Answer
$$
\int_{0}^{1} 4 x e^{-x^{2}} d x
$$
Here $a=0, b=1,$ and $n=4,$ with $(b-a) / n=(1-0) / 4=1/4 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1/4 , x_{2}=1/2 , x_{3}=3/4 ,$ and $x_{4}=1 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=1, a=0, f(x)=4 x e^{-x^{2}}\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 0 & 0 \\
1 & \frac{1}{4} & e^{-1 / 16} \\
2 & \frac{1}{2} & 2 e^{-1 / 4} \\
3 & \frac{3}{4} & 3 e^{-9 / 16} \\
4 & 1 & 4 e^{-1} \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{0}^{1} 4 x e^{-x^{2}} d x \\
\approx \frac{1-0}{4} \mid \frac{1}{2}(0)+e^{-1 / 16}+2 e^{-1 / 4} \\
\left.\quad+3 e^{-9 / 16}+\frac{1}{2}\left(4 e^{-1}\right)\right] \\
=\frac{1}{4}\left(e^{-1 / 16}+2 e^{-1 / 4}+3 e^{-9 / 16}+2 e^{-1}\right) \\
\approx 1.236
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{0}^{1} 4 x e^{-x^{2}} d x \\
\approx \frac{1-0}{3(4)}\left[0+4\left(e^{-1 / 16}\right)+2\left(2 e^{-1 / 4}\right)\right. \\
\left.\quad+4\left(3 e^{-9 / 16}\right)+4 e^{-1}\right] \\
=\frac{1}{12}\left(4 e^{-1 / 16}+4 e^{-1 / 4}+12 e^{-9 / 16}+4 e^{-1}\right) \\
\approx 1.265
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{0}^{1} 4 x e^{-x^{2}} d x & =\int_{0}^{1} 2 e^{-x^{2}} (2 x d x) \\
&=\int_{0}^{1} -2 e^{-x^{2}} (-2 x d x) \\
&=\int_{0}^{1} -2 e^{-x^{2}} d (-x^{2})\\
&=-\left.2 e^{-x^{2}}\right|_{0} ^{1} \\
&=\left(-2 e^{-1}\right)-(-2) \\
&=2-2 e^{-1} \approx 1.264
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{1} 4 x e^{-x^{2}} d x
$$
Here $a=0, b=1,$ and $n=4,$ with $(b-a) / n=(1-0) / 4=1/4 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1/4 , x_{2}=1/2 , x_{3}=3/4 ,$ and $x_{4}=1 .$ Now find the corresponding function values. The work can be organized into a table, as follows.
$$
\begin{aligned}
&n=4, b=1, a=0, f(x)=4 x e^{-x^{2}}\\
&\begin{array}{c|c|l}
\hline i & x_{i} & f\left(x_{i}\right) \\
\hline 0 & 0 & 0 \\
1 & \frac{1}{4} & e^{-1 / 16} \\
2 & \frac{1}{2} & 2 e^{-1 / 4} \\
3 & \frac{3}{4} & 3 e^{-9 / 16} \\
4 & 1 & 4 e^{-1} \\
\hline
\end{array}
\end{aligned}
$$
(a) the trapezoidal rule:
$$
\begin{array}{l}
\int_{0}^{1} 4 x e^{-x^{2}} d x \\
\approx \frac{1-0}{4} \mid \frac{1}{2}(0)+e^{-1 / 16}+2 e^{-1 / 4} \\
\left.\quad+3 e^{-9 / 16}+\frac{1}{2}\left(4 e^{-1}\right)\right] \\
=\frac{1}{4}\left(e^{-1 / 16}+2 e^{-1 / 4}+3 e^{-9 / 16}+2 e^{-1}\right) \\
\approx 1.236
\end{array}
$$
(b) Simpson’s Rule:
$$
\begin{array}{l}
\int_{0}^{1} 4 x e^{-x^{2}} d x \\
\approx \frac{1-0}{3(4)}\left[0+4\left(e^{-1 / 16}\right)+2\left(2 e^{-1 / 4}\right)\right. \\
\left.\quad+4\left(3 e^{-9 / 16}\right)+4 e^{-1}\right] \\
=\frac{1}{12}\left(4 e^{-1 / 16}+4 e^{-1 / 4}+12 e^{-9 / 16}+4 e^{-1}\right) \\
\approx 1.265
\end{array}
$$
(c) Exact value:
$$
\begin{aligned}
\int_{0}^{1} 4 x e^{-x^{2}} d x & =\int_{0}^{1} 2 e^{-x^{2}} (2 x d x) \\
&=\int_{0}^{1} -2 e^{-x^{2}} (-2 x d x) \\
&=\int_{0}^{1} -2 e^{-x^{2}} d (-x^{2})\\
&=-\left.2 e^{-x^{2}}\right|_{0} ^{1} \\
&=\left(-2 e^{-1}\right)-(-2) \\
&=2-2 e^{-1} \approx 1.264
\end{aligned}
$$
