4x3 6x2 7, 11, 92

dx

6

dy

6

4x2 2 , 14, 12

dx

x

7

dy

8

, 19, 22

dx

2x

Find the particular solution, using the information given.

378

8

dy

t2 1t 4 3t 2 42, y 6 when t 1

dt

9

p3 4p5

dQ

, Q 2 when p 0

dp

32p

14 Integration 1

14.4 Basic results

Considering the basic results from differentiation, standard results for integration can

now be produced. For polynomials, the general rule is:

冮

xn dx

However, consider

冮 x dx 冮x

1

1

dx.

Using the above rule, we would obtain

that

xn1

c

n1

x0

, but this is not defined. However, it is known

0

d

1

1ln x2 .

x

dx

This provides the result that

冮 x dx ln冟x冟 c

1

Remembering that ln x is defined only for positive values of x, we recognize that

1

1

is defined for all x H ⺢, x 0

dx ln冟x冟 c, taking the absolute value of x. As

x

x

and ln x is defined only for x 7 0, the absolute value sign is needed so that we can

1

integrate

for all values of x for which it is defined.

x

冮

Similarly

d x

1e 2 ex so

dx

冮e dx e c

x

x

When differentiating sine and cosine functions the following diagram was used and is

now extended:

Differentiate

S

C

S

C

Integrate

The integrals of other trigonometric functions can be found by reversing the basic rules

for differentiation, and will be discussed further in Chapter 15.

Standard results

Function

Integral ⴙ c

f(x)

冮f1x2 dx

xn 1n 12

xn1

n1

1

x

ex

sin x

cos x

ln x

ex

cos x

sin x

379

14 Integration 1

Example

Integrate sin x ex.

冮sin x e dx

x

cos x ex c

Example

Integrate

3

4 cos x.

x

冮 x 4 cos x dx

3

3 ln冟x冟 4 sin x c

Exercise 4

Integrate these functions.

2

1 x3

x

4 6 sin x 6x4

7

ex

5

7 sin x

3

2x

5

cos x

x

2 4ex sin x

3

5 8 sin x 7ex

6 5ex 2 sin x

8

3

x

ex

152x cos x

15

14.5 Anti-chain rule

When functions of the type 12x 12 5, e8x and sin¢2x

p

≤ are differentiated, the

3

chain rule is applied. The chain rule states that we multiply the derivative of the outside

function by the derivative of the inside function. So to integrate functions of these types

we consider what we started with to obtain that derivative.

Example

dy

12x 12 3. Find y.

dx

冮

So y 12x 12 3 dx

When integrating, 1 is added to the power, so y must be connected to

12x 12 4. Since we multiply by the power and by the derivative of the inside

function when differentiating, we need to balance this when finding y.

380

So y

1#1

12x 12 4 c

4 2

1y

1

12x 12 4 c

8

14 Integration 1

Example

Find

冮 cos 3x dx.

S

C

Using Diff

Int this begins with sin 3x.

S

C

Balancing to obtain cos 3x when differentiating

1

冮 cos 3x dx 3 sin 3x c

1

Example

Find

冮6e

4x

dx.

This started with e4x as

冮

d 4x

1e 2 4e4x.

dx

冮

So 6e4x dx 6 e4x dx

6#

1 4x

e c

4

3 4x

e c

2

Example

Find y given that

So y

As

dy

1

p

sin¢4x ≤.

dx

3x 4

2

冮3x 4 sin¢4x 2 ≤ dx

1

p

1

13x 42 1

3x 4

we recognize that this comes from ln冟3x 4冟 as

So y

d

3

1ln冟3x 4冟 2

.

dx

3x 4

1

p

1

ln冟3x 4冟 cos¢4x ≤ c

3

4

2

For these simple cases of the “anti-chain rule”, we divide by the derivative of the

inside function each time. With more complicated integrals, which will be met in

the next chapter, this is not always the case, and at that point the results will be

formalized. This is why it is useful to consider these integrals as the reverse of

differentiation.

381

14 Integration 1

Exercise 5

Find these integrals.

