I am working on a question and I am stuck, wish to find some help.
The question ask me to find the error bound for approximating the following integral by consider $n=4$ non-composite Newton-Cotes formula: $$\int^{1.2}_0 x^3 \tanh(0.5x) dx$$ The error term of $n=4$ non-composite Newton-Cotes formula is $\frac{8h^7}{945}f^{(6)}(\zeta)$.
So I calculate out the 6th derivative for the integral, and get $$E=\frac{8\left(\frac{b-a}{n}\right)}{945}M = 0.0000277709$$ which $M=\max\limits_{x\in[a,b]}|f^{(6)}(\zeta)|$.
Then I am stuck, what can I do next?
The error bound is given by: \begin{equation*} E=\frac{8\left(\frac{1.2-0}{4}\right)^7}{945}M \end{equation*} Where $M=30$, $E=0.00005554285714$.
$f^{(6)}(x)$ over $x \in[0,1.2]$