Answer
$$
\int_{-1}^{3} \frac{3}{5-x} d x
$$
Here $a=-1, b=3,$ and $n=4,$ with $(b-a) / n=(3+1) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=-1, x_{1}=0, x_{2}=1, x_{3}=2,$ and $x_{4}=3 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=3, a=-1, f(x)=\frac{3}{5-x}\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & -1 & 0.5 \\
1 & 0 & 0.6 \\
2 & 1 & 0.75 \\
3 & 2 & 1 \\
4 & 3 & 1.5 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{-1}^{3} \frac{3}{5-x} d x \\
\quad \approx \frac{3-(-1)}{4}\left[\frac{1}{2}(0.5)+0.6+0.75+1+\frac{1}{2}(1.5)\right] \\
\quad=1(3.35) \\
\quad=3.35
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{-1}^{3} \frac{3}{5-x} d x \\
\quad \approx \frac{3-(-1)}{3(4)}[0.5+4(0.6)+2(0.75)+4(1)+1.5] \\
\quad=\frac{1}{3}\left(\frac{99}{10}\right) \\
\quad=\frac{33}{10} \approx 3.3
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{-1}^{3} \frac{3}{5-x} d x &=-\left.3 \ln |5-x|\right|_{-1} ^{3} \\
&=-3(\ln |2|-\ln |6|) \\
&=3 \ln 3 \approx 3.296
\end{aligned}
$$
Work Step by Step
$$
\int_{-1}^{3} \frac{3}{5-x} d x
$$
Here $a=-1, b=3,$ and $n=4,$ with $(b-a) / n=(3+1) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=-1, x_{1}=0, x_{2}=1, x_{3}=2,$ and $x_{4}=3 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=3, a=-1, f(x)=\frac{3}{5-x}\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & -1 & 0.5 \\
1 & 0 & 0.6 \\
2 & 1 & 0.75 \\
3 & 2 & 1 \\
4 & 3 & 1.5 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{-1}^{3} \frac{3}{5-x} d x \\
\quad \approx \frac{3-(-1)}{4}\left[\frac{1}{2}(0.5)+0.6+0.75+1+\frac{1}{2}(1.5)\right] \\
\quad=1(3.35) \\
\quad=3.35
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{-1}^{3} \frac{3}{5-x} d x \\
\quad \approx \frac{3-(-1)}{3(4)}[0.5+4(0.6)+2(0.75)+4(1)+1.5] \\
\quad=\frac{1}{3}\left(\frac{99}{10}\right) \\
\quad=\frac{33}{10} \approx 3.3
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{-1}^{3} \frac{3}{5-x} d x &=-\left.3 \ln |5-x|\right|_{-1} ^{3} \\
&=-3(\ln |2|-\ln |6|) \\
&=3 \ln 3 \approx 3.296
\end{aligned}
$$