1

5

9

13

冮 sin 5x dx

冮8 cos 4x dx

冮e dx

冮8x e dx

5x

2x

2

6

10

14

冮 cos 6x dx

冮6 sin 3x dx

冮4e dx

冮4e dx

4x

3

7

11

冮 sin 2x dx

冮5 cos 2x dx

冮8e dt

6t

4

8

12

冮 sin 2 x dx

冮e dx

冮5e dp

1

6x

6p

2x

Find y if:

dy

1

dx

2x 3

dy

13x 12 5

18

dx

dy

21

13 2x2 4

dx

15

dy

4

dx

3x 1

dy

3

27

dt

6t

24

dy

1

dx

8x 7

dy

14x 72 6

19

dx

dy

4

22

dx

13x 22 3

dy

4

dx

2x 5

dy

14x 32 3

20

dx

dy

3

23

dt

12t 12 2

16

25

17

dy

6

dx

3x 5

26

dy

8

dp

4p

Integrate these functions.

28 6e4x

1

13x 42 5

31

2x 1

30 4e8x 4 cos 2x

sin 3x 4x

2

32 6x2

3x 2

29

14.6 Definite integration

Definite integration is where the integration is performed between limits, and

this produces a numerical answer.

Upper limit 1x b2

b

A definite integral is of the form

冮 f1x2 dx

a

Lower limit 1x a2

Example

3

冮 2x 1 dx

2

c x2 x d

3

2

382

This notation means that the

integration has taken place.

The two values are now

substituted into the function

and subtracted.

When a definite integral

is created, the lower

limit is always smaller

than the upper limit.

14 Integration 1

132 32 122 22

62

4

There is no constant of integration used here. This is because it cancels itself out

and so does not need to be included.

c x2 x c d

3

2

132 3 c2 122 2 c2

6c2c

4

Example

4

冮 12x 32

2

dx

0

4

1

B 12x 32 3R

6

0

1

1

18 32 3 ¢ 10 32 3≤

6

6

125

27

6

6

152

6

76

3

Example

p

4

0

冮 sin 2x 1 dx

p

4

1

B cos 2x xR

2

0

1

p

p

1

¢ cos ≤ ¢ cos 0 0≤

2

2

4

2

p

1

4

2

p2

4

To differentiate trigonometric

functions, we always use

radians, and the same is true

in integration.

383

14 Integration 1

Example

2

冮 x 4 dx

2

2

c 2 ln冟x 4冟 d

2

2

2 ln冟2冟2 ln冟6冟

2 ln 2 2 ln 6

ln 4 ln 36

An answer could have

been approximated earlier,

but if an exact answer

were required, this form

would need to be given.

4

ln

36

ln

1

9

2.20

Example

Although there is no value for

the upper limit, an answer

can still be found that is an

expression in a.

a

冮e

4x

dx where a 7 2

2

a

1

B e4xR

4

2

1 4a 1 8

e e

4

4

1 4a

1e e8 2

4

Exercise 6

Find the value of these definite integrals

2

1

3

冮 2x dx

2

1

冮 8x 4x

dx

5

冮e

dx

8

0

冮

冮 5 dx

0

1

4

dx

x2

6

冮 x dx

6

3

3

冮 sin 3x dx

1

冮 4e

3x

dx

9

冮 cos 2u du

0

4x

p

6

p

2

11

冮 2e

dx

2

2

p

3

384

3

4

2x

0

dx

2

3

10

2

3

3

0

7

冮 6x

2

2

4

4

12

冮 cos 3t 6 dt

0

14 Integration 1

3

13

0

冮 12x 12

2

dx

14

冮

2

dx

12x 12 3

1

17

2

冮 13 2x2

15

20

4

dx

5

冮

冮

4

1

冮 2x 1 dx

1

18

2

1

4

dx

3x 4

p

22

dx

3

dx

13x 42 2

2

4

冮

3

1

4

19

冮 13x 12

2

0

16

2

2

5

dt

t2

冮 6p 4p 1 dp

3

21

1

k

冮 sin 4x 6x dx

p

23

冮 2x 1 dx

4

0

14.7 Geometric significance of integration

When we met differentiation, it was considered as a technique for finding the gradient

of a function at any point. We now consider the geometric significance of integration.

5

Consider

冮 3 dx.

Remember the limits

are values of x.

y

1

5

冮 3 dx

3

y3

1

c 3x d

5

0

1

1

5

x

15 3

12

This is the same as the area enclosed by the function, the x-axis and the vertical lines

x 1 and x 5.

