Double Integration Check: Evaluating a Double Integral for a Triangle

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More of a check

Homework Statement


Evaluate \int\int (x^{2}(y+1) + e^{y}) for a triangle with vertices at (0,0) (1,1) (-1,1)


Homework Equations


Integration

The Attempt at a Solution


The only this I wanted to check, is that integrating the square at y=0 to 1 and x=0 to 1 will give the correct answer (as the triangle is reflected on the y axis) and if not, how you get the limits.
 
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I don't see how "integrating the square at y = 0 to 1 and x = 0 to 1" relates to this problem, even if it happens to produce the right answer.

If you integrate first with respect to x and then with respect to y, your limits of integration will be x = - y to x = y, and then y = 0 to y = 1.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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