Exercise: trapezoid and midpoint

Compute $I=\int_0^1 x^3\,dx$. Compare composite trapezoid with $n=2$, simple midpoint and composite midpoint with $n=2$.

1. The exact value is:

   $$I=\int_0^1x^3\,dx=\left[\frac{x^4}{4}\right]_0^1=\frac14$$

2. For composite trapezoid with $n=2$, $h=1/2$ and the nodes are $0,1/2,1$:

   $$\begin{aligned}T_2&=\frac14\left[f(0)+2f\!\left(\frac12\right)+f(1)\right]\\&=\frac14\left[0+2\cdot\frac18+1\right]=\frac{5}{16}\end{aligned}$$

3. For simple midpoint, the centre of the full interval is $m=1/2$:

   $$M_1=(1-0)f\!\left(\frac12\right)=\frac18$$

4. For composite midpoint with $n=2$, the centres are $1/4$ and $3/4$:

   $$\begin{aligned}M_2&=\frac12\left[f\!\left(\frac14\right)+f\!\left(\frac34\right)\right]\\&=\frac12\left(\frac{1}{64}+\frac{27}{64}\right)=\frac{7}{32}\end{aligned}$$

5. The absolute and relative errors are:

   $$\begin{array}{c|c|c} \text{método} & |I-Q| & |I-Q|/I \\ \hline T_2 & 1/16 & 25\% \\ M_1 & 1/8 & 50\% \\ M_2 & 1/32 & 12.5\% \end{array}$$

Compute I=∫_0^{π/2} sin(x)e^{-x} dx with composite midpoint using n=4 and n=8.

1. The exact value is the same as in the previous exercise:

   $$I=0.396060211824619$$

2. The midpoint approximations are:

   $$M_4=0.409710138362011,\qquad M_8=0.401008515076530$$

3. Their absolute errors are:

   $$|E_4|=0.013649926537392,\qquad |E_8|=0.004948303251911$$