This suggests that the geometric interpretation of integration is the area between the

curve and the x-axis.

Consider y 3x2.

y

y 3x2

0

1

3 x

In the previous example where y 3, it was easy to find the area enclosed by the

function, the x-axis and the limits. However, to find the area under a curve is less obvious.

This area could be approximated by splitting it into rectangles.

385

14 Integration 1

y

27

y 3x2

A ⯝ 1 3 1 12

12

15 square units

3

0

1 2

3 x

This is clearly not a very

accurate approximation,

so to make it more

accurate thinner

rectangles are used.

Four rectangles would make this more accurate.

y

27

75

4

y 3x2

A⯝

1

1

27

1

1

75

3

12

2

2

4

2

2

4

A⯝

81

4

12

27

4

3

0

1 2 3 x

1.5

2.5

As the rectangles become thinner, the approximation becomes more accurate.

1

x

3

x

This can be considered in a more formal way. Each strip has width dx with height y and

area dA.

So dA ⯝ y # dx

1 A a dA

1 A ⯝ a y dx

As dx gets smaller, the approximation improves

xb

and so A lim a ydx.

dxS0

xa

dA

Now y ⯝

dx

1 y lim

dxS0

1y

dA

dx

Integrating gives

386

dA

dx

冮y dx 冮 dx dx 冮dA A c

dA

Limits are needed here to

specify the boundaries

of the area.

14 Integration 1

With the boundary conditions,

c can now be ignored.

b

冮 y dx

A

This is the basic formula

for finding the area

between the curve and

the x-axis.

a

b

xb

Hence lim a y # dx

dxS0

xa

冮 y dx.

a

This now shows more formally that the geometric significance of integration is that it

finds the area between the curve and the x-axis.

If the two notations for summation are compared, we find that sigma notation is used

for a discrete variable and that integral notation is used for a continuous variable.

So the

冮 sign actually means

“sum of” (it is an elongated S).

3

3

2

a 3x

and

1

冮 3x

2

This is the sum of a

continuous variable.

dx

1

c x3 d

This is a sum of a

discrete variable.

3

1

27 1

The area required is

26 square units.

26

Example

4

Find the area given by

冮 2x 1 dx.

2

4

冮 2x 1 dx

2

cx xd

4

2

2

116 42 14 22

12 2

10 square units

y 2x 1

y

0

1

2

4

x

387

14 Integration 1

This integration can be performed on a calculator. Although a calculator

cannot perform calculus algebraically, it can calculate definite integrals.

This is shown in the diagram below, and we find the area is still 10 square

units.

Example

p

Find the area enclosed by y cos 2x and the x axis, between x

and

4

p

x .

2

A

p

2

p

4

冮 cos 2x dx

p

2

1

B sin 2xR

p

2

4

1

1

p

sin p sin

2

2

2

0

1

2

1

2

The answer to this definite integral is negative. However, area is a scalar quantity

1

(it has no direction, only magnitude) and so the area required is

square unit.

2

As the negative sign has no effect on the area, the area can be considered to be

b

冮

the absolute value of the integral. So A 2 f1x2 dx2. Whether the calculation

a

is done using the absolute value sign or whether we do the calculation and then

ignore the negative sign at the end does not matter. As can be seen from the

graph, the significance of the negative sign is that the area is contained below

the x-axis.

388

Some definite integration

questions require the use of

a calculator. Where an exact

value is required, one of the

limits is a variable or the

question is in a non-calculator

paper, algebraic methods

must be employed.

14 Integration 1

Example

1

Find the area given by

冮 x dx.

1

2

1

A2

冮 x dx2

1

2

2 c ln冟x冟 d

1

2

2

冟ln冟1冟 ln冟2冟冟

冟ln 1 ln 2冟

冟ln 2冟

0.693

So the required area is 0.693 square units.

1

It should be noted that it is actually not possible to find the area given by

冮 x dx as

1

2

there is a vertical asymptote at x 0. It is not possible to find the definite integral over

an asymptote of any curve, as technically the area would be infinite.

This example also provides another explanation for the need for the modulus signs in

冮 x dx ln冟x冟. Although logarithms are not defined for negative values of x, in order to

1

1

find the area under a hyperbola like y , which clearly exists, negative values need to

x

be substituted into a logarithm, and hence the absolute value is required. This was

shown in the above example.

