BA201 Engineering Mathematic UNIT 5 AppOfIntegration

TOPIC 5

APPLICATION OF INTEGRATION

1

PREPARED BY SITI AMINAH BINTI ABD HALIM

5.0 AREA AND THE DEFINITE INTEGRAL

An intuitive look

Suppose a certain states annual rate of petroleum consumption over a 4-year

period is constant and given by the function f(t) = 1.2t for (0

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5.1 AREA BOUNDED BETWEEN FUNCTION AND THE X-AXIS.

If f is a nonnegative continuous function on [a, b], then the area under the

graph of f(x) on the interval is

A= b

a

dxxf )(

Suppose now f(x) 0 for all x in [a, b] as shown in fig. Since f(x)0 we

define the area bounded by the graph of y = f(x) and the x-xis from x=a to

x=b to be the area A under the graph of y = -f(x) on [a, b], Then

A = b

a

dxxf )(

= - b

adxxf )(

x

y

b a

y =f(x)

y

x b a

f(x)

A

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Note:

The area of the region above the x-axis is positive while the area of the region

below the x- axis is negative

5.2 AREA BOUNDED BETWEEN FUNCTION AND THE Y-AXIS.

(a)

Area of the shaded region

.

(+)

(-)

a

b

y

x

x=f(y)

The area of the region to the right of the y-axis is positive

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(b)

Example 5.1

Find the area bounded by the graph of y = x2 and the x-axis on [0, 2]

Solution

We have A=- b

adxxf )( , then

A = - 2

0)( dxxf =-

2

0

2 dxx

= 2

0

3

3

x =

3

8 unit2

b

a

y

x

x=f(y)

The area of the region to the left of the y-axis is negative

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Example 5.2

Find the area bounded by the graph and x-axis

Solution

a. Shaded area = 6

7

6

1

6

8

62

2

1

32

1

2

xdx

x unit2

b. Shaded area = 1273

4

2

3

22

3

1

2

3

3

1

2

1

x

dxx unit2

a.

b.

2 1 1 3 x x

y y

Fig 5.2a Fig 5.2b

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Example 5.3

Find the area of the shaded region

Solution

a. Area of shaded region = dxxxdxxx )4

0

24(4

0 )4(

=

4

03

322

xx =

3

6432 =

3

32 unit2

b. Area of shaded region = 5

1 )5)(1.( dxxx

= dxxx )56(5

1

2

=

5

1

5233

3

xx

x

=

53

3

12575

3

125 =

3

32 unit2

a.

y = x(4 x)

0 4 x

b.

y = (x-1)(x-5)

0 1 5

4 0

x

y y

x

Fig 5.3a Fig 5.3b

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For (b) we can also calculate using the negative formulae,

Area of the shaded region A=- 5

1 )5)(1.( dxxx =

3

32 unit2

Example 5.4

Find the area bounded by the graph of y = x2 + 2x and the x-axis on [-2, 2]

Solution The graphs of y = f(x) and y = )(xf on the interval [-2, 2] are given

in Fig 5.4

Now

)(xf = )2(

2

2

2{xx

xx

Therefore,

A = 2

2)( dxxf

= 2

0

0

2)()( dxxfdxxf

= 2

0

20

2

2 )2()2( dxxxdxxx

= 2

0

23

0

2

23

)3

)3

xx

xx

= 0 ( )43

8()4

3

8( -0 = 8 unit2

x

y

y = x2 + 2x

-

2 2

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Example 5.5

Find the area bounded by the graph y = sin x and the x-axis on [0, 2]

Solution

The area = dxx2

0sin

As the graph consists of areas above and below the x-axis, then,

A =

2

0)sin(sin dxxxdx

= 20 coscos xx

= -cos + cos 0 + (cos 2 cos )

= -(-1) +1 + (1 - (-1)) = 4 unit2

5.3 Area between Two Graphs

Sometimes, we have to break parts of the diagram (shaded region) to find its area.

Example 5.6


2xy

2 xy

0 1 2

Fig. 5.5

x

y

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Solution

In fig.5.5, the shaded region lies under two different functions. To find its area, we

have to break the region into two under the two functions.

