Answer
$$
\int_{0}^{4}x^{3} \sin x d x,\quad n=8
$$
Here
$$
f(x)=x^{3} \sin x, \quad \Delta x=\frac{4-0}{8}=\frac{1}{2}
$$
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{8}&=\frac{1}{2 \cdot 2}\left[f(0)+2 f\left(\frac{1}{2}\right)+2 f(1)+2 f\left(\frac{3}{2}\right)+\\
+2 f(2)+2 f\left(\frac{5}{2}\right)+2 f(3)+2 f\left(\frac{7}{2}\right)+f(4)\right] \\
&\approx-7.276910
\end{aligned}
$$
(b) The Midpoints Rule is given by:
$$
\begin{aligned}
M_{8} &=\frac{1}{2}\left[f\left(\frac{1}{4}\right)+f\left(\frac{3}{4}\right)+f\left(\frac{5}{4}\right)+f\left(\frac{7}{4}\right)+f\left(\frac{9}{4}\right)+\\
+f\left(\frac{11}{4}\right)+f\left(\frac{13}{4}\right)+f\left(\frac{15}{4}\right)\right] \\
& \approx-4.818251
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
S_{8} &=\frac{1}{2 \cdot 3}\left[f(0)+4 f\left(\frac{1}{2}\right)+2 f(1)+4 f\left(\frac{3}{2}\right)+2 f(2)+4 f\left(\frac{5}{2}\right)+\\ +2 f(3)
+4 f\left(\frac{7}{2}\right)+f(4)\right] \\
& \approx-5.605350
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{4}x^{3} \sin x d x,\quad n=8
$$
Here
$$
f(x)=x^{3} \sin x, \quad \Delta x=\frac{4-0}{8}=\frac{1}{2}
$$
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
$$
\begin{aligned}
T_{8}&=\frac{1}{2 \cdot 2}\left[f(0)+2 f\left(\frac{1}{2}\right)+2 f(1)+2 f\left(\frac{3}{2}\right)+\\
+2 f(2)+2 f\left(\frac{5}{2}\right)+2 f(3)+2 f\left(\frac{7}{2}\right)+f(4)\right] \\
&\approx-7.276910
\end{aligned}
$$
(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:
$$
\begin{aligned}
M_{8} &=\frac{1}{2}\left[f\left(\frac{1}{4}\right)+f\left(\frac{3}{4}\right)+f\left(\frac{5}{4}\right)+f\left(\frac{7}{4}\right)+f\left(\frac{9}{4}\right)+\\
+f\left(\frac{11}{4}\right)+f\left(\frac{13}{4}\right)+f\left(\frac{15}{4}\right)\right] \\
& \approx-4.818251
\end{aligned}
$$
(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:
$$
\begin{aligned}
S_{8} &=\frac{1}{2 \cdot 3}\left[f(0)+4 f\left(\frac{1}{2}\right)+2 f(1)+4 f\left(\frac{3}{2}\right)+2 f(2)+4 f\left(\frac{5}{2}\right)+\\ +2 f(3)
+4 f\left(\frac{7}{2}\right)+f(4)\right] \\
& \approx-5.605350
\end{aligned}
$$