$$ ${\int }_{-1}^2{\int }_{x^2}^{x\+2}\left(x\+2 y\right) \mathit{\ⅆ}y \mathit{\ⅆ}x\=\frac{333}{20}$ 

${\int }_0^1{\int }_{-\sqrt{y}}^{\sqrt{y}}\left(x\+2y\right)\mathit{\ⅆ}x\mathit{\ⅆ}y\+{\int }_1^4{\int }_{y-2}^{\sqrt{y}}\left(x\+2y\right)\mathit{\ⅆ}x\mathit{\ⅆ}y\=\frac{333}{20}$ 

Evaluate the given integral

${\int }_{-1}^2{\int }_{x^2}^{x\+2}\left(x\+2 y\right) \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{333}{20}$$$

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Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻ 

Table 5.4.4(a)   Integration in the order $\mathrm{dy} \mathrm{dx}$ 

Guided by Table 5.4.4(a), reverse the order of integration

${\int }_0^1{\int }_{-\sqrt{y}}^{{\sqrt{y}}_{}}\left(x\+2 y\right) \mathit{\ⅆ}x \mathit{\ⅆ}y\+{\int }_1^4{\int }_{y-2}^{{\sqrt{y}}_{}}\left(x\+2 y\right) \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{333}{20}$ 

Evaluate the given integral

$\mathrm{int}\left(x\+2 y\,\left[y\=x^2..x\+2\,x\=-1..2\right]\right)$ = $\frac{333}{20}$$$

Graph the region of integration

$\mathrm{plots}:-\mathrm{inequal}\left(\left\{y\ge x^2\,y\le x\+2\,x\ge -1\,x\le 4\right\}\,x\=-1..2\,y\=0..4\right)$

Reverse the order of integration

$q≔\mathrm{Int}\left(x\+2 y\,\left[x\=-\sqrt{y}..\sqrt{y}\,y\=0..1\right]\right)\+\mathrm{Int}\left(x \+2 y\,\left[x\=y-2..\sqrt{y}\,y\=1..4\right]\right)$

${\∫}_0^1{\∫}_{-\sqrt{y}}^{\sqrt{y}}\left(x\+2y\right)\ⅆx\ⅆy\+{\∫}_1^4{\∫}_{y-2}^{\sqrt{y}}\left(x\+2y\right)\ⅆx\ⅆy$

$\mathrm{value}\left(q\right)$ = $\frac{333}{20}$$$