Chapter 6.3, Problem 45E

To determine

The average value of the function f on [a,b] that lies between f(a) and f(b).

Answer to Problem 45E

It is not necessary that average value of a function f on [a, b] always lies between f(a) and f(b).

Explanation of Solution

Given information:

The average value of the function f on [a,b] that lies between f(a) and f(b).

Formula used:

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

Calculation:

Statement is false.

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

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

If max f and min f are the maximum and minimum values of f on [a, b], then

  $\min f\left(a-b\right)\le {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx\le \max f\left(b-a\right)$

Now, divide the inequality by (b - a)

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

So,

  $\min f\le av\left(f\right)\le \max f$

If function is monotonously increasing on [a, b]

  $\begin{array}{l}\min f=f\left(a\right)\\ \min f=f\left(b\right)\end{array}$

So,

  $f\left(a\right)\le av\left(f\right)\le f\left(b\right)$

If function is monotonously decreasing on [a, b]

  $\begin{array}{l}\min f=f\left(b\right)\\ \max f=f\left(a\right)\end{array}$

So,

  $f\left(b\right)\le av\left(f\right)\le f\left(a\right)$

But, if function is not monotonously increasing or decreasing on [a, b], then it is not necessary that average value of a function f on [a, b] always lies between f(a) and f(b).

Conclusion:

If the function is not monotonously increasing or decreasing on [a, b], then it is not necessary that average value of a function f on [a, b] always lies between f(a) and f(b).