Chap 2 Applications of Integration

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MTH1134/LN/02 Chap 2 Application of Integration

Mohd Nasir Mahmud@MICET Page 1 of 18 Wednesday, January 25, 2006

2.1 Integration of gradient functions

In the previous chapter, we have learnt the method of finding the gradient of tangent functions by

using differentiation. Now we will learn on how to find the function by using the reverse method

which is by using the integration.

Example 24:

Given that the gradient function 42 += xdx

dy intercepts the y -axis at 2. Find the function of y .

Solution:

Given that

42 += xdx

dy

So,

dx xdxdx

dy y 42∫∫ +==

c x x ++= 42

The function intercepts the y -axis at 2. Therefore,

( ) ( ) c++= 04022

2=c

∴ 242++= x x y

2.4.3 Kinematics problems

Kinematics is a portion of physics concerned with motion in the abstract, such as of points or

space figures, and separated from its dynamic properties. This section is intended for students

who have already learned a few physics concepts in kinematics and dynamics.

i. Velocity

dt

dx

t

xv

t =

∆

∆=

→∆ 0lim

where ν = velocity

x∆ = range of displacement

t ∆

= time interval





dt

dx= derivative of x with

respect to t .

We can reverse the process to get the value of displacement, x.

From the above equation,

∫=−

=

2

1

12

t

t

vdt x x

vdt dx

ii. Acceleration

When the velocity of a moving body changes with time, we say that the body has

acceleration. Just as velocity describes the rate of change of position with time, acceleration

describes the rate of change of velocity with time.

2

2

0lim

dt

xd

dt

dx

dt

d

dt

dv

t

va

t =

==

∆

∆=

→∆

where a = acceleration

v∆ = change in velocity

t ∆

= time interval

dt

dv= derivative of v with respect to t .

As the above method, we can reverse the process

to get the value of velocity, v.

From the above equation,

∫=−

=

2

1

12

t

t

adt vv

adt dv

Example 25:

You are driving along a straight highway in your new Mustang. At time 0=t , when you

are moving at 10 m/s, you pass a signpost at m x 50= . Your acceleration is a function of

time:

( )t smsma

32

/ 10.0 / 0.2 −=

Derive expressions for your velocity and position as functions of time.





Solution:

The position at time 0=t is m x 500 = , and the velocity at time

0=t is smv

/ 100 = .

∫=−

2

1

12

t

t

adt vv *Assume that 1v and 1t are at

initial condition i.e. 01 vv =

and 01 t t = .

( )[ ]

( ) ( ) 23

212

2

0

32

2

/ 10.0 / 0.210

/ 10.0 / 0.210

t smt smv

dt t smsmv

t

−+=

−=−∴

∫

Then we use equation ∫=−

2

1

12

t

t

vdt x x to find x as a function of t :

* vv =2 = velocity at final condition.

( ) ( )[ ]( ) ( ) ( )( )33

6122

21

2

0

232122

/ 10.0 / 0.2 / 1050

/ 10.0 / 0.2 / 1050

t smt smt smm x

dt t smt smsmm x

t

−++=

−+=−∴ ∫

iii. Work done by a variable force(W)

It turns out that the work-energy theorem translates directly into the case of variable force.

In this case, however, we must be careful with the calculus since it requires a simple

integral.

Consider an x-component of force F x( x) that varies as a function of position along the x-

axis of a 1D trajectory in a non-trivial way, as indicated in the following figure stolen from

Tipler:





As the number of segments becomes very large and the size of each becomes very small,

this sum becomes (in the limit) the integral of F from 1 x to 2 x .

∫=

2

1

x

x

FdxW

iv. Impulse (J)

When a constant force, F acts on a body during a tie interval t ∆ , from 1t to 2t , the

impulse of the force , denoted by J , is defined to be

( )

∫=∴

∆=−=

2

1

12t

t

Fdt J

t F t t F J

2.4.4 Area under a curve and area between two curves

Introduction

One of the important applications of integration is finding the area bounded by a curve.

Often such an area can have a physical significance like the work done by a motor, or the

distance traveled by a vehicle. In this section we explain how such an area is calculated.

1. Calculating the Area under a Curve

Let us denote the area under y = f(x) between a fixed point a and a variable point x by

A(x):





It is clearly a function of x since as the upper limit changes, so does the area. How does

the area change if we change the upper limit by a very small amount xδ ? See the figure

below.

To a good approximation the change in the area ( ) ( ) x A x x A −+δ

is:

( ) ( ) ( ) x x f x A x x A δ δ ≈−+

an approximation which gets better and better as xδ gets smaller and smaller, and so:

( ) ( ) ( )

x

x A x x A x f

δ

δ −+≈

Clearly, in the limit as 0→ xδ we have

( )

( ) ( )

x

x A x x A

x f x δ

δ

δ

−+

= →0lim

But this limit on the right-hand side is the derivative of A(x) with respect to x, so

( )dx

dA x f =

Thus A(x) is an indefinite integral of f(x) and we can therefore write:

( ) ( )dx x f x A ∫=

Now the area under the curve from a to b is clearly A(b)−A(a). But remembering our

shorthand notation for this differences, introduced in the last block we have, finally





( ) ( ) ( )[ ] ( )∫=≡−

b

a

b

a dx x f x Aa Ab A

We conclude that the area under the curve y = f(x) from a to b is given by the definite

integral of f(x) from a to b .

