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Page 1: Applications of Integrations

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Page 2: Applications of Integrations

Applications Of The Definite IntegralApplications Of The Definite IntegralThe Area under the curve of a functionThe area between two curvesThe Volume of the Solid of revolution

In calculus, the integral of a function is an extension of the concept of a sum. The process of finding integrals is called integration. The process is usually used to find a measure of totality such as area, volume, mass, displacement, etc.

The integral would be written f(x) . The ∫ sign represents integration, a and b are the endpoints of the interval, f(x) is the function we are integrating known as the integrand, and dx is a notation for the variable of integration. Integrals discussed in this project are termed definite integrals. © iTutor. 2000-2013. All Rights Reserved

Page 3: Applications of Integrations

Area under a CurveArea under a Curve

)()()()( aFbFxFdxxf ba

b

a

y= f(x)

Area = b

a

dxxf )(

x

Y

x = a x= b© iTutor. 2000-2013. All Rights Reserved

Page 4: Applications of Integrations

But when the graph line is below the ‘x’ axis, the definite integral is negative. The area is then given by:

y= f(x)

Area = b

a

dxxf )(

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Page 5: Applications of Integrations

(Positive)

(Negative)

210

21

21

1

0

1

0

2

xxdx

1

1

11

21

210

21

0

1

0

1

2

xxdx

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Page 6: Applications of Integrations

Example 1:let f (x)=2-x .

Find the area bounded by the curve of f , the x-axis and the lines x =a and x=b for each of the following cases:

a = -2 b = 2a = 2 b = 3a = -2 b = 3

The graph:Is a straight line y=2-x:F (x) is positive on the interval [-2, 2)

F (x) is negative on the interval (2, 3]

2

2 3-2

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Page 7: Applications of Integrations

Case 1:The area A1 between f, the x-axis and the lines x = -2 and x = 2 is:

f(x)>0; x [-2,2)

862

)244()

244( )

22(

)2(

2

)(

2

2

2

2

2

2

2

2

21

xx

dxx

dxx

dxxfA

2

32-2

A1

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Page 8: Applications of Integrations

f(x)<0; x (2, 3]

Case2:The area A2 between f, the x-axis and the Lines x=2 and x=3 is:

2/1

)244()

296(

)2

2( )2(

2

)(

32

23

2

3

2

3

21

dxxxdxx

dxx

dxxfA

3

2

2-2

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Page 9: Applications of Integrations

Case3: The area a between f, the X-axis and the lines x = -2 and x = 3 is :

2/172/18

)2()2(

2

3

2

2

2

3

2

dxxdxx

dxx2

2

3-2

3

2

)( dxxf

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Page 10: Applications of Integrations

Area Bounded by 2 CurvesArea Bounded by 2 Curves

Area under f(x) =                               

Area under g(x) =                               

b

a

dxxf )(

b

a

dxxg )(

Say you have 2 curves y = f(x) and y = g(x)

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Page 11: Applications of Integrations

Superimposing the two graphs, Area bounded by f(x) and g(x)

b

a

b

a

b

a

dxxgxf

dxxgdxxf

)()(

)()(

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Page 12: Applications of Integrations

Example (2) Let f (x) = x , g (x) = x5

Find the area between f and g from x = a to x = b Following casesa = -1 b = 0a = 0 b = 1a = -1 b = 1

g(x)>f (x) on (-1,0) and hence on this interval, we have: g (x) –f (x)>0

So |g (x) – f (x)| = g (x) - f (x) = x5 - x

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Page 13: Applications of Integrations

Case (1):The area A1 between f and g from X= -1 and x=0 is:

g (x)>f (x) on (-1,0) and hence on this interval, we have :

g (x) –f (x) > 0 So

|g (x) –f (x)| = g (x) - f (x) = x5 - x

3/1 )2/16/1()00(

)2/6/(

)(

)()(

0

1

26

0

1

5

0

11

xx

dxxx

dxxfxgA 1

1

11

gf

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Page 14: Applications of Integrations

Case (2) The area A between f and g from x = 0 to x = 1

f(x) > g (x) on(0,1) and hence on this interval,we have f (x) –g (x)>0 so

|g (x) –f (x)| =f (x) –g (x) = x - x5

31 00

61

21

6/2/

)()(

1

0

62

1

0

5

1

02

xx

dxxx

dxxfxgA1

1

11

gf

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Page 15: Applications of Integrations

Case (3) The area A between f and g from x = -1 to x =1

3/2 3/13/1

)()(

)()(

1

0

50

1

5

1

13

dxxxxx

dxxfxgA

1

1

11

gf

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Page 16: Applications of Integrations

Volumes of Revolution :Volumes of Revolution :V=Π ∫fV=Π ∫f22(x) (x) dxdx A solid of revolution is formed when a region bounded

by part of a curve is rotated about a straight line.

Rotation about x-axis:

Rotation about y-axis:

dxyVb

a 2

dyxVd

c 2

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Page 17: Applications of Integrations

Example: Find the volume of the solid generated by revolving the region bounded by the graph of

y = x, y = 0, x = 0 and x = 2. At the solidSolution:

Volume

3/8

]3/[(

)(

20

32

0

2

2

0

2

22

1

xdxx

dxxdxxfx

x

we shall now use definite integrals to find the volume defined above. If we let f(x) = x according to 1 above, the volume is given by the definite integral

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Page 18: Applications of Integrations

Example 1:1Consider the area bounded by the graph of the function f(x) = x – x2 and x-axis:

The volume of solid is:

30/)5/04/03/0()5/14/23/1(

)5/23/(

)2(1

0

533

431

0

2

xxx

dxxxx

1

1

0

22 )( dxxx

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Page 19: Applications of Integrations

• In conclusion, an integral is a mathematical object that can be interpreted as an area or a generalization of area. Integrals, together with derivatives, are the fundamental objects of calculus. Other words for integral include anti-derivative and primitive.

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