Answer
$$
\int_{1}^{2} \sqrt{x^{3}-1} d x,\quad n=10
$$
Here
$$
f(x)=\sqrt{x^{3}-1}, \quad \Delta x=\frac{b-a}{n}=\frac{2-1}{10}=\frac{1}{10}
$$
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{10} &=\frac{1}{10 \cdot 2}[f(1)+2 f(1.1)+2 f(1.2)+2 f(1.3)+2 f(1.4)+2 f(1.5)\\
& \quad\quad +2 f(1.6)+2 f(1.7)+2 f(1.8)+2 f(1.9)+f(2)]\\
& \approx 1.506361
\end{aligned} \\
$$
(b) The midpoints Rule :
$$
\begin{aligned}
M_{10} &=\frac{1}{10}[f(1.05)+f(1.15)+f(1.25)+f(1.35)+f(1.45) \\
&\quad \quad +f(1.55)+f(1.65)+f(1.75)+f(1.85)+f(1.95)] \\
& \approx 1.518362
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
S_{10}=& \frac{1}{10 \cdot 3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+2 f(1.4)\\
&+4 f(1.5)+2 f(1.6)+4 f(1.7)+2 f(1.8)+4 f(1.9)+f(2)] \\
& \approx 1.511519
\end{aligned}
$$
Work Step by Step
$$
\int_{1}^{2} \sqrt{x^{3}-1} d x,\quad n=10
$$
Here
$$
f(x)=\sqrt{x^{3}-1}, \quad \Delta x=\frac{b-a}{n}=\frac{2-1}{10}=\frac{1}{10}
$$
and so the Trapezoidal Rule gives:
$$
\int_{a}^{b} f(x) d x \approx T_{n}=\frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+2 f\left(x_{2}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]
$$
Therefore,
(a) the Trapezoidal Rule
$$
\begin{aligned}
T_{10} &=\frac{1}{10 \cdot 2}[f(1)+2 f(1.1)+2 f(1.2)+2 f(1.3)+2 f(1.4)+2 f(1.5)\\
& \quad\quad +2 f(1.6)+2 f(1.7)+2 f(1.8)+2 f(1.9)+f(2)]\\
& \approx 1.506361
\end{aligned} \\
$$
(b) The midpoints of the ten subintervals are 1.1,1.2, 1.3,1.4, 1.5,1.6, 1.7,1.8, 1.9,and 2, so the Midpoint Rule gives
$$
\int_{a}^{b} f(x) d x \approx M_{n}=\Delta x\left[f\left(\bar{x}_{1}\right)+f\left(\bar{x}_{2}\right)+\cdots+f\left(\bar{x}_{n}\right)\right]
$$
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:
$$
\begin{aligned}
M_{10} &=\frac{1}{10}[f(1.05)+f(1.15)+f(1.25)+f(1.35)+f(1.45) \\
&\quad \quad +f(1.55)+f(1.65)+f(1.75)+f(1.85)+f(1.95)] \\
& \approx 1.518362
\end{aligned}
$$
(c) Simpson’s Rule:
$$
\begin{aligned}
\int_{a}^{b} f(x) d x & \approx S_{n} \\
&=\frac{\Delta x}{3}\left[f\left(x_{0}\right)+4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)\\
+\cdots\right. \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]
\end{aligned}
$$
In Simpson’s Rule, we obtain:
$$
\begin{aligned}
S_{10}=& \frac{1}{10 \cdot 3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+2 f(1.4)\\
&+4 f(1.5)+2 f(1.6)+4 f(1.7)+2 f(1.8)+4 f(1.9)+f(2)] \\
& \approx 1.511519
\end{aligned}
$$