Shaded area = 1

0

2dxx Shaded area = 2

1)2( x dx

Therefore,

The area of the shaded region = sum of two regions under different functions

The area of the shaded region = 2

1

1

0

2 )2( dxxdxx

` =

2

1

21

0

3

223

x

xx

=

2

2

1420

3

1

= 6

5 unit2

x

y = -x + 2

y

y

2xy

x

= dxx1

0

2

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0

4

2yx

5.4 Area Bounded between Function and the y-axis

Figure 5.6 a Figure 5.6 b

Fig. 5.6a shows a region bounded by the function and y-axis on {a b].

Area of shaded region =

i

i

ix

dyxhad

10

= b

adyx

Example 5.7

Find the area of the shaded region.

y

b

a

0 x

y

b

yi+y

yi

a

x

x i

b.

y

x=y2-4y+5

3

1

x

0

a.

y

2 y 2 = x

y2=xx4x

1

0 x

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1

a. y

x=y2-4y+3

0 x

Solution

a. Area of the shaded region = dyy

dyx 2

1

22

1 3

=

2

1

3

3

y=

3

1

3

8 =

3

7 unit2

b. Area of the shaded region = 3

1

23

1 )54( dyyydyx

= 3

1

23

523

yy

y

=

52

3

115189 =

3

8 unit2

Example 5.8


Solution

a. Area of shaded region = 2

1

2 )34( dyyy

=

2

1

23

323

yy

y

=

32

3

168

3

8 =

3

2 unit2

b. y

x =-y2+4y-5

0 x

3

3

1

2

1

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2

b. Area of shaded region = 3

1

2 )54( dyyy

=

3

1

23

523

yy

y

=

52

3

115189

= 3

8 unit2

From the examples of 5.6a and 5.6b the area on the left of the y-axis is a negative

value.

ATTENTION :

5.5 AREA BETWEEN TWO GRAPHS

The area under the graph of a continuous nonnegative function y = f(x) on [a,b] is the area

of the region bounded between its graph and the graph of the function y=0 from x = a and

x=b . Suppose f(x) and g(x) are continuous on [a, b] and f(x) >g(x) for all x in the interval [a,

b], the area of the region A between the two graphs x=a and x=b is given by

A = dxxgxfb

a )()(

If an area is valued negative, the region is situated on the left

of the y-axis

Area between 2 functions

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3

Example 5.9

Find the area of the region bounded by the graphs of y = x and y = x2

Solution :

As shown in above figure, the area in question is located in the first quadrant, and the

graph intersect at the points (0, 0) and (1, 1); (that is , 0 and 1 are the solutions of x2 = x ).

Since y = x is the upper graph on the interval (0, 1) it follows that

A = dxxx 1

0

2

= ( 10

3

2

3

]33

2 xx

= 03

1

3

2

= 3

1sq units

Shaded region

y =

x

y

(0,0)

(1,1)

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4

5.6 Find The Generated Volume.

Figure 5.7

For some objects as above, we know how to find the volume by using certain

formula.

What is the formula to find the volume of a cylinder?

What is the formula to find the volume of a cone?..........

How much is the volume of the pot?

So, we will discuss how to find the volume of rotating solids by using integration. For example, to find the volume of the pot

Identify the axis symmetry and assume it is represented by the x-axis.

Choose a function f(x) which represents the side of the pot.

Find the approximation of the volume by dividing an object into many disc circles.

y y=f(x)

x

a

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5

Lets say, y = f(x) is the equation that represents the sides of the pot and the stripe

disc has radius y.

The volume of the stripe disc = yi2 x

The total volume of all discs is one approximation to the volume of the pot.

The volume of the pot yi2 x

The total of discs volume will approximate the real volume of the pot, if the width

x of each disc becomes smaller or the number of discs approximate to .

The volume of the pot = dxydxyita

i

i

i

ix

0

2

1

2

0lim

yi

x

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6

5.7 The Generated Volume When a Region Is Rotated 360o around the X-Axis

X-axis as symmetry axis.

Figure 5.8

The volume of the pot is the same as the volume revolved when the shaded region

in Figure 13.2 is rotated 360o around the x-axis.

