${\int }_0^1{\int }_{{\sqrt{y}}_{}}^2\left(2 x\+3 y\right) \mathit{\ⅆ}x \mathit{\ⅆ}y\=\frac{53}{10}$ 

${\int }_0^1{\int }_0^{x^2}\left(2x\+3y\right)\mathit{\ⅆ}y\mathit{\ⅆ}x\+{\int }_1^2{\int }_0^1\left(2x\+3y\right)\mathit{\ⅆ}y\mathit{\ⅆ}x\=\frac{53}{10}$ 

Evaluate the given integral

${\int }_0^1{\int }_{{\sqrt{y}}_{}}^2\left(2 x\+3 y\right) \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{53}{10}$$$

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Calculus - Multivariate≻Integration≻Visualizing Regions of Integration≻ 

Table 5.4.5(a)   Integration in the order $\mathrm{dy} \mathrm{dx}$ 

Guided by Table 5.4.5(a), reverse the order of integration

${\int }_0^1{\int }_0^{x^2}\left(2 x\+3 y\right) \ⅆy \ⅆx\+{\int }_1^2{\int }_0^1\left(2 x\+3 y\right) \ⅆy \ⅆx$ = $\frac{53}{10}$ 

Evaluate the given integral

$\mathrm{int}\left(2 x\+3 y\,\left[x\=\sqrt{y}..2\,y\=0..1\right]\right)$ = $\frac{53}{10}$$$

Graph the region of integration

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

Reverse the order of integration

$q≔\mathrm{Int}\left(2 x\+3 y\,\left[y\=0..x^2\,x\=0..1\right]\right)\+\mathrm{Int}\left(2 x\+3 y\,\left[y\=0..1\,x\=1..2\right]\right)$

${\∫}_0^1{\∫}_0^{x^2}\left(2x\+3y\right)\ⅆy\ⅆx\+{\∫}_1^2{\∫}_0^1\left(2x\+3y\right)\ⅆy\ⅆx$

$\mathrm{value}\left(q\right)$ = $\frac{53}{10}$$$