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I have the double integral ∫∫x dA bounded by the curves y=1, y=-x, and y = √x

I drew out the graph, but I'm having trouble determining what the bounds for the integrals are. Is x from -y to y^2, and y from 0 to 1, or is x from 0 to 1 and y from -x to √x (or vice versa)?

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    The problem is that I don't believe the top bound for both integrals is 1 as you said, so I'm confused if you're right with your hints or not2011-12-14

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$\int_0^1 \int_{-y}^{y^2} dx \ dy$ is what you would get with horizontal "representative rectangles". So in this case, the inner integration gives:

$\int_0^1 (y^2 - (-y)) dy$, which is the integral I hinted at.

If you want to integrate dx,

$A = \int_{-1}^0 (1 - (-x)) dx + \int_0^1 (1 - \sqrt x) dx$

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    @TheChaz =) That was my reaction to that. I discovered that many people tend to get a little worked up when discussing maths, kind of looking for the other's mistake but not providing their insight, correction or proof. I guess it's an aspect to work on.2012-02-21