Example

Find the area enclosed by y 12x 12 2 4 and the x-axis.

First the limits need to be found.

389

14 Integration 1

These are the roots of the graph (they can be found algebraically or by using a

calculator)

12x 12 2 4 0

1 2x 1 ;2

1x

3

1

or x

2

2

So the area is given by

3

2

冮 12x 12

2

4 dx

12

3

2

1

B 12x 12 3 4xR

6

12

16

3

Exercise 7

Find the area given by these definite integrals.

6

1

3

冮 8 dx

2

2

冮 13x 22

4

dx

5

8

1

2

10

冮

4

冮 sin x dx

冮

2

6

11

冮

冮 e dx

x

2

2

dx

2x 3

9

e2x1 4x dx

冮 3x 2 dx

4

p

3

12

0

13 y x2 2 and the lines x 1 and x 4

14 y e6x and the lines x 2 and x 1

15 y sin x and the lines x 0 and x p

4

and the lines x 5 and x 2

3x 4

17 y 1 x2 and the lines x 1 and x 1

18 y 14x 12 2 9 and the lines x 1 and x

19 y x3 2x2 and the lines x 0 and x 2

p

20 y e2x sin 2x and the line x and the y-axis

2

390

冮 cos 3x 4x dx

0

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

16 y

1 dx

0

1

4

dx

2x 1

2

2

5

4

dx

x

冮 6x

1

0

4

冮

3

p

0

7

冮 2x 3 dx

0

2

4

3

1

2

14 Integration 1

21 Find an expression in terms of p for this area.

y

2

3x 5

y

0

p x

1

22 Find an expression in terms of p for this area.

y e2x x2

y

x

p

0

k

23 Find k 1k 7 02 given that

˛

冮 3x

1

2

dx 16.

0

a

24 Find a given that

冮 19 x2

25

a

2

5

dx .

8

14.8 Areas above and below the x-axis

In this case the formula needs to be applied carefully. Consider y x 1x 12 1x 22 and the

area enclosed by this curve and the x-axis.

˛

y x3 3x2 2x

2

If the definite integral

冮x

3

3x2 2x dx is calculated, we obtain an answer of 0 (remember

0

the calculator uses a numerical process to calculate an integral) and so this result is interpreted

as zero.

391

14 Integration 1

However, it is clear that the area is not zero.

We know that a definite integral for an area below the x-axis provides a negative answer.

This explains the zero answer – the two (identical) areas have cancelled each other out.

So although the answer to the definite integral is zero, the area is not zero.

To find an area that has parts above and below the x-axis, consider the parts separately.

1

So in the above example, area 2

冮x

3

3x2 2x dx

0

1

2

4

1

unit2

2

This demonstrates an important point. The answer to finding an area and to finding the

value of the definite integral may actually be different.

The method used in the example below shows how to avoid such problems.

Method

1 Sketch the curve to find the relevant roots of the graph.

2 Calculate the areas above and below the x-axis separately.

3 Add together the areas (ignoring the negative sign).

Example

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

y

A

x

0

B

1. Sketch the curve and shade the areas required.

The roots of the graph are given by x2 4x 3 0

1 1x 12 1x 32 0

1 x 1 or x 3

392

These results can be

found using a

calculator.

14 Integration 1

2. Work out the areas separately.

1

冮

3

冮

A 2 x 4x 3 dx2

B 2 x2 4x 3 dx2

2

0

1

1

1

2 B x3 2x2 3xR 2

3

0

3

1

2 B x3 2x2 3xR 2

3

1

1

2 ¢ 2 3≤ 102 2

3

4

2 19 18 92 ¢ ≤ 2

3

4

4

2 2

3

3

4

3

3. These areas can be calculated on a calculator (separately) and then added.

8

So the total area square units.

3

Example

Find the area bounded by y sin x, the y-axis and the line x

5p

.

3

Graphing this on a calculator,

5

3

This area can be split into:

5p

3

p

冮 sin x dx

and

p

0

c cos x d

冮 sin x dx

c cos x d

p

5p

3

p

0

1cos p2 1cos 02

1 112

2

¢cos

1

¢ ≤ 112

2

So the total area is

5p

≤ 1cos p2

3

3

2

7

square units.

