Average value of a function
The average of a list of numbers is easy: add them, and divide by how many there are. But how do we find the average temperature during a day, when the temperature changes every moment? There are infinitely many values. An integral solves this.
The formula
Let $f$ be continuous on $[a,b]$. Pick $n$ evenly spaced numbers $x_1,\ldots,x_n$ in the interval, and average the values of $f$ there:
$\dfrac{f(x_1)+f(x_2)+\cdots+f(x_n)}{n}$
The spacing is $\Delta x=\dfrac{b-a}{n}$, so $\dfrac{1}{n}=\dfrac{\Delta x}{b-a}$. The average becomes
$\displaystyle\dfrac{1}{b-a}\sum_{i=1}^{n}f(x_i)\,\Delta x$.
This is a Riemann sum divided by $b-a$. As $n$ gets larger and larger, it becomes an integral. So the average value of $f$ on $[a,b]$ is
$\displaystyle f_{\text{ave}}=\dfrac{1}{b-a}\int_a^b f(x)\,dx$.
In words: the integral divided by the length of the interval.
Example 1: Find the average value of
$f(x)=1+x^2$
on the interval $[-1,2]$.
Solution:
The interval has length $3$.
$\begin{align*}&f_{\text{ave}}\\&=\dfrac{1}{3}\int_{-1}^{2}(1+x^2)\,dx\\&=\dfrac{1}{3}\left[x+\dfrac{x^3}{3}\right]_{-1}^{2}\\&=\dfrac{1}{3}\left[\left(2+\dfrac{8}{3}\right)-\left(-1-\dfrac{1}{3}\right)\right]\\&=\dfrac{1}{3}\cdot 6\\&=2\end{align*}$
The Mean Value Theorem for Integrals
Multiply the formula by $b-a$:
$\displaystyle\int_a^b f(x)\,dx=f_{\text{ave}}\cdot(b-a)$
So a rectangle with height $f_{\text{ave}}$ and width $b-a$ has the same area as the region under the curve. The parts of the curve above the rectangle balance the parts below it.
Does $f$ actually reach its average value somewhere? Yes, if it is continuous.
The Mean Value Theorem for Integrals: If $f$ is continuous on $[a,b]$, then there is a number $c$ in $[a,b]$ where
$f(c)=f_{\text{ave}}$.
Why: $f$ has a smallest value $m$ and a largest value $M$ on $[a,b]$ (by the Extreme Value Theorem). The average is between them. By the Intermediate Value Theorem (see continuity), $f$ takes every value between $m$ and $M$, including the average.
Example 2: For the function in Example 1, find every number $c$ where
$f(c)=f_{\text{ave}}$.
Solution:
From Example 1, $f_{\text{ave}}=2$. Solve:
$\begin{align*}1+c^2&=2\\c^2&=1\\c&=\pm 1\end{align*}$
Both $-1$ and $1$ are in $[-1,2]$. So there are two such numbers, as the figure shows.
Applications
Example 3: On a spring day, the temperature $t$ hours after 9 a.m. is
$T(t)=50+14\sin\dfrac{\pi t}{12}$
degrees Fahrenheit. Find the average temperature from 9 a.m. to 9 p.m.
Solution:
From 9 a.m. to 9 p.m. is $t=0$ to $t=12$. By the substitution rule, an antiderivative of $\sin\dfrac{\pi t}{12}$ is $-\dfrac{12}{\pi}\cos\dfrac{\pi t}{12}$. So
$\begin{align*}&T_{\text{ave}}\\&=\dfrac{1}{12}\int_0^{12}\left(50+14\sin\dfrac{\pi t}{12}\right)dt\\&=\dfrac{1}{12}\left[50t-\dfrac{168}{\pi}\cos\dfrac{\pi t}{12}\right]_0^{12}\\&=\dfrac{1}{12}\left[600-\dfrac{168}{\pi}(\cos\pi-\cos 0)\right]\\&=\dfrac{1}{12}\left(600+\dfrac{336}{\pi}\right)\\&=50+\dfrac{28}{\pi}\approx 58.9\end{align*}$
The average temperature is about $58.9^\circ$F. (Angles are in radians here.)
Example 4: A ball's velocity is
$v(t)=19.6-9.8t$ m/s.
Find its average velocity for the first $2$ seconds.
Solution:
$\begin{align*}&v_{\text{ave}}\\&=\dfrac{1}{2}\int_0^2(19.6-9.8t)\,dt\\&=\dfrac{1}{2}\Big[19.6t-4.9t^2\Big]_0^2\\&=\dfrac{1}{2}(39.2-19.6)\\&=9.8\ \text{m/s}\end{align*}$
This is the same as the change in position divided by the time. The ball from antiderivatives (Example 5) rises from $2$ m to $21.6$ m in these $2$ seconds. So
$\dfrac{21.6-2}{2}=9.8$ m/s.
By the net change theorem, the average value of a rate is always the average rate of change.
Example 5: Find the average value of $\sin x$ on $[0,\pi]$.
Solution:
In the Fundamental Theorem of Calculus (Example 6), we found $\displaystyle\int_0^\pi\sin x\,dx=2$. So
$\displaystyle(\sin x)_{\text{ave}}=\dfrac{1}{\pi}\cdot 2=\dfrac{2}{\pi}\approx 0.637$.
The values of $\sin x$ go from $0$ up to $1$ and back. Their average is a little less than $\dfrac{2}{3}$.
Summary
- The average value of $f$ on $[a,b]$ is $\displaystyle\dfrac{1}{b-a}\int_a^b f(x)\,dx$.
- A rectangle with height $f_{\text{ave}}$ has the same area as the region under the curve.
- If $f$ is continuous, it reaches its average value at some $c$ in $[a,b]$.
This is the last section of Calculus I. You have seen the two big ideas, the derivative and the integral, and how the Fundamental Theorem connects them.