Initialize

Loading Student:-MultivariateCalculus

$f\=1\+2 x^2\+3 y^2$$\stackrel{\text{assign}}{\to }$

Access the MultiInt command via the Context Panel

$f$ = $2x^2\+3y^2\+1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^1{\∫}_{x^2}^{\sqrt{x}}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx$$\=$$\frac{16}{21}$$$

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻

Table 5.3.1(a)   Visualizing $R$ and the resulting volume for iteration in the order $\mathrm{dy} \mathrm{dx}$ 

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻

Table 5.3.1(b)   Visualizing $R$ and the resulting volume for iteration in the order $\mathrm{dx} \mathrm{dy}$ 

Iterate $\mathrm{dy} \mathrm{dx}$ and $\mathrm{dx} \mathrm{dy}$ 

${\int }_0^1{\int }_{x^2}^{{\sqrt{x}}_{}}f \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{16}{21}$$$

${\int }_0^1{\int }_{y^2}^{{\sqrt{y}}_{}}f \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{16}{21}$$$

Display the iterated integrals

${\int }_0^1{\int }_{x^2}^{{\sqrt{x}}_{}}f \mathit{\ⅆ}y \mathit{\ⅆ}x$

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

$\=$

$\frac{16}{21}$

${\int }_0^1{\int }_{y^2}^{{\sqrt{y}}_{}}f \mathit{\ⅆ}x \mathit{\ⅆ}y$

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

$\=$

$\frac{16}{21}$

Initialize

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

$f≔1\+2 x^2\+3 y^2\:$

Iterate in the order $\mathrm{dy} \mathrm{dx}$

$\mathrm{Int}\left(f\,\left[y\=x^2..\sqrt{x}\,x\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[y\=x^2..\sqrt{x}\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_{x^2}^{\sqrt{x}}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx\=\frac{16}{21}$

Iterate in the order $\mathrm{dx} \mathrm{dy}$ 

$\mathrm{Int}\left(f\,\left[x\=y^2..\sqrt{y}\,y\=0..1\right]\right)\=\mathrm{int}\left(f\,\left[x\=y^2..\sqrt{y}\,y\=0..1\right]\right)$

${\∫}_0^1{\∫}_{y^2}^{\sqrt{y}}\left(2x^2\+3y^2\+1\right)\ⅆx\ⅆy\=\frac{16}{21}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(f\,y\=x^2..\sqrt{x}\,x\=0..1\,\mathrm{output}\=\mathrm{integral}\right)$

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

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

$\frac{16}{21}$

$\mathrm{MultiInt}\left(f\,x\=y^2..\sqrt{y}\,y\=0..1\,\mathrm{output}\=\mathrm{integral}\right)$

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

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

$\frac{16}{21}$

Obtain stepwise evaluations via the MultiInt command

$\mathrm{MultiInt}\left(f\,y\=x^2..\sqrt{x}\,x\=0..1\,\mathrm{output}\=\mathrm{steps}\right)$

$\begin{array}{ccc}\multicolumn{3}{c}{{\int }_0^1{\int }_{x^2}^{\sqrt{x}}\left(2x^2+3y^2+1\right)\ⅆy\ⅆx}\\ & \text{=}& {\int }_0^1\left(\genfrac{}{}{0ex}{}{\left(2x^{2}y+y^3+y\right)}{\phantom{y=x^2..\sqrt{x}}}|\genfrac{}{}{0ex}{}{\phantom{\left(2x^{2}y+y^3+y\right)}}{y=x^2..\sqrt{x}}\right)\ⅆx\hfill \\ & \text{=}& {\int }_0^1\left(2x^2\left(\sqrt{x}-x^2\right)+x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-x^6+\sqrt{x}-x^2\right)\ⅆx\hfill \\ & \text{=}& \genfrac{}{}{0ex}{}{\left(-\frac{2x^5}{5}+\frac{4x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{7}+\frac{2x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{5}-\frac{x^7}{7}+\frac{2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}-\frac{x^3}{3}\right)}{\phantom{x=0..1}}|\genfrac{}{}{0ex}{}{\phantom{\left(-\frac{2x^5}{5}+\frac{4x^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{7}+\frac{2x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{5}-\frac{x^7}{7}+\frac{2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}-\frac{x^3}{3}\right)}}{x=0..1}\hfill \end{array}$

$\mathrm{MultiInt}\left(f\,x\=y^2..\sqrt{y}\,y\=0..1\,\mathrm{output}\=\mathrm{steps}\right)$

$\begin{array}{ccc}\multicolumn{3}{c}{{\int }_0^1{\int }_{y^2}^{\sqrt{y}}\left(2x^2+3y^2+1\right)\ⅆx\ⅆy}\\ & \text{=}& {\int }_0^1\left(\genfrac{}{}{0ex}{}{\left(\frac{2}{3}{x}^3+3y^{2}x+x\right)}{\phantom{x=y^2..\sqrt{y}}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{2}{3}{x}^3+3y^{2}x+x\right)}}{x=y^2..\sqrt{y}}\right)\ⅆy\hfill \\ & \text{=}& {\int }_0^1\left(\frac{2y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}-\frac{2y^6}{3}+3y^2\left(\sqrt{y}-y^2\right)+\sqrt{y}-y^2\right)\ⅆy\hfill \\ & \text{=}& \genfrac{}{}{0ex}{}{\left(\frac{4y^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{15}-\frac{2y^7}{21}-\frac{3y^5}{5}+\frac{6y^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{7}+\frac{2y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}-\frac{y^3}{3}\right)}{\phantom{y=0..1}}|\genfrac{}{}{0ex}{}{\phantom{\left(\frac{4y^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{15}-\frac{2y^7}{21}-\frac{3y^5}{5}+\frac{6y^{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{7}+\frac{2y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}-\frac{y^3}{3}\right)}}{y=0..1}\hfill \end{array}$