Answer
$$
\int_{1}^{2}\sqrt {x^{3}-1}dx, \quad n=10
$$
(a) The approximation of the given integral by using the Trapezoidal Rule is $\approx 1.506361$
(b) The approximation of the given integral by using the midpoints Rule is $\approx 1.518362 $
(c) The approximation of the given integral by using Simpson’s Rule is $\approx 1.511519 $
Work Step by Step
$$
\int_{1}^{2}\sqrt {x^{3}-1}dx, \quad n=10
$$
(a) Use the Trapezoidal Rule to approximate the given integral with the
specified value of n.
With $ n =10 , a = 1$, and $b = 2$ we have
$$
\Delta x=\frac{b-a}{n}=\frac{2-1}{10}=\frac{1}{10}
$$ and so the Trapezoidal Rule gives:
$$
\begin{aligned} \int_{1}^{2}\sqrt {x^{3}-1}dx &= \frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]\\
& \approx T_{10} \\
& =\frac{1}{10.2}[f(1)+2 f(1.1)+2 f(1.2)+ 2 f(1.3)+\\
& \quad\quad\quad+ 2 f(1.4)+2 f(1.5)+2 f(1.6)+2 f(1.7) +\\
& \quad\quad\quad+2 f(1.8)+2 f(1.9)+f(2)] \\
&\approx 1.506361 \end{aligned}
$$
(b) The midpoints Rule to approximate the given integral with the
specified value of $ n=10$ gives :
$$
\begin{aligned} \int_{1}^{2}\sqrt {x^{3}-1}dx &=\Delta x\left[f\left(\overline{x}_{1}\right)+f\left(\overline{x}_{2}\right)+\cdots+f\left(\overline{x}_{n}\right)\right] \\
& = M_{10} \\
& \approx
\frac{1}{10}[f(1.05)+f(1.15)+f(1.25)+f(1.35)+\\
&\quad\quad+f(1.45)+f(1.55)+f(1.65)+f(1.75)+ \\
&\quad\quad+f(1.85)+f(1.95)] \\
&\approx 1.518362 \end{aligned}
$$
(c) The Simpson’s Rule to approximate the given integral with the
specified value of $ n=10$ gives :
$$
\begin{aligned} \int_{1}^{2}\sqrt {x^{3}-1}dx & = \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\ & \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\
&= S_{10}\\
& \approx \frac{1}{10 \cdot 3}[f(1)+4 f(1.1)+2 f(1.2)+4 f(1.3)+\\
& \quad\quad+2 f(1.4)+4 f(1.5)+2 f(1.6)+4 f(1.7)+ \\
& \quad\quad +2 f(1.8)+4 f(1.9)+f(2)] \\ & \approx 1.511519 \end{aligned}
$$
