We have the following integral $$\int_{1}^{3} \frac x{x^2+4}\; dx$$ and $n=6$. I have to approximate it using composite trapezoidal rule and then composite simpson rule.
Using trapezoidal rule:
$$\int_a^b f(x)\; dx = \frac h2 \left[f(a)+2\sum_{j=1}^{n-1}f(x_j)+f(b)\right]-\frac{b-a}{12}h^2f''(\mu)$$
where $h=\frac{3-1}6=1/3$.
Here is my solution for composite trapezoidal rule and error formula:
$$ h=\frac{b-a}n=\frac{3-1}6=\frac13 \\ \int_a^bf(x)\;dx = \frac h2 \left[f(a)+2\sum_{j=1}^{n-1}f(x_1)+f(b)\right] $$
$$\begin{array}{c|l} x & y \\ \hline 1 & 0.2 \\ \frac43 & 0.2343 \\ \frac53 & 0.2466 \\ 2 & 0.25 \\ \frac73 & 0.2470 \\ \frac83 & 0.2400 \\ 3 & 0.2307 \end{array}$$
$$ \begin{align} \int_1^3\frac x{x^2+4}\;dx & = \frac16[0.2+0.2307+2(0.2343+0.2466+0.25+0.2470+0.24)] \\ & = 0.48065 \end{align} $$
We see that $n=2$.
$$\begin{align} E & =-\frac{b-a}{12}\cdot h^2f''(n) \\ & =-\frac2{12}\cdot\left(\frac13\right)^2\cdot\frac{2n\left(n^2-12\right)}{\left(n^2+n\right)^3} \end{align}$$
Here $n=2$.
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