358. Calculate the integral $\int_{1}^{9} \sqrt{6 x-5} d x$ using the Newton-Leibniz formula and approximate formulas for rectangles and trapezoids, dividing the interval of integration into 8 equal parts. Estimate the error of the results.
Solution. According to the Newton-Leibniz formula, we find

$$
I=\int_{1}^{9} \sqrt{6 x-5} d x=\frac{1}{6} \int_{1}^{9}(6 x-5)^{1 / 2} d(6 x-5)=\left.\frac{1}{9}(6 x-5)^{3 / 2}\right|_{1} ^{9}=\frac{7^{3}-1}{9}=38
$$

Since the interval $[1,9]$ is divided into 8 equal parts, then $\Delta x=1$. We find the values $y_{i}(i=1,2, \ldots, 8)$ of the integrand function $y=\sqrt{6 x-5}$ at the division points $x_{i}: x_{0}=1, y_{0}=\sqrt{1}=1,0000 ; x_{1}=2, y_{1}=\sqrt{7}=$ $=2,6458 ; \quad x_{2}=3, \quad y_{2}=\sqrt{13}=3,6056 ; \quad x_{3}=4, \quad y_{3}=\sqrt{19}=4,3589 ; \quad x_{4}=5$, $y_{4}=\sqrt{25}=5,0000 ; \quad x_{5}=6, \quad y_{5}=\sqrt{31}=5,5678 ; \quad x_{6}=7, \quad y_{6}=\sqrt{37}=6,0828 ;$ $x_{7}=8, y_{7}=\sqrt{43}=6,5574 ; x_{8}=9, y=\sqrt{49}=7,0000$.

We will compute this integral using approximate formulas. Using the rectangle formula (10), we find

$$
I=\sum_{i=0}^{7} y_{i}=34,8183
$$

The absolute error of this approximate value (by deficiency) is 3.1817, and the relative error is $\frac{3.1817 \cdot 100 \%}{38} \approx 8.37 \%$.

According to the rectangle formula (11), we get

$$
I=\sum_{i=1}^{8} y_{i}=40,8183
$$

Here the absolute error (by excess) is 2.8183, and the relative error is $\frac{2.8183 \cdot 100 \%}{38} \approx 7.42 \%$.

Using the trapezoidal formula (12), we find

$$
I \approx 4+\sum_{i=1}^{7} y_{i}=37,8183
$$

The absolute error of this result is 0.1817, and the relative error is $\frac{0.1817 \cdot 100 \%}{38} \approx 0.48 \%$.