Chapter 5, Problem 9RCC

To determine

To state : the rules for approximating the definite integral.

Answer to Problem 9RCC

trapezoidal rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=T_n=\frac{\Delta x}{2}[f\left(x_0\right)+2f(x_1)+2f\left(x_2\right)+\mathrm{...}+2f\left(x_{n-1}\right)+f\left(x_n\right)]$

Midpoint rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx=M_n=\Delta x[f\left({\overline{x}}_1\right)}+f\left({\overline{x}}_2\right)+\mathrm{.....}+f\left({\overline{x}}_n\right)]$

Simpson’s rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=S_n=\frac{\Delta x}{3}[f\left(x_0\right)+4f(x_1)+2f\left(x_2\right)+\mathrm{...}+4f\left(x_{n-1}\right)+f\left(x_n\right)]$

Explanation of Solution

Given information :

Definite integral : ${\displaystyle {\int }_a^{\infty }f\left(x\right)}dx$

trapezoidal rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=T_n=\frac{\Delta x}{2}[f\left(x_0\right)+2f(x_1)+2f\left(x_2\right)+\mathrm{...}+2f\left(x_{n-1}\right)+f\left(x_n\right)]$

Where $\Delta x=\frac{\left(b-a\right)}{n}\text{ and }{x}_i=a+i\Delta x$ .

Midpoint rule:

  $\begin{array}{l}{\displaystyle {\int }_a^{b}f\left(x\right)dx=M_n=\Delta x[f\left({\overline{x}}_1\right)}+f\left({\overline{x}}_2\right)+\mathrm{.....}+f\left({\overline{x}}_n\right)]\\ \Delta x=\frac{b-a}{n}\\ {\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)=\text{midpoint of }[x_{i-1},x_i]\end{array}$

Simpson’s rule:

  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=S_n=\frac{\Delta x}{3}[f\left(x_0\right)+4f(x_1)+2f\left(x_2\right)+\mathrm{...}+4f\left(x_{n-1}\right)+f\left(x_n\right)]$

Where n is even and $\Delta x=\frac{\left(b-a\right)}{n}$ .

The trapezoidal rule give the best estimate.

Error bound by trapezoidal rule: if $|f\text{'}\text{'}\left(x\right)|\le K$ for $a\le x\le b$ , then the error is

  $\left|E_T\right|\le \frac{K{\left(b-a\right)}^3}{12n^2}$ .

Error bound by trapezoidal rule: if $|f\text{'}\text{'}\left(x\right)|\le K$ for $a\le x\le b$ , then the error is

  $\left|E_M\right|\le \frac{K{\left(b-a\right)}^3}{24n^2}$

Error bound by trapezoidal rule: if $|f\text{'}\text{'}\left(x\right)|\le K$ for $a\le x\le b$ , then the error is

  $\left|E_S\right|\le \frac{K{\left(b-a\right)}^5}{180n^4}$