1.5. Evaluate the double integral

$$
\iint_{\Omega}(2 x+3 y+1) d x d y
$$

over the region $\Omega$, bounded by the triangle with vertices $A(1,3), B(-1,-1), C(2,-4)$.
Solution. The specified triangle is depicted in Fig. 1.26. We draw the segment $B D$, parallel to the $O X$ axis. Thus, the original triangle $A B C$ will be divided into two triangles $A B D$ and $D B C$. Next, we write the equations of the lines containing the sides of the triangle $A B, B C$, and $C A$:

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-046.jpg?height=414&width=225&top_left_y=1043&top_left_x=911)

Fig. 1.26

$$
\begin{aligned}
& A B \subset l_{A B}: \frac{-1-3}{-1-1}=\frac{y-3}{x-1} \text { or } x=\frac{y}{2}-\frac{1}{2} \\
& B C \subset l_{B C}: \frac{-4+1}{2+1}=\frac{y+1}{x+1} \text { or } x=-y-2 \\
& C A \subset l_{C A}: \frac{3+4}{1-2}=\frac{y+4}{x-2} \text { or } x=-\frac{y}{7}+\frac{10}{7} .
\end{aligned}
$$

Now we can proceed with repeated integration (inner integration with respect to the variable $x$, outer integration with respect to the variable $y$):

$$
\begin{aligned}
& \iint_{\Omega 2}(2 x+3 y+1) d x d y=\int_{-4}^{1} d y \int_{-y-2}^{\frac{y}{7}-\frac{10}{7}}(2 x+3 y+1) d x+\int_{1}^{3} d y \int_{\frac{y}{2}-\frac{1}{2}}^{\frac{y}{7} \cdot \int^{7}}(2 x+3 y+1) d x= \\
& =\left.\int_{-4}^{-1}\left(x^{2}+3 y x+x\right)\right|_{y-2} ^{-\frac{y}{7}+\frac{10}{7}} d y+\left.\int_{-1}^{3}\left(x^{2}+3 y x+x\right)\right|_{\frac{y}{2}} ^{-\frac{y}{7}} \cdot \frac{10}{2} d y= \\
& =\int_{4}^{-1}\left(\left(-\frac{y}{7}+\frac{10}{7}\right)^{2}+3 y\left(-\frac{y}{7}+\frac{10}{7}\right)+\frac{10}{7}-\frac{y}{7}\right) d y+ \\
& +\int_{-1}^{3}\left(\left(-\frac{y}{7}+\frac{10}{7}\right)^{2}+3 y\left(-\frac{y}{7}+\frac{10}{7}\right)+\frac{10}{7}-\frac{y}{7}\right) d y-\int_{-4}^{1}\left((y+2)^{2}-3 y(y+2)-y-2\right) d y- \\
& -\int_{-1}^{3}\left(\left(\frac{y}{2}-\frac{1}{2}\right)^{2}+3 y\left(\frac{y}{2}-\frac{1}{2}\right)+\frac{y}{2}-\frac{1}{2}\right) d y=\int_{-4}^{3}\left(-\frac{20}{49} y^{2}+\frac{183}{49} y+\frac{170}{49}\right) d y- \\
& -\int_{-4}^{-1}\left(-2 y^{2}-3 y+2\right) d y-\int_{1}^{3}\left(\frac{7}{4} y^{2}-\frac{3}{2} y-\frac{1}{4}\right) d y=\left.\left(-\frac{20}{147} y^{3}-\frac{183}{98} y^{2}+\frac{170}{49} y\right)\right|_{-4} ^{3}- \\
& -\left.\left(-\frac{2}{3} y^{3}-\frac{3}{2} y^{2}+2 y\right)\right|_{-4} ^{-1}-\left.\left(\frac{7}{12} y^{3}-\frac{3}{4} y^{2}-\frac{1}{4} y\right)\right|_{-1} ^{3}=\frac{8065}{249}-\frac{37}{12}-\frac{28}{3}=-\frac{83}{294}
\end{aligned}
$$