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 September 18th, 2011, 05:28 PM #1 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 area between curves Find the area between the curves f(x)= x^2 -4 and g(x)= -x^2+4 and the vertical lines x=-3 and x=3.
 September 18th, 2011, 05:51 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,389 Thanks: 2015 That's a bit unclear. I think the intention is to omit the area between the curves for |x| < 2, leaving $2\int_2\,^3((x^2\,-\,4)\,-\,(-x^2\,+\,4))\,dx,$ which equals 28/3.
 September 18th, 2011, 10:47 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: area between curves If the intention is to include all area for |x| < 3, we would add to [color=#00BF00]skipjack[/color]'s area with (and using the same symmetry of an even function taken over an interval symmetric about the functions' axis of symmetry.): $2\int_0\,^2 $$\(-x^2+4$$-$$x^2-4$$\)\,dx=4\int_0\,^2 4-x^2\,dx=4$4x-\frac{x^3}{3}$_0^2=4$$8-\frac{8}{3}$$=32$$\frac{2}{3}$$=\frac{64}{3}$ Thus, the total area A would be: $A=\frac{64+28}{3}=\frac{92}{3}$
 October 2nd, 2011, 07:44 PM #4 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: area between curves where did you get the 28 (in 64+2
 October 2nd, 2011, 08:16 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: area between curves I was adding the result [color=#00BF00]skipjack[/color] obtained in the post above the one you are referring to.
 October 2nd, 2011, 08:32 PM #6 Senior Member   Joined: Jan 2010 Posts: 324 Thanks: 0 Re: area between curves Oh ok, I see.

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