221. Using the equality $S^{\prime}(x)=f(x)$, calculate the area of the curvilinear trapezoid bounded by the lines $y=x^{2}, x=1$, $x=2, y=0$.
Solution. Let $x \in [1,2]$ (Fig. 141). Since $S'(x) = f(x)$, then $S'(x) = x^2$. Therefore, $S(x)$ is an antiderivative of the function $f(x) = x^2$.

Let's find the set of all antiderivatives: $S(x) = \frac{x^3}{3} + C$. The value of $C$ can be found from the condition $S(1) = 0$; we have $0 = \frac{1}{3} + C$, from which $C = -\frac{1}{3}$. Thus, $S(x) = \frac{x^3}{3} - \frac{1}{3}$.

By setting $x = 2$, we obtain the area of the given trapezoid:

$$
S(2) = \frac{2^3}{3} - \frac{1}{3} = \frac{7}{3} = 2 \frac{1}{3} \text{ (sq. units) }
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-308.jpg?height=260&width=383&top_left_y=1539&top_left_x=214)

Fig. 140

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-308.jpg?height=306&width=342&top_left_y=1497&top_left_x=673)

Fig. 141