Generated volume = a

0

2dxy

y

x a

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7

Example 5.10 Find the volume generated by the areas bounded by the given curves if they are

revolved 3600 about the x-axis

Solution:

a. Generated volume = 2

0

2dxy

= 2

0 dx )x2(

= dx x 22

0

=

2

0

2

2

x2

= )04(

= 4 unit3

b. Generated volume = 2

1

2dxy

= dx2

x2

2

1

2

= 2

1

4dxx4

=

2

1

5

5

x

4

= 12

20

5

= 20

31 unit3

b. y

2

2xy

0 1 2 x

a. y y2 = 2x 0 2 x

Substitute y2

with 2x

Substitute y2

with

22

2

x

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8

5.8 The Generated Volume When a Region Is Revolved 360o around the Y-Axis

x= g(y)

To find the generated volume when a shaded region as above is rotated 360o around

the y-axis, we choose one function x = g(y) to represent the curve.

Generated Volume = a

0

2dyx

Example 5.11

Find the generated volume when these shaded regions are rotated 3600 around the

y-axis

a.

y

y = 4x2

2

x

0

x

b.

y

y = 2x - 2

3

0 x

-1

x

y

a

0

Figure 5.9

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9

Solution:

a. Volume =

2

0

2

0

2 dy4

ydyx

= 2

0ydy

4

=

2

0

2

2

y

4

=

0

2

2

4

2

= 2

unit3

b. Generated volume = dy2

2ydyx

23

1

3

1

2

= dy4y4y4

3

1

2

=

3

1

23

y42

y4

3

y

4

=

)1(4)1(2

3

)1()3(4)3(2

3

3

4

23

23

=

42

3

112189

4

= 3

31 unit3

Substitute x2

with

and 4 can be moved outside before we do the integration

Substitute x2

with

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0

5.9 Volume By Rotating The Area Enclosed Between a Curve and a Line.

y = f(x)

Figure 5.10

To find the generated volume when a shaded region as above is rotated 360o around

the x-axis, we choose one function y= g(y) to represent the curve and one function

y = f(x) to represent the line.

Generated volume = a

0

22 dx)x(f)x(g

Example 5.12

Find the generated volume when the shaded region bounded by a straight line y =

x, a curve y2 = x 2 , straight line x = 4 and x-axis is rotated 360o around the x-axis.

y

y = x

y2

= x-2

0 2 4 x

x

y y= g(x)

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1

Solution

The shaded region is shown as in Figure 5.11

Figure 5.11a Figure 5.11b Figure 5.11c

Lets say 4

0

2dxx is the generated volume of the shaded area in Figure 5.11b is

rotated 360o around the x-axis and 4

2)2( dxx is the generated volume when

the shaded region in Figure 5.11c is rotated 360o around the x-axis.

So, the generated volume when the shaded region in Figure 5.11a is rotated 360o

around the x-axis is the same as I1 I2

Generated volume = I1 I2

= 4

0

2dxx - 4

2dx)2x(

=

4

2

24

0

3

x22

x

3

x

=

)2(2

2

2)4(2

2

40

3

4 223

= 3

58 unit3

y

y = x

y2=x-2

0 2 4 x

y y = x

0 4 x

Isipadu janaan I1=

y

y2=x-2

0 2 4 x

Isipadu janaan I2=

4

2)2( dxx

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2

Activity 5(a)

1. Find the area of the shaded region in each graph below.

e. y

y = x(x+1)(2-x)

-1 0 2 x

a.

y = 2x(2-x)

0 2

x

y b.

y = x2-6x+8

y

x

c. y

y=x2-1

-1 1

x

d.

y y = 2x2 4x

-1 0 2 3

-1 0 2 3

x

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2

3

2. Calculate the area of the shaded region in each of the following diagrams.

a.

x = 6y-y2

0

x

6

b.

x =(y-1)(y-4)

x

4

1

c. y

x = y(y+1)(y-1)

-1

x

1

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2

4

Feedback for Activity 5(a)

1. a. 23

8unit

b. 23

4unit

c. 23

4unit

d. 23

16unit

e. 212

37unit

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2

5

y

24 xy

xy 3

0 1 2 x

Activity 5(b)

1. Find the area of the shaded region in each graph.

a.

Feedback for Activity 5(b)

a. 26

19unit

b. 26

45unit

b. y

x

A B

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2

6

Activity 5(c)

Find the area of the shaded region in each of the following.

a.

y y=x2-1

3

1

0 x

b.

y

2

1

0 x

(c) y

4

1

0 x

d.

y

y2=x+1

2

1

0 x

1

2

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2

7

Feedback for Activity 5(c)

a. 22 45.3)88(3

2unitunit

b. 23

13unit

c. 24

3unit

d. 2 unit 2

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2

8

Activity 5(d)

1. Find the coordinates of the intersection of the curve y2 = x and a linear graph

2

xy . And then, find the area bounded by the curve and the linear equation.

