$x^3 y$

$x^{3}y$

$x^2$

$x^2$

$x$

$x$

$0$

$0$

$1$

$1$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{MultiInt}\right]\left(\,y\=..\,x\=..\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_{x^2}^x{x}^{3}y\ⅆy\ⅆx$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{MultiInt}\right]\left(\,y\=..\,x\=..\right)$

$\frac{1}{48}$

$\mathrm{Student}\left[\mathrm{MultivariateCalculus}\right]\left[\mathrm{MultiInt}\right]\left(\,y\=..\,x\=..\,\mathrm{output}\=\mathrm{steps}\right)$

$\begin{array}{ccc}\multicolumn{3}{c}{{\int }_0^1{\int }_{x^2}^x{x}^{3}y\ⅆy\ⅆx}\\ & \text{=}& {\int }_0^1\left(\genfrac{}{}{0ex}{}{\frac{x^3{y}^2}{2}}{\phantom{y=x^2..x}}|\genfrac{}{}{0ex}{}{\phantom{\frac{x^3{y}^2}{2}}}{y=x^2..x}\right)\ⅆx\hfill \\ & \text{=}& {\int }_0^1\frac{x^3\left(x^2-x^4\right)}{2}\ⅆx\hfill \\ & \text{=}& \genfrac{}{}{0ex}{}{\left(-\frac{1}{16}{x}^8+\frac{1}{12}{x}^6\right)}{\phantom{x=0..1}}|\genfrac{}{}{0ex}{}{\phantom{\left(-\frac{1}{16}{x}^8+\frac{1}{12}{x}^6\right)}}{x=0..1}\hfill \end{array}$

$\frac{1}{48}$