Average Value of a Function

Computing the Average Value of a Continuous Function

Definition

If \(f\) is integrable on \([a, b]\), then the average value of \(f\) over \([a, b]\) is given by

\[\frac{1}{b-a}\int_a^b f(x)dx.\]

Example 1

Computing the average value

Find the average value of \(f(x) = 3x^2 + 4x^3\) over the interval \([-1, 1]\).

Step 1: The average value is given by

\[\begin{align*} \frac{1}{1-(-1)} \int_{-1}^1 3x^2 + 4x^3 ~dx &= \frac{1}{2} \int_{-1}^1 3x^2 + 4x^3 ~dx \\ &= \frac{1}{2}\left(x^3+x^4\right)\Big|_{-1}^1 \\ &= \frac{1}{2}\left[(1^3 + 1^4) - ((-1)^3+(-1)^4)\right]\\ &= \frac{1}{2}(2-0) \\ &= 1 \end{align*}\]

Example 2

Computing the average value

Find the average value of \(f(x) = \dfrac{12}{x}\) over the interval \([-9, -3]\).

Step 1: The average value is given by

\[\begin{align*} \frac{1}{-3-(-9)} \int_{-9}^{-3} \frac{12}{x} ~dx &= \frac{12}{6} \int_{-9}^{-3} \frac{1}{x} ~dx \\ &= 2\ln|x|\Biggr|_{-9}^{-3} \\ &= 2[\ln|-3| - \ln|-9|]\\ \\ &= 2[\ln(3) - \ln(9)]\\ \\ &= 2\ln(3/9) && \hbox{Since $\ln(m/n) = \ln(m) - \ln(n)$}\\ \\ &= 2\ln(3^{-1}) \\ \\ &= -\ln(9) && \hbox{Since $m\ln(n) = \ln(n^m)$} \end{align*}\]

Example 3

Computing average sales

Sales of the Penn State Learning Calculus tutorial software packages are approximated by

\[f(t) = \frac{t^2}{(t^3+5)^2},\]

where \(t\) is in years and \(f(t)\) is in millions of software packages. What are the average sales over the time interval \(0 \leq t \leq 3\) years?

Step 1: Use the definition of average value.

The average value is given by

\[\frac{1}{3-0}\int_0^3 \frac{t^2}{(t^3+5)^2}~dt = \frac{1}{3}\int_0^3 \frac{t^2}{(t^3+5)^2}~dt.\]

Step 2: Identify a suitable substitution.

Based on rewriting the integral in the following form

\[\frac{1}{3}\int_{t=0}^{t=3} (t^3+5)^{-2}~t^2 ~dt\]

let \(u = t^3 + 5\) and \(du = 3t^2 ~dt\), or equivalently \(\dfrac{1}{3} du = t^2 ~dt\).

Step 3: Determine the new limits of integration using the substitution \(u = t^3 + 5\).

\[\begin{align*} \text{When } t=3 ~~&\Longrightarrow~~ u = 3^3 + 5 = 32\\ \text{When } t=0 ~~&\Longrightarrow~~ u = 0^3 + 5 = 5 \end{align*}\]

Step 4: Rewrite the integral in terms of \(u\) and \(du\).

\[\begin{align*} \frac{1}{3}\int_{t=0}^{t=3} (t^3+5)^{-2}~t^2 ~dt &= \frac{1}{3} \int_{u=5}^{u=32} u^{-2} ~ \frac{1}{3} du\\ &= \frac{1}{9} \int_{u=5}^{u=32} u^{-2} ~du \end{align*}\]

Step 5: Evaluate the integral.

\[\begin{align*} \frac{1}{9} \int_{u=5}^{u=32} u^{-2} ~du &= \frac{1}{9}(-u^{-1}) \Big|_{5}^{32} && \hbox{Since $\dfrac{u^{-1}}{-1}$ is an antiderivative of $u^{-2}$}\\ &= \frac{1}{9}\left[\left( -\frac{1}{32}\right)- \left(-\frac{1}{5}\right)\right] && \hbox{Plug in limits of integration}\\ &= \frac{1}{9}\left(\frac{1}{5} - \frac{1}{32}\right) && \hbox{Simplify}\\ &= 3/160 \end{align*}\]

Therefore, the average sales over the time interval \(0\leq t \leq 3\) is \(3/160\) millions of software packages per year.