Double Integral over General Region : Hass Section 13.2 - Problem 5

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Homework Statement



Outer Integral: From zero to one dy
Inner Integral: from zero to y^2 dx

Function is: 3y^3 * e^(xy)


Homework Equations


None


The Attempt at a Solution



Have tried numerous u substitutions on e^(xy), but taking me nowhere. I am clearly doing something wrong. Assuming 3y^3 is a constant and does not need to be integrated when integrated with respect to x.

Solutions manual shows result of inner integral being [3y^2 * e^(xy)] from zero to y^2 - which appears to me that a y in the original 3y^3 simply disappeared! No idea how they are getting from 3y^3 to 3y^2 as the result of the first integration!

I am quite sure this is an easy problem and I am simply overlooking a very simple step.
 
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You need to use integration by parts.
 
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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