Example 1. Calculate the definite integral $\int_{1}^{4} x^{2} d x$.
Solution. By formula (5.3) we have

$$
\int_{1}^{4} x^{2} d x=\left.\frac{x^{3}}{3}\right|_{1} ^{4}=\frac{4^{3}}{3}-\frac{1^{3}}{3}=\frac{64}{3}-\frac{1}{3}=21
$$

Remark According to the Newton-Leibniz theorem, we can take any antiderivative of the integrand. In this case, instead of $\frac{x^{3}}{3}$, we could have taken $\frac{x^{3}}{3}+5$, which is also an antiderivative of the function $y=x^{2}$:

$$
\int_{1}^{4} x^{2} d x=\left(\frac{x^{3}}{3}+5\right)_{1}^{4}=\left(\frac{4^{3}}{3}+5\right)-\left(\frac{1^{3}}{3}+5\right)=\frac{64}{3}+5-\frac{1}{3}-5=\frac{63}{3}=21
$$

The same result is obtained in the general case:

$$
\int_{1}^{4} x^{2} d x=\left(\frac{x^{3}}{3}+C\right)_{1}^{4}=\left(\frac{4^{3}}{3}+C\right)-\left(\frac{1^{3}}{3}+C\right)=\frac{64}{3}+C-\frac{1}{3}-C=21
$$

In the future, we will consider only the antiderivative for which the constant term is zero.