Chapter 5.3, Problem 11E

To determine

To find: The average value of equation.

Answer to Problem 11E

The average value of $f(x)=x^2-1$ in equation is $1.00000061$ .

Explanation of Solution

Given:

The function given is $f(x)=x^2-1$ .

Formula used:

The right-hand side of the equation is an approximation of the left-hand side.

Calculator has a numerical integration capability, which will denote as NINT. In particular, will use $NINT(f(x),x,a,b)$ to denote a calculator approximation of ${\displaystyle {\int }_a^{b}f(x)dx}$ .

  ${\displaystyle {\int }_a^{b}f(x)dx}=NINT(f(x),x,a,b)$ .

The understanding that the right-hand side of the equation is an approximation of the left-hand side.

Calculation:

If $f$ is integrated on $[a,b]$ its average (mean) value on $[a,b]$ is

  $av(f)=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}$

  $\begin{array}{l}av(f)=\frac{1}{\sqrt{3}-0}{\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}(x^2-1)dx}\\ av(f)=\frac{1}{\sqrt{3}}{\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}(x^2-1)dx}\end{array}$

Given $f(x)=x^2-1$

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}(x^2-1)dx}=NINT(x^2-1,x,0,\sqrt{3})\\ {\displaystyle \underset{0}{\overset{\sqrt{3}}{\int }}(x^2-1)dx}\simeq 0.00000016\end{array}$

  $\begin{array}{l}av(f)=\frac{1}{\sqrt{3}}(0.0000016)\\ av(f)=0.00000122\end{array}$

Substitute the average value of $f(x)=x^2-1$ in equation of y.

  $\begin{array}{l}0.00000122=x^2-1\\ x^2=1.000000122\\ x=\mathrm{1.00000061.}\end{array}$