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? $$\sqrt[3]{x^2}\sqrt{|x|^3}=\left (x\right )^{\frac{2}{3}}\cdot \sqrt{\left (\sqrt{x^2}\right )^3}=\left (x\right )^{\frac{2}{3}}\cdot \left (\sqrt{x^2}\right )^\frac{3}{2}=\left (x\right )^{\frac{2}{3}}\cdot \left (x\right )^{\frac{6}{4}}=\left (x\right )^{\frac{13}{6}}$$

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    $\begingroup$ why do you say that the exact value is $\log(4/3)$? Using the symmetry of the interval and of the function to be integrated, the proposed integral is equal to $ 2 \int_0^1 x^{13/6} dx = \frac{12}{19}$. $\endgroup$ – PierreCarre Feb 6 '19 at 18:27
  • $\begingroup$ Is it correct to write the initial function in the form $x^{\frac{13}{6}}$ ? @PierreCarre $\endgroup$ – Mary Star Feb 6 '19 at 18:47
  • $\begingroup$ If you integrate over [0,1], where $x \ge 0$ you can get rid of the absolute value and simplify the function, sure. $\endgroup$ – PierreCarre Feb 6 '19 at 18:53
  • $\begingroup$ And to calculate the second derivative of $f$ ? How can we do that? @PierreCarre $\endgroup$ – Mary Star Feb 6 '19 at 18:53
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    $\begingroup$ What do you mean? It is a power of $x$... $f'' = \left(\frac{13}{6} x^{7/6}\right)' = \frac{13}{6} \cdot \frac{7}{6} x^{1/6}$ $\endgroup$ – PierreCarre Feb 6 '19 at 19:08

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