Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 7

Applications of the Definite Integral

Average Value of a Function

The “average value of a function” is the last application of the definite integral that we cover in this course.

To extend this concept to infinitely many values of ${\displaystyle f\left(x\right)}$ on a closed interval ${\displaystyle \left[a,b\right]\text{,}}$ we use the definite integral. In this case, the number of terms corresponds to the length of the interval ${\displaystyle \left[a,b\right]\text{,}}$ that is ${\displaystyle b-a\text{,}}$ and the sum of the values corresponds to the definite integral.

Hence,

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

Example 7.15. Find the average value of the function ${\displaystyle f\left(x\right)=\sqrt{x}}$ on the interval ${\displaystyle \left[1,5\right]\text{,}}$ and find the values of ${\displaystyle x}$ for which the value of ${\displaystyle f}$ is equal to this average.

${\displaystyle \begin{array}{r}{f}_{{}_{\text{avg}}}=\frac{1}{5-<!--\; -\; -->1}{\int <!--\; \int \; -->}_1^5\sqrt{x}dx=\frac{2}{12}{x}^{3/2}{|}_1^5=\frac{1}{6}(5^{3/2}-<!--\; -\; -->1)\approx <!--\; \approx \; -->1.7\text{.}\end{array}}$

The values of ${\displaystyle x}$ for which

${\displaystyle f\left(x\right)=\sqrt{x}=\frac{\left(5^{3∕2}-1\right)}{6}}$

is the solution of

${\displaystyle x=\frac{{\left(5\sqrt{5}-1\right)}^2}{36}}$

and this is

${\displaystyle x=\frac{7}{3}-\frac{5\sqrt{5}}{18}\approx 1.71.}$

For a continuous functions ${\displaystyle f}$ on an interval ${\displaystyle \left[a,b\right]}$ it is always possible to find a number ${\displaystyle c}$ such that ${\displaystyle f\left(c\right)=f_{{}_{\text{avg}}}\text{.}}$ This statement is known as the Mean Value Theorem for Integrals.

Theorem 7.4. Mean Value Theorem for Integrals

If ${\displaystyle f}$ is continuous on ${\displaystyle \left[a,b\right]\text{,}}$ then there exists a number ${\displaystyle c}$ in ${\displaystyle \left[a,b\right]\text{,}}$ such that

${\displaystyle {\int }_a^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right).}$

Proof. Let

${\displaystyle g\left(x\right)={\int }_a^{x}f\left(t\right)dt\text{ for }a\le x\le b.}$

By Part 1 of the Fundamental Theorem of Calculus, the function ${\displaystyle g}$ is continuous on ${\displaystyle \left[a,b\right]}$ and differentiable on ${\displaystyle \left(a,b\right)\text{.}}$ Hence, by the Mean Value Theorem, there is a number ${\displaystyle c\text{,}}$ ${\displaystyle a\le c\le b\text{,}}$ such that

${\displaystyle g\left(b\right)-g\left(a\right)=g^{\prime }\left(c\right)\left(b-a\right).}$

Since ${\displaystyle g^{\prime }\left(c\right)=f\left(c\right)\text{,}}$ we conclude that

${\displaystyle {\int }_a^{b}f\left(t\right)dt-{\int }_a^{a}f\left(t\right)dt=f\left(c\right)\left(b-a\right).}$

Therefore, for ${\displaystyle c}$ in ${\displaystyle \left[a,b\right]}$

${\displaystyle {\int }_a^{b}f\left(x\right)dx=f\left(c\right)\left(b-a\right).}$

Q.E.D

The geometric interpretation of this theorem is given on page 332 of the textbook.