Example 11. A point is thrown into a rectangle with vertices $K(-1,0), L(-1,5)$, $M(2,5), N(2,0)$. What is the probability that its coordinates $(x, y)$ will satisfy the inequalities $x^{2}+1 \leq y \leq x+3$?
Solution. The points whose coordinates satisfy the given inequalities belong to the region $g$, bounded by the parabola $y=x^{2}+1$ and the line $y=x+3$ (Fig. 1.15). These lines intersect at points $Q(-1,2)$ and $M(2,5)$. Let's calculate the area of region $g$:

$$
\begin{aligned}
& S_{g}=\int_{-1}^{2}\left(y_{2}-y_{1}\right) d x=\int_{-1}^{2}\left(x+3-\left(x^{2}+1\right)\right) d x=\int_{-1}^{2}\left(x-x^{2}+2\right) d x= \\
& =\left.\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}+2 x\right)\right|_{-1} ^{2}=\frac{1}{2}\left(2^{2}-(-1)^{2}\right)-\frac{1}{3}\left(2^{3}-(-1)^{3}\right)+2(2-(-1))= \\
& =\frac{1}{2} \cdot 3-\frac{1}{3} \cdot 9+2 \cdot 3=\frac{3}{2}-3+6=4.5 .
\end{aligned}
$$

## The region $G$ here

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-036.jpg?height=708&width=696&top_left_y=40&top_left_x=0)

Fig. 1.15 is a rectangle with vertices $K, L$, $M, \quad N . \quad$ Since $a=|K M|=|2-(-1)|=3$, $h=N M=|5-0|=5$, the area of region $G$ is determined by the formula

$$
S_{G}=a h=3 \cdot 5=15
$$

According to formula (1.5.1), we find the desired probability of event $A$ - "the point falls into region $g$":

$$
P(A)=\frac{S_{g}}{S_{G}}=\frac{4.5}{15}=0.3
$$