2. The Area Bounded by a Curve Lying Above the x-axis

Consider the graph of the function y = f ( x) shown in the figure below. Suppose we are

interested in calculating the area underneath the graph and above the x-axis, between the

points where x = a and x = b. When such an area lies entirely above the x-axis, as is

clearly the case here,

this area is given by the definite integral ( )∫b

adx x f .

The area under the curve y = f ( x), between x = a and x = b is given by ( )∫b

adx x f when

the curve lies entirely above the x-axis between a and b.

Example 26:

Calculate the area bounded x

y1

= and the x-axis, between x = 1 and x = 4.

Solution:

Ke Point





Below is a graph of x

y1

= . The area required is shaded; it lies entirely above the x-axis.

Area = [ ] place)(3decimal 386.14ln1ln4lnln1 4

1

4

1==−==∫ xdx

x

Example 27:

Find the area bounded by the curve y = sin x and the x-axis between x = 0 and x = π. The

required area is shown in the diagram below. Note that it lies entirely above the x-axis.

Solution:

Shaded area = [ ] 2cossin 00

=−=∫ π

π

x xdx





Example 28:

Find the area under f(x) = e2x

from x = 1 to x = 3 given that the exponential function e2x

is

always positive.

Solution:

(Note: Because e2x

is positive, the area will lie above the x-axis.)

Area = 1982

13

1

23

1

2=

=∫

x xedxe

3. The Area Bounded by a Curve, Parts of which Lie Below the x-axis

The diagram below shows a graph of 12

+−= x y .

The shaded area is bounded by the x-axis and the curve, but lies entirely below the x-axis.

Let us evaluate the integral ( )∫ +−2

1

2 1 dx x .

( )∫

+−=+−

2

1

2

1

32

31 x

xdx x

+−−

+−= 1

3

12

3

233

3

4

13

7

−=

+−=





The evaluation of the area yields a negative quantity. There is no such thing as a negative

area. The area is actually 4/3 , and the negative sign is an indication that the area lies

below the x-axis. If an area contains parts both above and below the horizontal axis, care

must be taken when trying to calculate this area. It is necessary to determine which parts

of the graph lie above the horizontal axis and which lie below. Separate integrals need to

be calculated for each ‘piece’ of the graph. This idea is illustrated in the next example.

Example 29:

Find the total area enclosed by the curve y = x3 − 5x

2 + 4x and the x-axis

between x = 0 and x = 3.

Solution:

We need to determine which parts of the graph, if any, lie above and which lie below the

x-axis. To do this it is helpful to consider where the graph cuts the x-axis. So we consider

the function x3 − 5x

2 + 4x and look for its zeros

x3 − 5x

2 + 4x = x(x

2 − 5x + 4)

= x(x − 1)(x − 4)

So the graph cuts the x-axis when x = 0, x = 1 and x = 4. Also, when x is large and

positive, y is large and positive since the term involving x3 dominates. When x is large

and negative, y is large and negative for the same reason. With this information we can

sketch a graph showing

the required area. If you have access to a graphics calculator or computer package this is

a trivial matter. The graph is shown below.

From the graph we see that the required area lies partly above the x-axis (when 0 ≤ x ≤ 1)

and partly below (when 1 ≤ x ≤ 3). So we evaluate the integral in two parts:

First:

( ) ( )12

702

3

5

4

1

2

4

3

5

445

1

0

2341

0

23=−

+−=

+−=+−∫

x x xdx x x x

This is the part of the required area which lies above the x-axis.





Second:

( )3

1

2343

1

23

2

4

3

5

445

+−=+−∫

x x xdx x x x

3

222

3

5

4

118

3

135

4

81−=

+−−

+−=

This represents the part of the required area which lies below the x-axis. The actual area

is 22/3. Combining the results of the two separate calculations we can determine the total

area bounded by the curve:

Area =12

95

3

22

12

7=+

4. The Area Bounded Between Two Curves

The area between two curves ( ) x y1 and ( ) x y2 is

( ) ( )∫ −= dx x y x y A 12





or

Example 30:

Find the area A between ( ) 642

1 +−= x x x y and ( ) 2

2 4 x x x y −=

Solution:

The curves intersect where

( )( )( )

3or1

0132

0342

464

2

22

==

=−−

=+−

−=+−

x x

x x

x x

x x x x





Hence

( ) ( )[ ] ( )3

8682644

3

1

23

1

22=−+−=+−−−= ∫∫ dx x xdx x x x x A

Example 31:

The diagram shows the graphs of y = sin x and y = cos x for 0 ≤ x ≤ π 21 . The two graphs

intersect at the point where π 41= x . Find the shaded area.

