${\int }_0^1{\int }_0^{1-x}F \mathrm{dy} \mathrm{dx}$ = $\frac{25}{12}$ 

${\int }_0^1{\int }_0^{1-y}F \mathrm{dx} \mathrm{dy}$ = $\frac{25}{12}$ 

Loading Student:-Precalculus

$\left[0\,1\right]\,\left[1\,0\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-x\+1$$\stackrel{\text{isolate for x}}{\to }$$x\=-y\+1$

Table 6.2.5(a)   Obtaining the equation of the hypotenuse for the triangle defining region $R$

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

${\int }_0^1{\int }_0^{1-x}F \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{25}{12}$$$

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

${\int }_0^1{\int }_0^{1-y}F \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{25}{12}$$$

Table 6.2.5(b)   Iterated double-integrals for finding the volume "under" $F$ and over $R$  

Initialize

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

$F≔5-3 x^2-2 y^2\:$

Obtain the equation of the hypotenuse in region $R$ 

$\mathrm{GetRepresentation}\left(\mathrm{Line}\left(\left[0\,1\right]\,\left[1\,0\right]\right)\,\mathrm{form}\=\mathrm{equation}\right)$

$x\+y\=1$

Top-level, using the Int and int commands

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

${\∫}_0^1{\∫}_0^{-x\+1}\left(-3x^2-2y^2\+5\right)\ⅆy\ⅆx\=\frac{25}{12}$

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

${\∫}_0^1{\∫}_0^{-y\+1}\left(-3x^2-2y^2\+5\right)\ⅆx\ⅆy\=\frac{25}{12}$

Use the MultiInt command from the Student MultivariateCalculus package

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

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

${\∫}_0^1{\∫}_0^{1-x}F\ⅆy\ⅆx$

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

$\frac{1}{2}F$

Use the MultiInt command with a pre-defined domain option

$\mathrm{MultiInt}\left(F\,\left[x\,y\right]\=\mathrm{Region}\left(0..1\,0.. 1-x\right)\right)$ = $\frac{1}{2}F$$$

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

${\∫}_0^1{\∫}_0^{1-x}F\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(F\,\left[y\,x\right]\=\mathrm{Region}\left(0..1\,0..1-y\right)\right)$ = $\frac{1}{2}F$$$

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

${\∫}_0^1{\∫}_0^{1-y}F\ⅆx\ⅆy$