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July 30, 2014

July 30, 2014

Posted by **sol** on Thursday, April 5, 2012 at 10:11am.

f(x,y)=2x^2 y;R is the region bounded by the graphs of y=x and y=x^2

- calculus -
**Steve**, Thursday, April 5, 2012 at 2:45pmv = ∫∫R f(x,y) dA

= ∫[0,1]∫[x,x^2] 2x^2 y dy dx

= ∫[0,1] (x^2 y^2)[x,x^2] dx

= ∫[0,1] (x^6 - x^4) dx

= (1/7 x^7 - 1/5 x^5)[0,1]

= 1/7 - 1/5

= -2/35

- calculus - PS -
**Steve**, Thursday, April 5, 2012 at 2:47pmOops. That sign is reversed, since y f goes from x^2 to x.

Answer is 2/35

Couldn't figure how I got a minus value for a function that is positive over the region.

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