Answer
$$
\int_{0}^{2}\left(3 x^{2}+2\right) d x\
$$
Here $a=0, b=2,$ and $n=4,$ with $(b-a) / n=(2-0) / 4=1 / 2$ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1 / 2, x_{2}=1, x_{3}=3 / 2,$ and $x_{4}=2 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=2, a=0, f(x)=3 x^{2}+2\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 0 & 2 \\
1 & \frac{1}{2} & 2.75 \\
2 & 1 & 5 \\
3 & \frac{3}{2} & 8.75 \\
4 & 2 & 14 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{0}^{2}\left(3 x^{2}+2\right) d x \\
\quad\quad\quad\quad \approx \frac{2-0}{4}\left[\frac{1}{2}(2)+2.75+5+8.75+\frac{1}{2}(14)\right] \\
\quad\quad\quad\quad =0.5(24.5) \\
\quad\quad\quad\quad =12.25
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{0}^{2}\left(3 x^{2}+2\right) d x \\
\quad\quad\quad\quad \approx \frac{2-0}{3(4)}[2+4(2.75)+2(5)+4(8.75)+14] \\
\quad\quad\quad\quad =\frac{2}{12}(72) \\
\quad\quad\quad\quad =12
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{0}^{2}\left(3 x^{2}+2\right) d x &=\left.\left(x^{3}+2 x\right)\right|_{0} ^{2} \\
&=(8+4)-0 \\
&=12
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{2}\left(3 x^{2}+2\right) d x\
$$
Here $a=0, b=2,$ and $n=4,$ with $(b-a) / n=(2-0) / 4=1 / 2$ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1 / 2, x_{2}=1, x_{3}=3 / 2,$ and $x_{4}=2 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=2, a=0, f(x)=3 x^{2}+2\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 0 & 2 \\
1 & \frac{1}{2} & 2.75 \\
2 & 1 & 5 \\
3 & \frac{3}{2} & 8.75 \\
4 & 2 & 14 \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{0}^{2}\left(3 x^{2}+2\right) d x \\
\quad\quad\quad\quad \approx \frac{2-0}{4}\left[\frac{1}{2}(2)+2.75+5+8.75+\frac{1}{2}(14)\right] \\
\quad\quad\quad\quad =0.5(24.5) \\
\quad\quad\quad\quad =12.25
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{0}^{2}\left(3 x^{2}+2\right) d x \\
\quad\quad\quad\quad \approx \frac{2-0}{3(4)}[2+4(2.75)+2(5)+4(8.75)+14] \\
\quad\quad\quad\quad =\frac{2}{12}(72) \\
\quad\quad\quad\quad =12
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{0}^{2}\left(3 x^{2}+2\right) d x &=\left.\left(x^{3}+2 x\right)\right|_{0} ^{2} \\
&=(8+4)-0 \\
&=12
\end{aligned}
$$
