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Calculus AB · Unit 8 · 8.1
AP Calculus · Applications of Integration
Finding the Average Value of a Function
Sum becomes an integral, count becomes the interval's length. That is the whole formula.
key idea$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$
The rule
The average of finitely many numbers is their sum divided by how many there are. A function has infinitely many values, so the sum becomes an integral and the count becomes the length of the interval. That single substitution is the whole formula.
$f(x)=x^{2}$ on $[0,3]$ has average value $3$. The dashed line at height $3$ cuts the region so that the shortfall on the left exactly cancels the excess on the right — the two shaded pieces have equal area. The crossing happens at $x=\sqrt{3}$, marked in rose.| Setting | “Sum” | “Count” | Average |
|---|
| $n$ data values | $\sum y_i$ | $n$ | $\dfrac{1}{n}\sum y_i$ |
| a function on $[a,b]$ | $\displaystyle\int_a^b f$ | $b-a$ | $\dfrac{1}{b-a}\displaystyle\int_a^b f$ |
| velocity over $[a,b]$ | total displacement | elapsed time | average velocity |
| a rate $r(t)$ | total accumulated | duration | average rate |
Row 3 is worth pausing on: the average value of the velocity function is displacement over elapsed time — which is precisely the everyday definition of average velocity. The formula is not a new idea, only a continuous version of an old one.
four facts that prevent most errors
- Divide by $b-a$, the length of the interval — never by the number of terms in anything.
- The average value is a height, in the units of $f$, not in the units of the integral.
- $f_{\text{avg}}\cdot(b-a)=\displaystyle\int_a^b f$: the mean height times the width recovers the area exactly.
- For a continuous $f$, some $c$ in $(a,b)$ actually attains the average value.
Pattern recognition
The integral alonean area or total
Divided by $b-a$an average height
Set $f(c)=f_{\text{avg}}$the point where it is attained
Worked examples
EX 1
Straight from the formula
$f(x)=x^{2}$ on $[0,3]$Find the average value of $f(x)=x^{2}$ on $[0,3]$.
1
Write the formula
$f_{\text{avg}}=\dfrac{1}{3-0}\displaystyle\int_{0}^{3}x^{2}\,dx$.
2
Integrate
$\displaystyle\int_{0}^{3}x^{2}dx=\left[\dfrac{x^{3}}{3}\right]_{0}^{3}=9$.
3
Divide by the width
$f_{\text{avg}}=\dfrac{9}{3}=3$.
4
Sanity-check the size
$f$ runs from $0$ to $9$, so an average of $3$ is plausible — and low, as it should be for a curve that stays flat near the left end.
5
Interpret it
A rectangle of height $3$ over $[0,3]$ has area $9$, the same as the region under the parabola.
EX 2
Find where it is attained
solve $f(c)=f_{\text{avg}}$For $f(x)=x^{2}$ on $[0,3]$, find the $c$ guaranteed by the Mean Value Theorem for Integrals.
1
Recall the average
$f_{\text{avg}}=3$.
2
Set up the equation
$f(c)=3$ means $c^{2}=3$.
4
Discard the outsider
$-\sqrt3$ is not in $[0,3]$, so $c=\sqrt3\approx1.732$.
5
Read the guarantee
Continuity on a closed interval guarantees at least one such $c$ exists. Finding it is a separate step from computing the average.
EX 3
Average value of a rate
units matterWater flows into a tank at $r(t)=4+2t$ litres per minute for $0\le t\le6$. Find the average flow rate.
1
Identify the interval length
$6-0=6$ minutes.
2
Integrate the rate
$\displaystyle\int_{0}^{6}(4+2t)\,dt=\left[4t+t^{2}\right]_{0}^{6}=24+36=60$ litres.
3
Divide
$r_{\text{avg}}=\dfrac{60}{6}=10$ litres per minute.
4
Check the units
Litres divided by minutes gives litres per minute — the units of $r$, not of the integral. ✓
5
Note the two different answers
$60$ litres is the total; $10$ L/min is the average rate. Reporting the integral when the average was asked is the standard slip.
EX 4
A signed average
cancellationFind the average value of $f(x)=\sin x$ on $[0,2\pi]$, and explain the result.
1
Set up
$f_{\text{avg}}=\dfrac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\sin x\,dx$.
2
Integrate
$\left[-\cos x\right]_{0}^{2\pi}=-1-(-1)=0$.
3
Divide
$f_{\text{avg}}=0$.
4
Explain
The hump above the axis and the hump below it have equal area, and the integral counts area as signed, so they cancel exactly.
5
Contrast with a common misreading
The average value is $0$, even though $\lvert\sin x\rvert$ averages $\dfrac{2}{\pi}$. Signed and unsigned averages answer different questions.
Watch out
Divide by $b-a$, not by anything else
The denominator is the length of the interval. Dividing by the number of subintervals, or forgetting to divide at all, leaves you with a total instead of an average.
The answer is a height, in the units of $f$
If $f$ is a rate in litres per minute, the average value is in litres per minute — the integral's litres have been divided back out.
Signed area can cancel to zero
A function that spends equal area above and below the axis has average value $0$. That is correct, not a mistake.
Average value is not the midpoint of the range
For $x^{2}$ on $[0,3]$ the range midpoint is $4.5$, but the average value is $3$. The formula weights by how much time the function spends at each height.