Is this integral setup correctly

y= 1-2x^2 y=|x|

A=2 integral(1/2,0) [1-2x^2-x]

getting the limits by

1-2x^2 = x x>0

(2x-1)(x+1)

x = +- 1/2

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- December 15th 2009, 05:39 AMswatpup32Area between two curves with abs value.
Is this integral setup correctly

y= 1-2x^2 y=|x|

A=2 integral(1/2,0) [1-2x^2-x]

getting the limits by

1-2x^2 = x x>0

(2x-1)(x+1)

x = +- 1/2 - December 15th 2009, 05:55 AMJester