Answer
$$
\int_{0}^{2} x e^{-x^{2}} d x
$$
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)=x e^{-x^{2}}\\
&\begin{array}{l|l|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 0 & 0 \\
1 & 0.5 & 0.3894 \\
2 & 1 & 0.3679 \\
3 & 1.5 & 0.1581 \\
4 & 2 & 0.0366 \\
\hline
\end{array}
\end{aligned}
Simpson's Rule:
$$
\begin{array}{l}
\int_{0}^{2} x e^{-x^{2}} d x \\
\approx \frac{2-0}{3(4)}[0+4(0.3894)+2(0.3679) \\
\quad+4(0.1581)+0.0366] \\
\approx 0.4937
\end{array}
$$
The answer is close to the exact value obtained in
Exercise 70, which is approximately 0.4908.
Where the exact value can be obtained by:
\begin{aligned}
\int_{0}^{2} x e^{-x^{2}} d x &=-\frac{1}{2} \int_{0}^{2} e^{-x^{2}}(-2 x d x) \\
&=-\left.\frac{1}{2} e^{-x^{2}}\right|_{0} ^{2} \\
&=-\frac{1}{2} e^{-4}+\frac{1}{2} e^{0} \\
&=\frac{1}{2}\left(1-e^{-4}\right) \\
& \approx 0.4908
\end{aligned}
Work Step by Step
$$
\int_{0}^{2} x e^{-x^{2}} d x
$$
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)=x e^{-x^{2}}\\
&\begin{array}{l|l|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 0 & 0 \\
1 & 0.5 & 0.3894 \\
2 & 1 & 0.3679 \\
3 & 1.5 & 0.1581 \\
4 & 2 & 0.0366 \\
\hline
\end{array}
\end{aligned}
Simpson's Rule:
$$
\begin{array}{l}
\int_{0}^{2} x e^{-x^{2}} d x \\
\approx \frac{2-0}{3(4)}[0+4(0.3894)+2(0.3679) \\
\quad+4(0.1581)+0.0366] \\
\approx 0.4937
\end{array}
$$
The answer is close to the exact value obtained in
Exercise 70, which is approximately 0.4908.
Where the exact value can be obtained by:
\begin{aligned}
\int_{0}^{2} x e^{-x^{2}} d x &=-\frac{1}{2} \int_{0}^{2} e^{-x^{2}}(-2 x d x) \\
&=-\left.\frac{1}{2} e^{-x^{2}}\right|_{0} ^{2} \\
&=-\frac{1}{2} e^{-4}+\frac{1}{2} e^{0} \\
&=\frac{1}{2}\left(1-e^{-4}\right) \\
& \approx 0.4908
\end{aligned}