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$f\=1\+2 x^2\+3 y^2$$\stackrel{\text{assign}}{\to }$

Define the three vertices of the triangle

$p_1\=\left[0\,0\right]$$\stackrel{\text{assign}}{\to }$

$p_2\=\left[3\,1\right]$$\stackrel{\text{assign}}{\to }$

$p_3\=\left[2\,5\right]$$\stackrel{\text{assign}}{\to }$

Draw the triangle

$\left[p_1\,p_2\,p_3\,p_1\right]$ = $\left[\left[0\,0\right]\,\left[3\,1\right]\,\left[2\,5\right]\,\left[0\,0\right]\right]$$\to$

Obtain the equations for each side of the triangle

$p_1\,p_2$ = $\left[0\,0\right]\,\left[3\,1\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{1}{3}x$$\stackrel{\text{right hand side}}{\to }$$\frac{1}{3}x$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Y12}$

$p_2\,p_3$ = $\left[3\,1\right]\,\left[2\,5\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-4x\+13$$\stackrel{\text{right hand side}}{\to }$$-4x\+13$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Y23}$

$p_3\,p_1$ = $\left[2\,5\right]\,\left[0\,0\right]$$\stackrel{\text{equation of line}}{\to }$$y\=\frac{5}{2}x$$\stackrel{\text{right hand side}}{\to }$$\frac{5}{2}x$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Y31}$

Construct and evaluate the iterated integral(s)

${\int }_0^2{\int }_{\mathrm{Y12}}^{\mathrm{Y31}}f \mathit{\ⅆ}y \mathit{\ⅆ}x\+{\int }_2^3{\int }_{\mathrm{Y12}}^{\mathrm{Y23}}f \mathit{\ⅆ}y \mathit{\ⅆ}x$

${\∫}_0^2{\∫}_{\frac{1}{3}x}^{\frac{5}{2}x}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx\+{\∫}_2^3{\∫}_{\frac{1}{3}x}^{-4x\+13}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx$

$\=$

$\frac{1781}{12}$

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$$\begin{gathered} \mathrm{with}\left(\mathrm{Student}:-\mathrm{Precalculus}\right)\: \\ \mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\: \end{gathered}$$

$f≔1\+2 x^2\+3 y^2\:$

${\mathbf{v}}_1\,{\mathbf{v}}_2\,{\mathbf{v}}_3≔⟨0\,0⟩\,⟨3\,1⟩\,⟨2\,5⟩\:$

$P≔\mathrm{convert}~\left(\left[{\mathbf{v}}_1\,{\mathbf{v}}_2\,{\mathbf{v}}_3\right]\,\mathrm{list}\right)\:$

Integrate over the triangle

$\mathrm{int}\left(f\,\left[x\,y\right]\=\mathrm{Triangle}\left({\mathbf{v}}_1\,{\mathbf{v}}_2\,{\mathbf{v}}_3\right)\,\mathrm{inert}\right)$

$-\left({\∫}_0^2{\∫}_{\frac{5}{2}x}^{\frac{1}{3}x}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx\right)-\left({\∫}_2^3{\∫}_{-4x\+13}^{\frac{1}{3}x}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx\right)$

$\mathrm{int}\left(f\,\left[x\,y\right]\=\mathrm{Triangle}\left({\mathbf{v}}_1\,{\mathbf{v}}_2\,{\mathbf{v}}_3\right)\right)$ = $\frac{1781}{12}$

Draw the triangle

$\mathrm{plot}\left(\left[\left[P_1\,P_2\right]\,\left[P_2\,P_3\right]\,\left[P_3\,P_1\right]\right]\, \mathrm{color}\=\left[\mathrm{black}\,\mathrm{red}\,\mathrm{green}\right]\right)$

Towards a solution from first principles

$$\begin{gathered} Y_{12}≔\mathrm{rhs}\left(\mathrm{Line}\left(P_1\,P_2\right)\left[1\right]\right)\; \\ {Y}_{23}≔\mathrm{rhs}\left(\mathrm{Line}\left(P_2\,P_3\right)\left[1\right]\right)\; \\ {Y}_{31}≔\mathrm{rhs}\left(\mathrm{Line}\left(P_3\,P_1\right)\left[1\right]\right) \end{gathered}$$

$\frac{1}{3}x$

$\mathrm{Int}\left(f\,\left[y\=Y_{12}..Y_{31}\,x\=0..2\right]\right)\+\mathrm{Int}\left(f\,\left[y\=Y_{12}..Y_{23}\,x\=2..3\right]\right)$

${\∫}_0^2{\∫}_{\frac{1}{3}x}^{\frac{5}{2}x}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx\+{\∫}_2^3{\∫}_{\frac{1}{3}x}^{-4x\+13}\left(2x^2\+3y^2\+1\right)\ⅆy\ⅆx$

$:-\mathrm{int}\left(f\,\left[y\=Y_{12}..Y_{31}\,x\=0..2\right]\right)\+:-\mathrm{int}\left(f\,\left[y\=Y_{12}..Y_{23}\,x\=2..3\right]\right)$ = $\frac{1781}{12}$