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I need help starting with this. I can't find an example like this anywhere in my book

$\int_0^4 \int_{\sqrt {x}}^2 \frac{x^2 e^{y^2}}{y^5}\mathrm dy\,\mathrm dx$

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Hint: $ \int_{x = 0}^4 {\int_{y = \sqrt x }^2 {\frac{{x^2 e^{y^2 } }}{{y^5 }}\,dy} \,dx} = \int_{y = 0}^? {\int_{x = 0}^? {\frac{{x^2 e^{y^2 } }}{{y^5 }}\,dx} \,dy}. $

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    Indeed, this is right.2011-05-15
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Hint: Try the substitution $y^2=u$, then $2ydy=du$. Hence you get $\int_{\sqrt{x}}^2\frac{e^{y^2}}{y^5}dy=\int_{x}^4 \frac{e^{u}}{2u^3}du$

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    From here on, things start to become ugly; the integrand doesn't seem to have a nice primitive, unlike functions of the form $u^n e^u$ for $n \geq 0.$2011-05-15
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So you can do this. Multiply $y$ in the numerator and the denominator so that your function becomes $\frac{x^{2} e^{y^{2}} y}{y^{6}}$ and then substitute $y^{2}=t$ to integrate.