Solution:

To find the shaded area we could calculate the area under the graph of y = sin x for x

between 0 and π 41 , and subtract this from the area under the graph of y = cos x between

the same limits. Alternatively the two processes can be combined into one and we can

write





Shaded area

( )

[ ]

( ) ( )0cos0sincossin

cossin

sincos

41

41

4 /

0

4 /

0

+−+=

+=

−= ∫

π π

π

π

x x

dx x x

If you are aware of the standard triangles you will know that

2

1cossin

41

41 == π π

in which case the value of integral is 414.012

2 =− . Alternatively you can use your

calculator to obtain this result directly.

Similarly, to find the area of a region between a curve and the y -axis as shown, the area

of strip yδ is

y x A δ δ ≅

so that the area of the region is

∑ ∑≅= y x A A δ δ

Thus as 0→ yδ we get

∫=d

c xdy A

Example 32:

Find the area A shown





Here2 x y = , so that y x = . So

[ ] ( )3

1418

3

2

3

2 4

1

4

1

2

3

=−=== ∫ ydy y A

2.4.5 Volume of solid of revolution

Volume of revolution about the x - axis

Consider the area under the curve

2

x y = between

1= x and 2= x , as shown on the right.

Now consider the solid formed when this area is

rotated through π 2 radians about the x - axis. The

volume of this solid can be calculated using calculus

methods.

Consider a small strip of width x.δ under the curve

( ) x f . When this small area is rotated through π 2

radians about the x - axis, a disc is formed of radius

y and thickness x.δ . The volume, V δ , of the disc

is given by

x yV δ π δ 2

=





To find the volume, V , of the total solid, we must

find the sum of all such disc from a x = to b x = .

Therefore,

∑=

=

b

a x

x yV δ π 2

As 0→ xδ , this summation approaches a limiting

value, namely V . Therefore,

∑=

→

=

b

a x x

x yV δ π δ

2

0lim

which gives

∫= ba

dx yV 2π

Example 33:

Find the volume of the solid formed when the area

between the curve 22+= x y and the x - axis

from 1= x to 3= x is rotated through π 2

radians about the x - axis.

Solution:

The volume V is given by

∫=3

1

2dx yV π

Now the ( ) 442 24222++=+= x x x y . Therefore,

( )dx x xV 44243

1++= ∫π

3

1

35

43

4

5

++= x

x xπ

( ) ( )( )

++−

++= 4

3

4

5

134

3

34

5

335

π

−=

15

83

5

483π





∴ 15

1366π =V

Example 34:

The area enclosed between the curve2

4 x y −= and the

line x y 24 −= is rotated through π 2 radians about the

x - axis. Find the volume of the solid generated.

Solution:

The sketch of both the curve and the line on the same set of axes show the area to be rotated.

The required volume V is given by

( ) ( ) dx xdx xV 22

0

22

0

2244 ∫∫ −−−= π π

( ) ( ) dx x x∫

−−−=2

0

222 244π

( )dx x x x∫ +−=2

0

24 1612π

2

0

235

845

+−= x x

xπ

∴ 5

32π =V

Volume of revolution about the y - axis

The volume of the solid of revolution formed by rotating an area through π 2 radians about the

y - axis can be found in a way similar to that the x - axis. The volume of such a solid of

revolution is given by

∫=b

ady xV

2π

Remember that dy implies that the limits a and b are y limits.

Example 35:





Find the volume of the solid formed when the area between the curve

3 x y = and the y - axis from 1= y to 8= y is rotated through π 2

radians about the y - axis.

Solution:

The required volume V is given by

∫=8

1

2dy xV π

Now3

x y = . Therefore,

( )3

2

3

2 3 x y =

∴ 23

2

x y =

So,

∫=8

1

3

2

dy yV π

8

1

35

5

3

= yπ

( ) ( )

−= 3

5

3

5

1

5

38

5

3π

∴ 5

93=V

Example 36:

Find the volume generated when the region bounded by the curve2

x y = , the x - axis and

the line 2= x , is rotated througho

180 about the line 3= x .

Solution:

The sketches show the regions to be rotated.

The required volume, V , is found by first calculating the

volume, 1V , of the solid generated when the area between

the curve

2

x y =

, the x - axis and the line 2=

x , is





rotated througho180 about the line 3= x , and

subtracting the volume, 2V , of the half–cylinder with

radius (3-2) = 1 and the height 4.

Volume 1V is given by

( )∫ −=4

0

21 3

2dy xV

π

( )∫ −=4

0

23

2dy y

( )∫ +−=4

0 69

2

dy y y

( )∫ +−=4

0 69

2dy y y

π

4

0

2

249

22

3

+−=

y y y

π

( ) ( )[ ]0832362

−+−=

∴ 61 =V

The volume, 2V ,of the half – cylinder is given by

( ) π π

2412

22 ==V

Therefore, the required volume, V , is given by

21 V V V −=

π π 26 −=

π 4=

The volume of the solid generated is π 4 .

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Chap 2 Applications of Integration