Problem $2(60$ pts$).$ Consider the integration of the function $f(x)=1+e^{-x}sin(4x)$ over the interval $[a,b]=[0,1].$ Apply the trapezoid rule and Composite Simpson's rule. Refer $h$ as $1$ and $\frac{1}{4}$ for trapezoid and Composite Simpson's rule, respectively. Use $5$ decimal places in all calculations. Compare the results with the true value of the integral. Shortly comment on the results obtained.

Note$-1$: Use Radians.

Note$-2$: No round$-$off is allowed.

Hint$-1$:

${\displaystyle \underset{a}{\overset{b}{}}}f(x)dx=[f(x_0)+f(x_1)]\frac{x_0}{2}+[f(x_1)+f(x_2)]\frac{x_1}{2}+$cdots$+[f(x_{n-1})+f(x_n)]\frac{x_{n-1}}{2}$

${\displaystyle \underset{a}{\overset{b}{}}}f(x)dx=\frac{h[f_1+2f_2+2f_3+\text{c}\text{d}\text{o}\text{t}\text{s}+2f_{n-1}+f_n]}{2}$

Hint$-2$:

${\displaystyle \underset{a}{\overset{b}{}}}f(x)dx={\displaystyle \underset{x_0}{\overset{x_2}{}}}f(x)dx+{\displaystyle \underset{x_2}{\overset{x_4}{}}}f(x)dx+{\displaystyle \underset{x_4}{\overset{x_6}{}}}f(x)dx+$dots$+{\displaystyle \underset{x_{n-2}}{\overset{x_n}{}}}f(x)dx$

~~$h\frac{\{f(x_0)+4f(x_1)+f(x_2)\}}{3}+h\frac{\{f(x_2)+4f(x_3)+f(x_4)\}}{3}+h\frac{\{f(x_4)+4f(x_5)+f(x_6)\}}{3}+h\frac{\{f(x_{n-2})+4f(x_{n-1})+f(x_n)\}}{3}$

~~$h\frac{\{f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+\text{c}\text{d}\text{o}\text{t}\text{s}+4f(x_{n-1})+f(x_n)\}}{3}$