Example 5.26. Compute $\int_{0}^{1} \frac{\sin x}{x} d x$ with an accuracy of 0.01.
Solution. This definite integral can only be calculated approximately. For this, we expand the integrand function into a Taylor series:

$$
\frac{\sin x}{x}=1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+\ldots
$$

Hence,

$$
\begin{aligned}
& \int_{0}^{1} \frac{\sin x}{x} d x=\int_{0}^{1} 1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+\ldots d x= \\
= & \left.x\right|_{0} ^{1}-\left.\frac{1}{3!} \cdot \frac{x^{3}}{3}\right|_{0} ^{1}+\left.\frac{1}{5!} \frac{x^{5}}{5}\right|_{0} ^{1}-\left.\frac{1}{7!} \cdot \frac{x^{7}}{7}\right|_{0} ^{1}+\ldots= \\
& =1-\frac{1}{3!3}+\frac{1}{5!5}-\frac{1}{7!7}+\ldots \approx 1-\frac{1}{3!3}=0.94
\end{aligned}
$$

(here we limited ourselves to the first two terms of this alternating series, as the third term $\frac{1}{5!5}$ is less than 0.01).

## SOLUTIONS TO TYPICAL PROBLEMS OF THE CONTROL WORK ON SECTIONS 3, 4, AND 5

## 4. INTEGRAL CALCULUS

## 4.1. Indefinite Integral