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# Applications of Integrations

Date post: 16-Apr-2017
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• T- 1-855-694-8886Email- [email protected] iTutor.com

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• Applications 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.

• Area under a CurveTo find the area under a curve. This expression gives us a definite value (a number) at the end of the calculation.

When the curve is above the x axis, the area is the same as the definite integral :

y= f(x)

• 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 =

• (Positive) (Negative)

• 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] 223-2 iTutor. 2000-2013. All Rights Reserved

• 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)

232-2

• f(x)
• Case3: The area a between f, the X-axis and the lines x = -2 and x = 3 is :223-2

• Area Bounded by 2 Curves

Area under f(x) =

Area under g(x) = Say you have 2 curves y = f(x) and y = g(x) iTutor. 2000-2013. All Rights Reserved

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

• 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 iTutor. 2000-2013. All Rights Reserved

• 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

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

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

• Volumes of Revolution :V= f2(x) dxA solid of revolution is formed when a region bounded by part of a curve is rotated about a straight line.

• 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

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 iTutor. 2000-2013. All Rights Reserved

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

The volume of solid is:

• 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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