2. Find the area of y the region bounded by the graphs of y = x2 + 2x and y = -x + 4 on [-4, 2]

Feedback for Activity 5(d)

1. Points of intersection (0, 0) and (4, 2)

Area of the bounded region = 23

4unit

1. Points of intersection (-4, 8) and (1, 3)

Area of the bounded region = 3

71sq units

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2

9

Activity 5(e)

1. Find the generated volume when these shaded regions are rotated 3600 around

the x-axis

2. A region bounded by a curve y2 = x + 1, y-axis and a line x = 4 is rotated 360o

around the x-axis. Find the generated volume in .

a. y y = x +1 0 1 2 x

b. y

x

2y

0 1 4 x

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3

0

Feedback for Activity 5(e)

1. a. 2

1

2dxy

= 2

1

2 dx)1x(

=

3

)1x( 3

=

3

)11(

3

)12( 33

=

3

8

3

27 3

= 3

19 unit 3

b. 3 unit 3

2. Volume = 4

0

2dxy

= 4

0

dx)1x(

=

4

0

2

2

)1x(

=

2

)10(

2

)14( 22

= 12 unit 3

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3

1

Activity 5(f)

1. Find the generated volume when these shaded regions are rotated 3600 around the

y-axis

2. A sphere is formed when the area bounded by a curve y = 2x4 and the x-axis as shown below is rotated around the x-axis. Find the volume of the sphere if the radius is 2.

y

y = 2x4 -2 2 x

a. y

y = x2 - 1

2

0

b.

y

3

y = 3 x2

0

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3

2

Feedback for Activity 5(f)

1. a. Volume = 2

0

2dyx

= 2

0

dy1y

=

2

0

2

2

)1y(

=

2

)10(

2

)12( 22

= 4 unit 3

b. 4.5 unit 3

2. 2

3 unit 3

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3

3

Activity 5(g)

1. Find the generated volume when this shaded region are rotated through 3600 around

the x-axis.

2. Find the generated volume when this shaded region are rotated through 3600 around

the y-axis

y y =2x-2

y2=2x

0 1 2 x

y

y=x

y2=x-1

1

0 x

-1

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3

4

Feedback for Activity 5(g)

1. Volume =

2

1

2

2

0

dx)2x2(xdx2

=

2

1

232

0

2 x4x4x3

4x

=

)1(4)1(4)1

3

4)2(4)2(4)2

3

44 2323

=

3

44

= 3

8 unit 3

2. Generated volume =

1

0

2

1

0

22 dyydy)1y( +

1

0

2dyy

=

1

0

2

1

0

24 dyydy)1y2y( +

1

0

3

3

y

=

1

0

31

0

35

3

yy

3

y2

5

y+

1

0

3

3

y

= 15

19 unit 3 +

3

unit 3

= 8 unit 3

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3

5

SELF ASSESSMENT 5 (a)

1. Find the area of the bounded region

d.

y

y = x2 -4x + 5

y

x

0 1 3

c.

y

y = 2x

y = 4x - x2

x

0 2

b.

y

y = 6 - 2x

x

a.

y

y = 6 x

x

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3

6

SOLUTION : SELF ASSESSMENT 5 (a)

1. a. 26

1unit

b. 23

7unit

c. 23

4unit

d. 23

4unit

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3

7

SELF ASSESSMENT 5 (b)

1. Find the generated volume when these shaded regions are rotated 3600 around the x-axis

a. b.

y = x2

y = 2x - x2

x

1

2. Sketch and determine the volume of the revolving solids when these areas are rotated around

the x-axis.

a. area under a graph y = 3x + 3 from line x = -1 and x = 2.

b. area under a graph y = x(3 x) from line x = 0 and x = 3.

3. Find the generated volume when these shaded regions are rotated 3600 around the x-axis :

y

0 2

y

x

x

y

y = -2x

y

1 3

y=1/x

x

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3

8

SOLUTION : SELF ASSESSMENT 5 (b)

1. a. 35

unit

b. 315

16unit

2. a. 72 unit3 b. 37

4unit

3. a. 33

1unit b. 3

3

8unit

Documents

BA201 Engineering Mathematic UNIT 5 AppOfIntegration