Solve the integral $\int_0^1\int^1_xy^4e^{xy^2}dydx$. 
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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\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}
A: 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:
$$\mathcal{I}_\text{n}\left(1\right):=\int_0^1\int_x^1\text{y}^\text{n}\cdot\exp\left(x\cdot\text{y}^{\text{n}-2}\right)\space\text{d}\text{y}\space\text{d}x=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\left\{\frac{1}{1+\text{k}}-\frac{1}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{k}}\cdot\frac{1}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\tag4$$
