Chapter 6, Problem 6RCC

(a)

To determine

To Find: The average value of a function f on an interval [a, b]

Answer to Problem 6RCC

The average value of a function f on an interval [a, b] is $\underset{\_}{f_{\text{ave}}=\frac{1}{b-a}{\displaystyle {\int }_a^{b}f\left(x\right)dx}}$.

Explanation of Solution

Consider the function $y=f\left(x\right)$ with large number of the closely spaced value with an interval a to b.

The definite interval of average value is the area under the curve divided by the width of the interval.

Express the average value of the function $\left(f_{\text{avg}}\right)$ as follows:

$f_{avg}=\frac{1}{b-a}{\displaystyle {\int }_a^{b}f\left(x\right)dx}$

Here, $\left(b-a\right)$ is width of the closely spaced interval and ${\displaystyle {\int }_a^{b}f\left(x\right)dx}$ is area under the curve.

Thus, the average value of a function f on an interval [a, b] is $\underset{\_}{f_{\text{ave}}=\frac{1}{b-a}{\displaystyle {\int }_a^{b}f\left(x\right)dx}}$.

(b)

To determine

To define: The Mean Value Theorem for Integrals and its geometric interpretation.

Explanation of Solution

Mean value theorem states that, “if a function f continuous on closed interval [a, b] in which a number c exist in the interval, then the value of f is equal to the average value of the given function”.

The geometric interpretation refers that there is a number c in the positive function f such that the area of rectangle with base [a, b] and height $f\left(c\right)$ has the same area as the region under the graph of f from a to b.

Thus, the Mean Value Theorem and its geometric interpretation is explained.