Answer
$$
\int_{0}^{2}\frac{1}{1+x^{6}}dx, \quad n=8
$$
(a) The approximation of the given integral by using the Trapezoidal Rule is $\approx 1.040756 $
(b) The approximation of the given integral by using the midpoints Rule is $\approx 1.041109 $
(c) The approximation of the given integral by using Simpson’s Rule is $\approx 1.042172 $
Work Step by Step
$$
\int_{0}^{2}\frac{1}{1+x^{6}}dx, \quad n=8
$$
(a) Use the Trapezoidal Rule to approximate the given integral with the
specified value of n.
With $ n =8, a = 0$, and $b = 2$ we have
$$
\Delta x=\frac{b-a}{n}=\frac{2-0}{8}=\frac{1}{4}
$$ and so the Trapezoidal Rule gives:
$$
\begin{aligned} \int_{0}^{2}\frac{1}{1+x^{6}}dx &= \frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]\\
& \approx T_{8} \\
& =\frac{1}{4 \cdot 2}[f(0)+2 f(0.25)+2 f(0.5)+2 f(0.75)+\\
&\quad\quad+2 f(1)+2 f(1.25)+2 f(1.5)+\\
&\quad\quad+2 f(1.75)+f(2)] \\
& \approx 1.040756 \end{aligned}
$$
(b) The midpoints Rule to approximate the given integral with the
specified value of $ n=8$ gives :
$$
\begin{aligned}\int_{0}^{2}\frac{1}{1+x^{6}}dx &=\Delta x\left[f\left(\overline{x}_{1}\right)+f\left(\overline{x}_{2}\right)+\cdots+f\left(\overline{x}_{n}\right)\right] \\
& = M_{8} \\
& \approx
\frac{1}{4}[f(0.125)+f(0.375)+f(0.625)+f(0.875)+ \\
& \quad\quad+ f(1.125)+f(1.625)+f(1.875)] \\
& \approx 1.041109 \end{aligned}
$$
(c) Simpson’s Rule to approximate the given integral with the
specified value of $ n=8$ gives :
$$
\begin{aligned} \int_{0}^{2}\frac{1}{1+x^{6}}dx & = \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\ & \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\
&= S_{8}\\
& \approx \frac{1}{4 \cdot 3}[f(0)+4 f(0.25)+2 f(0.5)+ 4 f(0.75)+\\
&\quad\quad+ 2 f(1)+4 f(1.25)+2 f(1.5)+4 f(1.75)+f(2)] \\
& \approx 1.042172 \end{aligned}
$$
