${\int }_0^{\mathrm{\pi }\/3}{\int }_{\sin \left(x\right)}^{\sin \left(2 x\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_{\mathrm{\pi }\/3}^{\mathrm{\pi }}{\int }_{\sin \left(2 x\right)}^{\sin \left(x\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{5}{2}$

Obtain the intersections of the curves bounding $R$ 

$\sin \left(x\right)\=\sin \left(2 x\right)$$\stackrel{\text{solve}}{\to }$$\left\{x\=0\right\}\,\left\{x\=\mathrm{\π}\right\}\,\left\{x\=\frac{1}{3}\mathrm{\π}\right\}\,\left\{x\=-\frac{1}{3}\mathrm{\π}\right\}$

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

${\int }_0^{\mathrm{\pi }\/3}{\int }_{\sin \left(x\right)}^{\sin \left(2 x\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_{\mathrm{\pi }\/3}^{\mathrm{\pi }}{\int }_{\sin \left(2 x\right)}^{\sin \left(x\right)}1 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{5}{2}$ 

Initialize

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

Top-level, using the Int and int commands

$$\begin{gathered} \mathrm{Int}\left(1\,\left[y\=\sin \left(x\right)..\sin \left(2 x\right)\,x\=0..\mathrm{\π}\/3\right]\right)\+\mathrm{Int}\left(1\,\left[y\=\sin \left(2 x\right)..\sin \left(x\right)\,x\=\mathrm{\π}\/3..\mathrm{\π}\right]\right)\= \\ \mathrm{int}\left(1\,\left[y\=\sin \left(x\right)..\sin \left(2 x\right)\,x\=0..\mathrm{\pi }\/3\right]\right)\+\mathrm{int}\left(1\,\left[y\=\sin \left(2 x\right)..\sin \left(x\right)\,x\=\mathrm{\pi }\/3..\mathrm{\pi }\right]\right) \end{gathered}$$

${\∫}_0^{\frac{1}{3}\mathrm{\π}}{\∫}_{\sin \left(x\right)}^{\sin \left(2x\right)}1\ⅆy\ⅆx\+{\∫}_{\frac{1}{3}\mathrm{\π}}^{\mathrm{\π}}{\∫}_{\sin \left(2x\right)}^{\sin \left(x\right)}1\ⅆy\ⅆx\=\frac{5}{2}$

Use the MultiInt command from the Student MultivariateCalculus package

$\mathrm{MultiInt}\left(1\,y\=\sin \left(x\right)..\sin \left(2 x\right)\,x\=0..\mathrm{\π}\/3\right)\+\mathrm{MultiInt}\left(1\,y\=\sin \left(2 x\right)..\sin \left(x\right)\,x\=\mathrm{\π}\/3..\mathrm{\π}\right)$ = $\frac{5}{2}$$$

$$\begin{gathered} \mathrm{MultiInt}\left(1\,y\=\sin \left(x\right)..\sin \left(2 x\right)\,x\=0..\mathrm{\π}\/3\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{MultiInt}\left(1\,y\=\sin \left(2 x\right)..\sin \left(x\right)\,x\=\mathrm{\π}\/3..\mathrm{\π}\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_0^{\frac{1}{3}\mathrm{\π}}{\∫}_{\sin \left(x\right)}^{\sin \left(2x\right)}1\ⅆy\ⅆx\+{\∫}_{\frac{1}{3}\mathrm{\π}}^{\mathrm{\π}}{\∫}_{\sin \left(2x\right)}^{\sin \left(x\right)}1\ⅆy\ⅆx$

Use the MultiInt command with a pre-defined domain option

$$\begin{gathered} \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..\mathrm{\π}\/3\,\sin \left(x\right)..\sin \left(2 x\right)\right)\right)\+ \\ \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(\mathrm{\π}\/3..\mathrm{\π}\,\sin \left(2 x\right)..\sin \left(x\right)\right)\right) \end{gathered}$$

$\frac{5}{2}$

$$\begin{gathered} \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(0..\mathrm{\pi }\/3\,\sin \left(x\right)..\sin \left(2 x\right)\right)\,\mathrm{output}\=\mathrm{integral}\right)\+ \\ \mathrm{MultiInt}\left(1\,\left[x\,y\right]\=\mathrm{Region}\left(\mathrm{\pi }\/3..\mathrm{\pi }\,\sin \left(2 x\right)..\sin \left(x\right)\right)\,\mathrm{output}\=\mathrm{integral}\right) \end{gathered}$$

${\∫}_0^{\frac{1}{3}\mathrm{\π}}{\∫}_{\sin \left(x\right)}^{\sin \left(2x\right)}1\ⅆy\ⅆx\+{\∫}_{\frac{1}{3}\mathrm{\π}}^{\mathrm{\π}}{\∫}_{\sin \left(2x\right)}^{\sin \left(x\right)}1\ⅆy\ⅆx$