Answer
$$
\int_{0}^{3}\left(2 x^{3}+1\right) d x
$$
Here $a=0, b=3,$ and $n=4,$ with $(b-a) / n=(3-0) / 4=3 / 4 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=3 / 4, x_{2}=3/2, x_{3}=9 / 4,$ and $x_{4}=3 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=3, a=0, f(x)=2 x^{3}+1\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f(x) \\
\hline 0 & 0 & 1 \\
1 & \frac{3}{4} & \frac{59}{32} \\
2 & \frac{3}{2} & \frac{31}{4} \\
3 & \frac{9}{4} & \frac{761}{32} \\
4 & 3 & 55 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{0}^{3}\left(2 x^{3}+1\right) d x \\
\quad \approx \frac{3-0}{4}\left[\frac{1}{2}(1)+\frac{59}{32}+\frac{31}{4}+\frac{761}{32}+\frac{1}{2}(55)\right] \\
\quad =\frac{3}{4}\left(\frac{491}{8}\right) \\
\quad \approx 46.03
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{0}^{3}\left(2 x^{3}+1\right) d x \\
\quad =\frac{3-0}{3(4)}\left[1+4\left(\frac{59}{32}\right)+2\left(\frac{31}{4}\right)\right. \\
\left.\quad\quad +4\left(\frac{761}{32}\right)+55\right] \\
\quad =\frac{1}{4}(174) \\
\quad =43.5
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{0}^{3}\left(2 x^{3}+1\right) d x &=\left.\left(\frac{x^{4}}{2}+x\right)\right|_{0} ^{3} \\
&=\left(\frac{81}{2}+3\right)-0 \\
&=\frac{87}{2} \\
&=43.5
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{3}\left(2 x^{3}+1\right) d x
$$
Here $a=0, b=3,$ and $n=4,$ with $(b-a) / n=(3-0) / 4=3 / 4 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=3 / 4, x_{2}=3/2, x_{3}=9 / 4,$ and $x_{4}=3 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=3, a=0, f(x)=2 x^{3}+1\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f(x) \\
\hline 0 & 0 & 1 \\
1 & \frac{3}{4} & \frac{59}{32} \\
2 & \frac{3}{2} & \frac{31}{4} \\
3 & \frac{9}{4} & \frac{761}{32} \\
4 & 3 & 55 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{0}^{3}\left(2 x^{3}+1\right) d x \\
\quad \approx \frac{3-0}{4}\left[\frac{1}{2}(1)+\frac{59}{32}+\frac{31}{4}+\frac{761}{32}+\frac{1}{2}(55)\right] \\
\quad =\frac{3}{4}\left(\frac{491}{8}\right) \\
\quad \approx 46.03
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{0}^{3}\left(2 x^{3}+1\right) d x \\
\quad =\frac{3-0}{3(4)}\left[1+4\left(\frac{59}{32}\right)+2\left(\frac{31}{4}\right)\right. \\
\left.\quad\quad +4\left(\frac{761}{32}\right)+55\right] \\
\quad =\frac{1}{4}(174) \\
\quad =43.5
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{0}^{3}\left(2 x^{3}+1\right) d x &=\left.\left(\frac{x^{4}}{2}+x\right)\right|_{0} ^{3} \\
&=\left(\frac{81}{2}+3\right)-0 \\
&=\frac{87}{2} \\
&=43.5
\end{aligned}
$$
