Example 5. Calculate the double integral $I=\int_{D} \int(x+y) d x d y$, where $D-$ is the region bounded by the lines $x=0, y=x^{2}+x-3$, $2 y=3 x(x \geqslant 0)$.
Solution. The region $D$ is shown in Fig. 3.7. It is bounded from above by the segment of the line $y=\frac{3}{2} x$, from below by the parabola $y=x^{2}+x-3$, and from the left by the $O y$ axis. Let's determine the abscissa of the point $A$ of intersection of the line and the parabola:

$$
\left\{\begin{array}{l}
y=\frac{3}{2} x, \\
y=x^{2}+x-3 ;
\end{array} \quad x^{2}+x-3=\frac{3}{2} x ; 2 x^{2}-x-6=0 ; x=2\right.
$$

Thus, the region $D$ can be described by the system of inequalities

$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant 2 \\
x^{2}+x-3 \leqslant y \leqslant \frac{3}{2} x
\end{array}\right.
$$

and the double integral can be written as an iterated integral:

$$
\begin{gathered}
I=\int_{0}^{2} d x \int_{x^{2}+x-3}^{3 x / 2}(x+y) d y=\left.\int_{0}^{2} d x\left(x y+\frac{y^{2}}{2}\right)\right|_{x^{2}+x-3} ^{3 x / 2}= \\
=\frac{1}{2} \int_{0}^{2}\left(-x^{4}-4 x^{3}+\frac{33}{4} x^{2}+12 x-9\right) d x= \\
=\left.\frac{1}{2}\left(-\frac{x^{5}}{5}-x^{4}+\frac{11}{4} x^{3}+6 x^{2}-9 x\right)\right|_{0} ^{2}=\frac{14}{5}
\end{gathered}
$$

Answer. $\frac{14}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-052.jpg?height=454&width=954&top_left_y=201&top_left_x=1432)

Fig. 3.8