${\int }_{x\=0}^{x\=1}{\int }_{y\=0}^{y\=1}\left(x^2\+y^2\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{2}{3}$ 

${\int }_{y\=0}^{y\=1}{\int }_{x\=0}^{x\=1}\left(x^2\+y^2\right) \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{2}{3}$$$

Initialize

$f\=x^2\+y^2$$\stackrel{\text{assign}}{\to }$

Iterate and evaluate

${\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{2}{3}$$$

${\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{2}{3}$$$

Inert iterated double-integral and its evaluation

${\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}y \mathit{\ⅆ}x$

${\∫}_0^1{\∫}_0^1\left(x^2\+y^2\right)\ⅆy\ⅆx$

$\=$

$\frac{2}{3}$

${\int }_0^1{\int }_0^{1}f \mathit{\ⅆ}x \mathit{\ⅆ}y$

${\∫}_0^1{\∫}_0^1\left(x^2\+y^2\right)\ⅆx\ⅆy$

$\=$

$\frac{2}{3}$

Initialize

$f≔x^2\+y^2\:$

Iterate and evaluate

$\mathrm{Int}\left(f\,\left[x\=0..1\,y\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[x\=0..1\,y\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1\left(x^2\+y^2\right)\ⅆx\ⅆy\=\frac{2}{3}$

$\mathrm{Int}\left(f\,\left[y\=0..1\,x\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[y\=0..1\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^1\left(x^2\+y^2\right)\ⅆy\ⅆx\=\frac{2}{3}$

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

Use the MultiInt command to return the unevaluated iterated integral

$\mathrm{MultiInt}\left(f\,x\=0..1\,y\=0..1\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^1\left(x^2\+y^2\right)\ⅆx\ⅆy$

$\mathrm{MultiInt}\left(f\,y\=0..1\,x\=0..1\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^1\left(x^2\+y^2\right)\ⅆy\ⅆx$

Use the MultiInt command to return the value of the iterated integral

$\mathrm{MultiInt}\left(f\,x\=0..1\,y\=0..1\right)$ = $\frac{2}{3}$$$

$\mathrm{MultiInt}\left(f\,y\=0..1\,x\=0..1\right)$ = $\frac{2}{3}$$$

Use the MultiInt command to obtain a stepwise evaluation of the iterated integral

$\mathrm{MultiInt}\left(f\,x\=0..1\,y\=0..1\,\mathrm{output}\=\mathrm{steps}\right)$

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

$\mathrm{MultiInt}\left(f\,y\=0..1\,x\=0..1\,\mathrm{output}\=\mathrm{steps}\right)$

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