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Solve the integral $\int_0^1\int^1_xy^4e^{xy^2}dydx$.

I think that variables substituation is neede here. I've substitute $$ \\ \left\{\begin{matrix} u=xy^2\\ v=y \end{matrix}\right. \ $$ and calculated $$\\J=\begin{vmatrix} y^2 & 2xy\\ 0 & 1 \end{vmatrix}=y^2\ $$ Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.

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    $\begingroup$ Maybe try reversing order of integration? $\endgroup$ – user608030 Jan 6 at 17:44
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    $\begingroup$ Obviously, $$x=u/v^2\qquad v=y$$ hence the domain $$0<x<y<1$$ translates as $$0<u/v^2<v<1$$ that is, $$0<u<v^3<1$$ $\endgroup$ – Did Jan 6 at 17:46
  • $\begingroup$ @ZacharySelk Ok, but then, what meaning does the $x$ in the lower bound of the external integral have: $\int_x^1\int_0^1 A dxdy$ (if $A$ is the integrand)? $\endgroup$ – J. Doe Jan 6 at 17:52
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    $\begingroup$ Draw a picture of the region. $\endgroup$ – user608030 Jan 6 at 17:53
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    $\begingroup$ Yeah now it's totally doable. $\endgroup$ – user608030 Jan 6 at 18:12
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{1}\int_{x}^{1}y^{4}\expo{xy^{2}}\dd y\,\dd x:\ {\LARGE ?}}$.

\begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}\int_{x}^{1}y^{4}\expo{xy^{2}}\dd y\,\dd x} = \int_{0}^{1}y^{4}\int_{0}^{y}\expo{xy^{2}}\dd x\,\dd y = \int_{0}^{1}y^{4}{\expo{y^{3}} - 1 \over y^{2}}\,\dd y \\[5mm] = &\ \int_{0}^{1}\pars{y^{2}\expo{y^{3}} - y^{2}}\dd y = \left.{\expo{y^{3}} - y^{3} \over 3}\,\right\vert_{0}^{1} = \bbx{\expo{} - 2 \over 3} \approx 0.2394 \end{align}

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Well, solving a much more general problem:

$$\mathcal{I}_\text{n}\left(\alpha\right):=\int_0^\alpha\int_x^\alpha\text{y}^\text{n}\cdot\exp\left(x\cdot\text{y}^{\text{n}-2}\right)\space\text{d}\text{y}\space\text{d}x\tag1$$

Using that (for all $x$):

$$\exp\left(x\right)=\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\tag2$$

We can write:

$$\mathcal{I}_\text{n}\left(\alpha\right)=\int_0^\alpha\int_x^\alpha\text{y}^\text{n}\cdot\sum_{\text{k}=0}^\infty\frac{\left(x\cdot\text{y}^{\text{n}-2}\right)^\text{k}}{\text{k}!}\space\text{d}\text{y}\space\text{d}x=$$ $$\int_0^\alpha\int_x^\alpha\text{y}^\text{n}\cdot\sum_{\text{k}=0}^\infty\frac{x^\text{k}\cdot\text{y}^{\text{k}\left(\text{n}-2\right)}}{\text{k}!}\space\text{d}\text{y}\space\text{d}x=$$ $$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\int_x^\alpha\text{y}^\text{n}\cdot\text{y}^{\text{k}\left(\text{n}-2\right)}\space\text{d}\text{y}\right\}\space\text{d}x=$$ $$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\int_x^\alpha\text{y}^{\text{n}+\text{k}\left(\text{n}-2\right)}\space\text{d}\text{y}\right\}\space\text{d}x=$$ $$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\cdot\left[\frac{\text{y}^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\right]_x^\alpha\right\}\space\text{d}x=$$ $$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\cdot\left(\frac{\alpha^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}-\frac{x^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\right)\right\}\space\text{d}x=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left\{\frac{\alpha^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\int_0^\alpha x^\text{k}\space\text{d}x-\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\int_0^\alpha x^{1+\text{n}+\text{k}\left(\text{n}-1\right)}\space\text{d}x\right\}=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left\{\frac{\alpha^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{\alpha^{1+\text{k}}}{1+\text{k}}-\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{\alpha^{1+1+\text{n}+\text{k}\left(\text{n}-1\right)}}{1+1+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left\{\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{1}{1+\text{k}}-\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\left\{\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{1+\text{k}}-\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}\tag3$$

When $\alpha=1$, we get:

$$\mathcal{I}_\text{n}\left(1\right):=\int_0^1\int_x^1\text{y}^\text{n}\cdot\exp\left(x\cdot\text{y}^{\text{n}-2}\right)\space\text{d}\text{y}\space\text{d}x=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\left\{\frac{1}{1+\text{k}}-\frac{1}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{k}}\cdot\frac{1}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\tag4$$

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