Chapter 6.3, Problem 39E

To determine

The number $av\left(f\right)$ having the same integral over [a,b].

Answer to Problem 39E

  ${\displaystyle \underset{a}{\overset{b}{\int }}av\left(f\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}\left(f\right)}dx$

Explanation of Solution

Given information:

  ${\displaystyle \underset{a}{\overset{b}{\int }}av\left(f\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$

Formula used:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}cdx}$

Calculation:

Yes, number $av\left(f\right)$ has the same integral over [a, b] that f does

  $av\left(f\right)$ , it is constant quantity

If

  $f\left(x\right)=c$

Where c is a constant, on the interval [a, b] then

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}cdx}\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=c\left(b-a\right)\end{array}$

So,

  ${\displaystyle \underset{a}{\overset{b}{\int }}av\left(f\right)}dx=av\left(f\right)\left[b-a\right]$

Now,

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

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

So,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}av\left(f\right)}dx=\left(\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}\right)\left(b-a\right)\\ {\displaystyle \underset{a}{\overset{b}{\int }}av\left(f\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}\left(f\right)}dx\end{array}$

Conclusion:

  ${\displaystyle \underset{a}{\overset{b}{\int }}av\left(f\right)}dx={\displaystyle \underset{a}{\overset{b}{\int }}\left(f\right)}dx$