701. Calculate the integral $\int_{1}^{9} \sqrt{6 x-5} d x$ using the Newton-Leibniz formula and approximate formulas for rectangles, trapezoids, and Simpson's method, dividing the integration interval into 8 equal parts. Then estimate the percentage error of the results obtained using the approximate formulas.
Solution. By the Newton-Leibniz formula

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

Next, we divide the integration interval $[1 ; 9]$ into 8 equal parts, find the length of one part $h=1$, the division points $x_{i}$, and the values $y_{i}$ of the integrand function $y=\sqrt{6 x-5}$ at these points:

$$
\begin{aligned}
x_{0}=1 & y_{0}=\sqrt{\bar{I}}=1,0000 \\
x_{1}=2 & y_{1}=\sqrt{7}=2,6458 \\
x_{2}=3 & y_{2}=\sqrt{13}=3,6056 \\
x_{3}=4 & y_{3}=\sqrt{19}=4,3589 \\
x_{4}=5 & y_{4}=\sqrt{25}=5,0000 \\
x_{5}=6 & y_{6}=\sqrt{3 \overline{1}}=5,5678 \\
x_{6}=7 & y_{6}=\sqrt{37}=6,0828 \\
x_{7}=8 & y_{7}=\sqrt{43}=6,5574 \\
x_{8}=9 & y_{8}=\sqrt{49}=7,0000
\end{aligned}
$$

and we calculate the integral using approximate formulas.

By the rectangle formula (1) $/ \approx \sum_{i=n}^{7} y_{l}=34,8183$.

The absolute error of this approximate value (by deficiency) is $38-34,8183=3,1817$, and the relative (percentage) error is $\frac{3,1817 \cdot 100}{38} \approx 8,37 \%$.

By the rectangle formula ( $1 a$ ) $I \approx \sum_{i=1}^{8} y_{i}=40,8183$.

Here, the absolute error (by excess) is 2,8183, and the relative error is $\frac{2,8183 \cdot 100}{38} \approx 7,42 \%$.

By the trapezoidal formula $I \approx 4+\sum_{i=1}^{?} y_{i}=37,8183$.

The absolute error of this result is 0,1817, and the relative error is $\frac{0,1817 \cdot 100}{38} \approx 0,48 \%$.

By the Simpson's formula

$$
I \approx \frac{1}{3}(8+4 \cdot 19,1299+2 \cdot 14,6884) \approx 37,9655
$$

The absolute error is only 0,0345, and the relative error is $\frac{0,0345 \cdot 100}{38} \approx 0,09 \%$.