223. Calculate the area of the figure bounded by the lines $y=$ $=2 x-x^{2}$ and $y=0$ (Fig. 143).
Solution. Let's find the points of intersection of the curve $2x - x^2$ with the x-axis: $2x - x^2 = 0; x(2 - x) = 0; x_1 = 0, x_2 = 2$. Therefore, $a = 0$, $b = 2$.

We find the antiderivative of the function $f(x) = 2x - x^2$; we have $f(x) = \int (2x - x^2) dx = x^2 - \frac{1}{3} x^3 + C$. When $C = 0$, we get $F(x) = x^2 - \frac{1}{3} x^3$.

The desired area is found using formula (1):

$$
S = F(b) - F(a) = F(2) - F(0) = 4 - \frac{8}{3} - 0 + 0 = \frac{4}{3} \text{ (sq. units). }
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-309.jpg?height=361&width=442&top_left_y=1433&top_left_x=177)

Fig. 142

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-309.jpg?height=271&width=389&top_left_y=1522&top_left_x=690)

Fig. 143

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-310.jpg?height=301&width=421&top_left_y=261&top_left_x=197)

Fig. 144

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-310.jpg?height=386&width=386&top_left_y=184&top_left_x=676)

Fig. 145