Chapter 6.5, Problem 22E

To determine

To show: The expression $f_{\text{ave}}\left[a,b\right]=\left(\frac{c-a}{b-a}\right)f_{\text{ave}}\left[a,c\right]+\left(\frac{b-c}{b-c}\right)f_{\text{ave}}\left[c,b\right]$.

Explanation of Solution

Formula:

The expression for the average value of f on the interval $f_{\text{ave}}\left[a,b\right]$ is,

$f_{\text{ave}}\left[a,b\right]=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ (1)

Given information:

The average value of the function, $a<c<b$ is the average value of f on interval $\left[a,b\right]$ and $f_{\text{ave}}\left[a,b\right]=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$.

Calculation:

Since $a<c<b$, modify integral limit a to b by the addition of integral limits a to c and c to b.

Modify equation (1).

$f_{\text{ave}}\left[a,b\right]=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx+\frac{1}{b-a}{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx$

Multiply and divide by terms $\left(c-a\right)$ and $\left(b-c\right)$.

$f_{\text{ave}}\left[a,b\right]=\frac{c-a}{b-a}\left[\frac{1}{c-a}{\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx\right]+\frac{b-c}{b-a}\left[\frac{1}{b-c}{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx\right]$ (2)

Express the function $f_{\text{ave}}\left[a,c\right]$ by the use of equation (1).

$f_{\text{ave}}\left[a,c\right]=\frac{1}{c-a}{\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)}dx$ (3)

Here, $f_{\text{ave}}\left[a,c\right]$ is average value of f on the interval $\left[a,c\right]$.

Express the function $f_{\text{ave}}\left[c,b\right]$ by using Equation (1).

$f_{\text{ave}}\left[c,b\right]=\frac{1}{b-c}{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx$ (4)

Here, $f_{\text{ave}}\left[c,b\right]$ is average value of f on the interval $\left[c,b\right]$.

Substitute equations (3) and (4) in (2).

$f_{\text{ave}}\left[a,b\right]=\frac{c-a}{b-a}{f}_{\text{ave}}\left[a,c\right]+\frac{b-c}{b-a}{f}_{\text{ave}}\left[a,c\right]$

Thus, the expression $f_{\text{ave}}\left[a,b\right]=\frac{c-a}{b-a}{f}_{\text{ave}}\left[a,c\right]+\frac{b-c}{b-a}{f}_{\text{ave}}\left[a,c\right]$ is derived.