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

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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Using this, you should easily find that the iterated integral evaluates to $\frac{{e^4 - 1}}{6}$. –  Shai Covo May 15 '11 at 17:04
thank you this helps. –  guge May 15 '11 at 17:40
$\int_{0}^{2}\int_{0}^{y^2}dx dy$ does this seem right? –  guge May 15 '11 at 20:38
Indeed, this is right. –  Shai Covo May 15 '11 at 20:47
Hint: Try the substitution $y^2=u$, then $2ydy=du$. Hence you get $$\int_{\sqrt{x}}^2\frac{e^{y^2}}{y^5}dy=\int_{x}^4 \frac{e^{u}}{2u^3}du$$
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 $n \geq 0.$ –  Gerben May 15 '11 at 16:29
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.