${\int }_0^1{\int }_0^{2-x}{\int }_0^{2-x-y}1 \mathrm{dz} \mathrm{dy} \mathrm{dx}$ = $\frac{7}{6}$

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Access the MultiInt command via the Context Panel

$1$$\stackrel{\text{MultiInt}}{\to }$${\∫}_0^1{\∫}_0^{2-x}{\∫}_0^{2-x-y}1\ⅆz\ⅆy\ⅆx$$\=$$\frac{7}{6}$

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Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻Cartesian 3-D

Table 8.1.29(a)   Task template integrating in Cartesian coordinates

${\int }_0^1{\int }_0^{2-x}{\int }_0^{2-x-y}1 \mathit{\ⅆ}z \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{7}{6}$$$

Table 8.1.29(b)   Integration via first principles

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

${\∫}_0^1{\∫}_0^{2-x}{\∫}_0^{2-x-y}1\ⅆz\ⅆy\ⅆx$

$\mathrm{MultiInt}\left(1\,z\=0..2-x-y\,y\=0..2-x\,x\=0..1\right)$ = $\frac{7}{6}$$$

Table 8.1.29(c)   MultiInt command iterating in Cartesian coordinates in the order $\mathrm{dy} \mathrm{dz} \mathrm{dx}$ 

$$\begin{gathered} \mathrm{Int}\left(1\,\left[z\=0..2-x-y\,y\=0..2-x\,x\=0..1\right]\right)\= \\ \mathrm{int}\left(1\,\left[z\=0..2-x-y\,y\=0..2-x\,x\=0..1\right]\right) \end{gathered}$$

${\∫}_0^1{\∫}_0^{2-x}{\∫}_0^{2-x-y}1\ⅆz\ⅆy\ⅆx\=\frac{7}{6}$

Table 8.1.29(d)   Top-level Int and int commands