320. $y=4x-x^2$, $y=0$ and $x=5$.
The parabola $y=4 x-x^{2}$ intersects the x-axis at points $x=0$ and $x=4$. The figure whose area we need to find is marked in color in Fig. 167. Let $S_{1}$ and $S_{2}$ be the areas of the parts of this figure corresponding to the segments $[0,4]$ and $[4, 5]$, and let $S$ be the required area; then $S=S_{1}+S_{2}$. Using formula (1), we get

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-326.jpg?height=252&width=423&top_left_y=305&top_left_x=198)

Fig. 166

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-326.jpg?height=321&width=394&top_left_y=770&top_left_x=226)

Fig. 168

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-326.jpg?height=381&width=329&top_left_y=188&top_left_x=689)

Fig. 167

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-326.jpg?height=448&width=327&top_left_y=641&top_left_x=688)

Fig. 169

$$
S_{1}=\int_{0}^{4}\left(4 x-x^{2}\right) d x=\left.\left(2 x^{2}-\frac{x^{3}}{3}\right)\right|_{0}^{4}=32-\frac{64}{3}=\frac{32}{3} \text { (sq. units) }
$$

and using formula (2) we find

$$
\begin{gathered}
S_{2}=\left|\int_{4}^{5}\left(4 x-x^{2}\right) d x\right|=\left|\left(\frac{x^{3}}{3}-2 x^{2}\right)\right|_{4}^{5}=\left(\frac{125}{3}-50\right)-\left(\frac{64}{3}-32\right)= \\
=\frac{7}{3} \text { (sq. units) }
\end{gathered}
$$

Therefore, $S=S_{1}+S_{2}=32 / 3+7 / 3=13$ (sq. units).