Answer
$$
\int_{0}^{2} \frac{e^{x}}{1+x^{2}} d x,\quad n=10
$$
Here
$$
f(x)=\frac{e^{x}}{1+x^{2}}, \quad \Delta x=\frac{b-a}{n}=\frac{2-0}{10}=\frac{1}{5}
$$
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{10} &=\frac{1}{5 \cdot 2}[f(0)+2 f(0.2)+2 f(0.4)+2 f(0.6)+2 f(0.8)+2 f(1) \\
& \quad +2 f(1.2)+2 f(1.4)+2 f(1.6)+2 f(1.8)+f(2)] \\
& \approx 2.660833
\end{aligned}
$$
(b) The Midpoints Rule:
$$
\begin{aligned}
M_{10}& =\frac{1}{5}[f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9)\\
&\quad +f(1.1)+f(1.3)+f(1.5)+f(1.7)+f(1.9)] \\
& \approx 2.664377 \\
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
S_{10} &=\frac{1}{5 \cdot 3}[f(0)+4 f(0.2)+2 f(0.4)+4 f(0.6)+2 f(0.8) \\
&\quad+4 f(1)+2 f(1.2)+4 f(1.4)+2 f(1.6)+4 f(1.8)+f(2)] \\
&\approx 2.663244
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{2} \frac{e^{x}}{1+x^{2}} d x,\quad n=10
$$
Here
$$
f(x)=\frac{e^{x}}{1+x^{2}}, \quad \Delta x=\frac{b-a}{n}=\frac{2-0}{10}=\frac{1}{5}
$$
and so the Trapezoidal Rule can be given by:
$$
\int_{a}^{b} f(x) d x \approx T_{n}=\frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+2 f\left(x_{2}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]
$$
Therefore,
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{10} &=\frac{1}{5 \cdot 2}[f(0)+2 f(0.2)+2 f(0.4)+2 f(0.6)+2 f(0.8)+2 f(1) \\
& \quad +2 f(1.2)+2 f(1.4)+2 f(1.6)+2 f(1.8)+f(2)] \\
& \approx 2.660833
\end{aligned}
$$
(b) The midpoints Rule is given by:
$$
\int_{a}^{b} f(x) d x \approx M_{n}=\Delta x\left[f\left(\bar{x}_{1}\right)+f\left(\bar{x}_{2}\right)+\cdots+f\left(\bar{x}_{n}\right)\right]
$$
where
$$
\bar{x}_{i}=\frac{1}{2}\left(x_{i-1}+x_{i}\right)=\text { midpoint of }\left[x_{i-1}, x_{i}\right]
$$
So, we have:
$$
\begin{aligned}
M_{10}& =\frac{1}{5}[f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9)\\
&\quad +f(1.1)+f(1.3)+f(1.5)+f(1.7)+f(1.9)] \\
& \approx 2.664377 \\
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
\int_{a}^{b} f(x) d x & \approx S_{n} \\
&=\frac{\Delta x}{3}\left[f\left(x_{0}\right)+4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)\\
+\cdots\right. \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]
\end{aligned}
$$
So, we obtain:
$$
\begin{aligned}
S_{10} &=\frac{1}{5 \cdot 3}[f(0)+4 f(0.2)+2 f(0.4)+4 f(0.6)+2 f(0.8) \\
&\quad+4 f(1)+2 f(1.2)+4 f(1.4)+2 f(1.6)+4 f(1.8)+f(2)] \\
&\approx 2.663244
\end{aligned}
$$