2

393

14 Integration 1

Exercise 8

Find the shaded area on the following diagrams.

1 y x2 8x 12

y

0

x

2 y x2 9x 8

y

x

0

3 y 4 sin x

y

4

0

x

2

4

4 y 5x3

y

1

x

0

2

Find the area bounded by the curve and the x-axis in the following cases.

5 y x 1x 32 1x 22

6 y 1x 42 12x 12 1x 32

˛

7 y x3 x2 16x 16

8 y 6x3 5x2 12x 4

394

14 Integration 1

Find the area bounded by the curve, the x-axis and the lines given.

9 y x2 x 6, x 2 and x 4

10 y 6x3, x 4 and x 2

11 y 3 sin 2x, x 0 and x

5p

6

p

≤, x 0 and x p

6

p

5p

13 y 4 cos 2x, x

and x

2

4

12 y cos¢2x

14 y x3 ex, x 0 and x 4

14.9 Area between two curves

The area contained between two curves can be found as follows.

b

The area under f(x) is given by

b

冮 f1x2 dx and under g(x) is given by 冮 g1x2 dx.

a

a

b

So the shaded area is

b

冮 f1x2 dx 冮 g1x2 dx.

a

y

g(x)

a

f (x)

b

Combining these gives

冮

f1x2 g1x2 dx.

0

a

b

This can be expressed as

冮 upper curve – lower curve dx.

a

b

x

As long as we always

take upper – lower, the

answer is positive and

hence it is not necessary

to worry about above

and below the x-axis.

a

Example

Find the shaded area.

y 2x2

y

y 4x

0

The functions intersect where 2x2 4x.

1 2x2 4x 0

1 2x 1x 22 0

1 x 0 or x 2

˛

x

These intersection

points can also be found

using a calculator.

395

14 Integration 1

2

So the area

冮 4x 2x

2

dx

0

B2x2

¢8

2 3 2

xR

3

0

16

≤ 102

3

8

3

This function can be drawn using Y1-Y2 and then the area calculated.

The two limits are roots of

the resulting function.

Example

Find the area contained between y ex and y sin x from x p to

x p.

Note that these graphs cross twice within the given interval. So, we need to find

the intersections and then treat each part separately (as the curve that is the

upper one changes within the interval).

396

14 Integration 1

So we

we need

need to

to find

find

So

3.096

3.096

冮

1exx 2

sin xx 1 e 2 dx

dx.

sin

p

p

p

0.589

冮

sin and

xdx and

x

e

p

x

sin xsinx1e1x 2 edx.

2 dx.

0.589

0.589

3.096

25.7 13 sf2

The total area 0.00976 ... +1.32 p 24.4 p

The total area 0.00976 ... 1.32 p 24.4 p 25.7 13 sf2

Area between a curve and the y-axis

Mostly we are concerned with the area bounded by a curve and the x-axis. However, for

some functions it is more relevant to consider the area between the curve and the y-axis.

This is particularly pertinent when volumes of revolution are considered in Chapter 16.

b

The area between a curve and the x-axis is

冮 y dx

a

To find the area between a curve and the y-axis we calculate

b

冮 x dy

This is where x is a function

of y.

a

This formula is proved in an identical way to the area between the curve and the x-axis,

except that thin horizontal rectangles of length x and thickness dy are used.

Example

Consider y2 9 x.

y

3

x

0

3

397

14 Integration 1

There is no choice here but to use horizontal strips as opposed to vertical strips,

as vertical strips would have the curve at both ends, and hence the length of the

rectangle would no longer be y and the formula would no longer work.

Area

3

冮 x dy

3

3

冮9y

2

dy

3

1 3 3

yR

3 3

127 92 127 92

18 18

36 square units

B9y

Although the integration is performed with respect to y, a calculator can still be

used to find the area (although of course it is not the correct graph).

y 9 x2

Exercise 9

Calculate the area enclosed by the two functions.

1 y x2, y x

2 y 6x2, y 3x

3 y x3, y x

4 y x2, y 2x

5 y 8 x2, y 2 x

6 y x3 24, y 3x2 10x

7 y 10 x2, y 19 2x2

8 y ex, y 4 x2

1

1

9 y x2 6x 10, y 4x x2 and the x-axis. In this case draw the

2

3

graphs and shade the area.

