Answer
$$
\int_{1}^{5} \frac{1}{x^{2}} d x
$$
Here $a=1, b=5,$ and $n=4,$ with $(b-a) / n=(5-1) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=1, x_{1}=2, x_{2}=3, x_{3}=4,$ and $x_{4}=5 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=5, a=1, f(x)=\frac{1}{x^{2}}\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 1 & 1 \\
1 & 2 & 0.25 \\
2 & 3 & 0.1111 \\
3 & 4 & 0.0625 \\
4 & 5 & 0.04 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{1}^{5} \frac{1}{x^{2}} d x \\
\quad \quad \approx \frac{5-1}{4}\left[\frac{1}{2}(1)+0.25+0.1111\right. \\
\left.\quad\quad\quad+0.0625+\frac{1}{2}(0.04)\right] \\
\quad\quad \approx 0.9436
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{aligned}
& \int_{1}^{5} \frac{1}{x^{2}} d x \\
& \quad \approx \frac{5-1}{12}[1+4(0.25)+2(0.1111)\\
&\quad\quad +4(0.0625)+0.04)] \\
& \quad \approx 0.8374
\end{aligned}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{1}^{5} x^{-2} d x &=-\left.x^{-1}\right|_{1} ^{5} \\
&=-\frac{1}{5}+1 \\
&=\frac{4}{5} \\
&=0.8
\end{aligned}
$$
Work Step by Step
$$
\int_{1}^{5} \frac{1}{x^{2}} d x
$$
Here $a=1, b=5,$ and $n=4,$ with $(b-a) / n=(5-1) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=1, x_{1}=2, x_{2}=3, x_{3}=4,$ and $x_{4}=5 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=5, a=1, f(x)=\frac{1}{x^{2}}\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 1 & 1 \\
1 & 2 & 0.25 \\
2 & 3 & 0.1111 \\
3 & 4 & 0.0625 \\
4 & 5 & 0.04 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{1}^{5} \frac{1}{x^{2}} d x \\
\quad \quad \approx \frac{5-1}{4}\left[\frac{1}{2}(1)+0.25+0.1111\right. \\
\left.\quad\quad\quad+0.0625+\frac{1}{2}(0.04)\right] \\
\quad\quad \approx 0.9436
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{aligned}
& \int_{1}^{5} \frac{1}{x^{2}} d x \\
& \quad \approx \frac{5-1}{12}[1+4(0.25)+2(0.1111)\\
&\quad\quad +4(0.0625)+0.04)] \\
& \quad \approx 0.8374
\end{aligned}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{1}^{5} x^{-2} d x &=-\left.x^{-1}\right|_{1} ^{5} \\
&=-\frac{1}{5}+1 \\
&=\frac{4}{5} \\
&=0.8
\end{aligned}
$$
