238. Calculate the area of the curvilinear trapezoid bounded by the $O x$ axis, the lines $x=-1, x=2$, and the parabola $y=9-x^{2}$ (Fig. 146).
Solution. Since the function $y=9-x^{2}$ takes positive values on the interval $[-1,2]$, to find the desired area $S$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-313.jpg?height=309&width=373&top_left_y=184&top_left_x=244)

Fig. 146

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-313.jpg?height=453&width=356&top_left_y=631&top_left_x=199)

Fig. 148

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-313.jpg?height=309&width=344&top_left_y=184&top_left_x=659)

Fig. 147

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-313.jpg?height=310&width=448&top_left_y=768&top_left_x=611)

Fig. 149

we will use formula (3):

$$
S=\int_{-1}^{2}\left(9-x^{2}\right) d x
$$

Applying the Newton-Leibniz formula, we find

$$
\begin{gathered}
S=\int_{-1}^{2}\left(9-x^{2}\right) d x=\left.\left(9 x-\frac{x^{3}}{3}\right)\right|_{-1} ^{2}=\left(9 \cdot 2-\frac{2^{3}}{3}\right)- \\
-\left(9 \cdot(-1)-\frac{(-1)^{3}}{3}\right)=24 \frac{2}{3} \text { (sq. units)}. 
\end{gathered}
$$