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 April 28th, 2012, 06:01 AM #1 Member   Joined: Mar 2012 Posts: 52 Thanks: 0 Find the area 1- Find the area of the region bounded by y=x^3-3x^2+2x , the x-axis, x = 0, and x = 2.
 April 28th, 2012, 06:40 AM #2 Newbie   Joined: Apr 2012 Posts: 11 Thanks: 0 Re: Find the area Calculate the integral from x= 0 to x=2 and you'll get that the integral equals 0.
 April 28th, 2012, 07:01 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Find the area Since the area is being asked for, you need to compute: $A=\int_0\,^2 \|x^3-3x^2+2x\|\,dx$ You need to find the roots of the integrand and determine on which intervals it is positive and on which it is negative. $x^3-3x^2+2x=0$ $x$$x^2-3x+2$$=0$ $x(x-1)(x-2)=0$ $(0,1)$ positive $(1,2)$ negative Hence: $A=\int_0\,^1 x^3-3x^2+2x\,dx-\int_1\,^2 x^3-3x^2+2x\,dx$ With a translation of axes, replacing x with x + 1, we find the integrand is odd about the line x = 1, and we may use this symmetry to write: $A=2\int_0\,^1 x^3-3x^2+2x\,dx$
 April 28th, 2012, 09:22 AM #4 Member   Joined: Mar 2012 Posts: 52 Thanks: 0 Re: Find the area Thanks! The Area= 1/2 U^2

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