398

14 Integration 1

10 y 3 cos 2x and y 1 produce an infinite pattern as shown.

y

3

y1

1

0

x

Find the area of each shaded part.

11 Find the area between y e2x, y 2 x, the x-axis and the y-axis as

shown.

y2x

y

y e2x

x

0

12 Find the area between y

2

, y 4 x, the x-axis and the y-axis as

4x

shown.

y

y

2

4x

0

x

4

y4x

13 Find the shaded area.

y

y sin x

1

0

1

2

x

y sin 2x

14 Find the shaded area.

y

3

y 3 cos 2x

1

0

1

x

y cos x

3

399

14 Integration 1

15 Find the area of the “curved triangle” shown below, the sides of which lie

1

4

on the curves with equations y x 1x 32, y x x2 and y 2 .

4

x

˛

y x(x 3)

y

y 42

x

0

x

y x 14 x 2

Find the area enclosed by the y-axis and the following curves.

16 x 4 y2

y

2

x 4 y2

x

2

17 y2 16 x

18 8y2 18 2x

19 Evaluate the shaded area i with respect to x and ii with respect to y.

y 25 x

y

5

y 5x

0

5

x

20 Find the shaded area.

y

1 x

y e

2

y 2 3ex

0

400

x

14 Integration 1

21 Find the shaded area.

y

e2x

y ex 2

y

3

x

0

22 Find the shaded area.

y

6

y 4 sin x

0

x

y 6 cos x

Review exercise

✗

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

✗

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

1 Integrate these functions.

=

ON

X

=

a 4x2 7

b 9x2 4x 5

2 Solve these equations.

dy

dy

x3 6

p2 13 p5 2

a

b

5

dx

dp

x

✗

3 Given

✗

4 Find these integrals.

M

M–

M+

C

CE

%

7

8

9

–

4

5

6

÷

1

2

3

+

0

M

C

7

4

1

ON

X

=

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

M

C

7

4

1

0

ON

X

=

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

dy

3t2 2t

dt

4t

冮4e sin x dx

x

b

冮7 cos x x dx

4

c

冮2e

6x

5

4 sin x dx

x

5 Find these integrals.

=

✗

✗

M

c

=

✗

C

d 13 2x2 2

8

x3

dy

3x 8 and the curve passes through (2, 8), find the

dx

equation of the curve.

a

M

c

ON

X

a

冮6 cos 2x dx

b

冮4e

d

冮13x 22

e

冮7e

6

dx

6 Let f 1x2 2x ¢2x

2x

dx

c

2

4

dx

13x 42 5

3x

冮4x 3 dx

冮f 1x2 dx.

3

Find

5≤.

x2

7 Find these definite integrals.

˛

˛

=

p

4

3

a

冮

1

3

4p

dp

12p 12 3

b

冮 sin 4u 1 du

0

k

c

冮 3 cos 2u 4u du

k

401

14 Integration 1

✗

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

8 Find the area given by these definite integrals.

=

5

a

✗

M

M–

M+

C

CE

%

7

8

9

–

4

5

6

÷

1

2

3

+

0

ON

X

=

p

6

1

冮 e dx

x

2

b

冮 4x 5 dx

冮 2 cos 3u du

c

3

0

9 Find an expression in terms of p for this shaded area.

y

y

0

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

5

2

3

2x 5

px

3

10 Find the total area of the two regions enclosed by the curve

=

y x3 3x2 9x 27 and the line y x 3.

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

=

[IB Nov 04 P1 Q14]

11 The figure below shows part of the curve y x 7x 14x 7.

The curve crosses the x-axis at the points A, B and C.

3

2

y

0

✗

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

=

A

B

C

x

a Find the x-coordinate of A.

b Find the x-coordinate of B.

c Find the area of the shaded region.

[IB May 02 P1 Q13]

12 Find the area bounded by the curve and the x-axis for:

3p

a y 3 cos 2u, u 0 and u

4

b y x4 2ex, x 1 and x 3

M

C

7

4

1

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

0

ON

X

13 Find the area enclosed by:

=

a y ex and y 6 x2

b y

M

C

7

4

1

0

402

M–

M+

CE

%

8

9

–

5

6

÷

2

3

+

ON

X

=

3

, y 2 x and the x- and y-axes

2x

14 Find the area between x 8 y2 and the y-axis.