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
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
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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(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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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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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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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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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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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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Superimposing the two graphs, Area bounded by f(x) and g(x)
b
a
b
a
b
a
dxxgxf
dxxgdxxf
)()(
)()(
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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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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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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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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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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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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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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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• 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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