Applications of Integrations

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Applications of Integrations

  1. 1. T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com
  2. 2. Applications Of The Definite IntegralApplications Of The Definite Integral The Area under the curve of a function The area between two curves The 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
  3. 3. Area under a CurveArea under a Curve [ ] )()()()( aFbFxFdxxf b a b a −==∫ To 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) Area = ∫ b a dxxf )( x Y x = a x= b © iTutor. 2000-2013. All Rights Reserved
  4. 4. 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 )( © iTutor. 2000-2013. All Rights Reserved
  5. 5. (Positive) (Negative) 2 1 0 2 1 2 1 1 0 1 0 2 =−=      =∫ xxdx 1− 1 11− 2 1 2 1 0 2 1 0 1 0 1 2 −=−=      = −− ∫ xxdx © iTutor. 2000-2013. All Rights Reserved
  6. 6. 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 = 2 a = 2 b = 3 a = -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 © iTutor. 2000-2013. All Rights Reserved
  7. 7. 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 ) 2 4 4() 2 4 4() 2 2( )2( 2 )( 2 2 2 2 2 2 2 2 2 1 =+= −−−−=−= −= −= = − − − − ∫ ∫ ∫ x x dxx dxx dxxfA 2 32-2 A1 © iTutor. 2000-2013. All Rights Reserved
  8. 8. 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 ) 2 4 4() 2 9 6( ) 2 2()2( 2 )( 3 2 23 2 3 2 3 2 1 = +−−+−= +−=−−= −= = ∫ ∫ ∫ dx x xdxx dxx dxxfA 3 2 2-2 © iTutor. 2000-2013. All Rights Reserved
  9. 9. Case3: The area a between f, the X-axis and the lines x = -2 and x = 3 is : 2/17 2/18 )2()2( 2 3 2 2 2 3 2 = += −−+−= −= ∫∫ ∫ − − dxxdxx dxx 2 2 3-2 ∫− 3 2 )( dxxf © iTutor. 2000-2013. All Rights Reserved
  10. 10. 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) © iTutor. 2000-2013. All Rights Reserved
  11. 11. Superimposing the two graphs, Area bounded by f(x) and g(x) ∫ ∫∫ −= − b a b a b a dxxgxf dxxgdxxf )()( )()( © iTutor. 2000-2013. All Rights Reserved
  12. 12. Example (2) Let f (x) = x , g (x) = x5 Find the area between f and g from x = a to x = b Following cases a = -1 b = 0 a = 0 b = 1 a = -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
  13. 13. 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 1 1 = −−−= −= −= −= − − − ∫ ∫ xx dxxx dxxfxgA 1− 1 11− g f © iTutor. 2000-2013. All Rights Reserved
  14. 14. 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 ( ) ( ) ( ) 3 1 00 6 1 2 1 6/2/ )()( 1 0 62 1 0 5 1 0 2 =−−      −= −= −= −= ∫ ∫ xx dxxx dxxfxgA 1− 1 11− g f © iTutor. 2000-2013. All Rights Reserved
  15. 15. Case (3) The area A between f and g from x = -1 to x =1 3/2 3/13/1 )()( )()( 1 0 5 0 1 5 1 1 3 = += −+−= −= ∫∫ ∫ − − dxxxxx dxxfxgA 1− 1 11− g f © iTutor. 2000-2013. All Rights Reserved
  16. 16. Volumes of Revolution :Volumes of Revolution :V=Π ∫fV=Π ∫f22 (x) dx(x) dx  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: dxyV b a ∫Π= 2 dyxV d c ∫Π= 2 © iTutor. 2000-2013. All Rights Reserved
  17. 17. 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 solid Solution: Volume [ ] 3/8 ]3/[( )( 2 0 3 2 0 2 2 0 2 2 2 1 π ππ ππ = == == ∫ ∫∫ xdxx dxxdxxf x 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 © iTutor. 2000-2013. All Rights Reserved
  18. 18. Example 1:1 Consider 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 43 1 0 2 π ππ π π = +−−+−= +−= +−= ∫ xxx dxxxx 1 ∫ − 1 0 22 )( dxxxπ © iTutor. 2000-2013. All Rights Reserved
  19. 19. • 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. Call us for more Information: www.iTutor.com 1-855-694-8886 Visit

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