Answer
$$
\int_{0}^{1}\frac{x^{2}}{1+x^{4}} dx ,\quad n=10
$$
Here
$$
f(x)=\frac{x^{2}}{1+x^{4}}, \quad \Delta x=\frac{1-0}{10}=\frac{1}{10}
$$
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{10} &=\frac{1}{10 \cdot 2}\{f(0)+2[f(0.1+f(0.2)+\cdots+f(0.9)]+f(1)\} \\
& \approx 0.243747
\end{aligned}
$$
(b) The midpoints Rule is given by:
$$
\begin{aligned}
M_{10}&=\frac{1}{10}[f(0.05)+f(0.15)+\cdots+f(0.85)+f(0.95)] \\
&\approx 0.243748
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
S_{10}&=\frac{1}{10 \cdot 3}[f(0)+4 f(0.1)+2 f(0.2)+4 f(0.3)+2 f(0.4)+4 f(0.5) \\
& \quad\quad +2 f(0.6) +4 f(0.7)+2 f(0.8)+4 f(0.9)+f(1)] \\&
\approx 0.243751
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{1}\frac{x^{2}}{1+x^{4}} dx ,\quad n=10
$$
Here
$$
f(x)=\frac{x^{2}}{1+x^{4}}, \quad \Delta x=\frac{1-0}{10}=\frac{1}{10}
$$
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
$$
\begin{aligned}
T_{10} &=\frac{1}{10 \cdot 2}\{f(0)+2[f(0.1+f(0.2)+\cdots+f(0.9)]+f(1)\} \\
& \approx 0.243747
\end{aligned}
$$
(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:
$$
\begin{aligned}
M_{10}&=\frac{1}{10}[f(0.05)+f(0.15)+\cdots+f(0.85)+f(0.95)] \\
&\approx 0.243748
\end{aligned}
$$
(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:
$$
\begin{aligned}
S_{10}&=\frac{1}{10 \cdot 3}[f(0)+4 f(0.1)+2 f(0.2)+4 f(0.3)+2 f(0.4)+4 f(0.5) \\
& \quad\quad +2 f(0.6) +4 f(0.7)+2 f(0.8)+4 f(0.9)+f(1)] \\&
\approx 0.243751
\end{aligned}
$$