We consider the integral $$\int_{-1}^1\sqrt[3]{x^2}\sqrt{|x|^3}\, dx$$
Using the trapezoid formula with $h=0.25$ we get the following:
The formula is $$T(h)=h\cdot \left (\frac{f(a)}{2}+\sum_{i=1}^{n-1}f(a+ih)+\frac{f(b)}{2}\right )$$
In this case we get \begin{align*}T(0.25)&=0.25\cdot \left (\frac{f(-1)}{2}+f(-0.75)+f(-0.5)+f(-0.25)+f(0)+f(0.25)+f(0.5)+f(0.75)+\frac{f(1)}{2}\right )\\ & \approx 0.65424865\end{align*}
The exact value of the integral is $\log\left (\frac{4}{3}\right )\approx 0.287682072$.
An upper bound for the error is $$\frac{h^2}{12}(b-a)\|f''\|_{\infty}=\frac{1}{96}\|f''\|_{\infty}$$ right?
To calculate the derivative can we write the funciton as follows? $$\sqrt[3]{x^2}\sqrt{|x|^3}=\left (x\right )^{\frac{2}{3}}\cdot \sqrt{\left (\sqrt{x^2}\right )^3}=\left (x\right )^{\frac{2}{3}}\cdot \left (\sqrt{x^2}\right )^\frac{3}{2}=\left (x\right )^{\frac{2}{3}}\cdot \left (x\right )^{\frac{6}{4}}=\left (x\right )^{\frac{13}{6}}$$