${\int }_0^1{\int }_0^{x \left(1-x\right)}1 \mathrm{dy} \mathrm{dx}$ = $\frac{1}{6}$ 

${\int }_0^{1\/4}{\int }_{\left(1 -\sqrt{1-4 y}\right)\/2}^{\left(1\+\sqrt{1-4 y}\right)\/2}1 \mathrm{dx} \mathrm{dy}$ = $\frac{1}{6}$ 

Initialize

Loading Student:-MultivariateCalculus

Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^1{\∫}_0^{x\left(1-x\right)}1\ⅆy\ⅆx$$\=$$\frac{1}{6}$

Initialize

$f\=x \left(1-x\right)$$\stackrel{\text{assign}}{\to }$

Solve $y\=f\left(x\right)$ for $x\=x\left(y\right)$ 

$y\=f$

$y\=x\left(1-x\right)$

$\stackrel{\text{solutions for x}}{\to }$

$\frac{1}{2}\+\frac{1}{2}\sqrt{1-4y}\,\frac{1}{2}-\frac{1}{2}\sqrt{1-4y}$

$\stackrel{\text{assign to a name}}{\to }$

$X$

Iterate in the order $\mathrm{dy} \mathrm{dx}$ via the template in the Calculus palette

${\int }_0^1{\int }_0^{f}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{1}{6}$$$

Iterate in the order $\mathrm{dx} \mathrm{dy}$ via the template in the Calculus palette

${\int }_0^{1\/4}{\int }_{X_2}^{X_1}1 \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{1}{6}$$$

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 2-D

Table 6.1.1(a)   Iteration in the order $\mathrm{dy} \mathrm{dx}$ via visualization task-template

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 2-D

Table 6.1.1(b)   Iteration in the order $\mathrm{dx} \mathrm{dy}$ via visualization task-template

Initialize

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

Top-level, using the Int and int commands

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

${\∫}_0^1{\∫}_0^{x\left(1-x\right)}1\ⅆy\ⅆx\=\frac{1}{6}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(1\,y\=0..x \left(1-x\right)\,x\=0..1\right)$ = $\frac{1}{6}$$$

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

${\∫}_0^1{\∫}_0^{x\left(1-x\right)}1\ⅆy\ⅆx$

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

$\frac{1}{6}$

Use the MultiInt command with pre-defined domain option

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..1\,0.. x \left(1-x\right)\right)\right)$ = $\frac{1}{6}$$$

$\mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..1\,0.. x \left(1-x\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^1{\∫}_0^{x\left(1-x\right)}1\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(1\,\left[y\,x\right]\=\mathrm{Region}\left(0..1\/4\,\frac{1-\sqrt{1-4 y}}{2}..\frac{1\+\sqrt{1-4 y}}{2}\right)\right)$ = $\frac{1}{6}$$$

$\mathrm{MultiInt}\left(1\,\left[y\,x\right]\=\mathrm{Region}\left(0..1\/4\,\frac{1-\sqrt{1-4 y}}{2}..\frac{1\+\sqrt{1-4 y}}{2}\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^{\frac{1}{4}}{\∫}_{\frac{1}{2}-\frac{1}{2}\sqrt{1-4y}}^{\frac{1}{2}\+\frac{1}{2}\sqrt{1-4y}}1\ⅆx\ⅆy$