What Is The Average Value Of A Function?
The average value of a function is the mean height of the curve $y = f(x)$ across an interval, found by integrating and dividing by the width. If $f$ is continuous on $[a, b]$, its average value is:
$$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x),dx$$
The integral $\int_a^b f(x),dx$ is the signed area under the curve, and dividing by the width $b - a$ spreads that area evenly to give one representative height. This mirrors how you average a finite list: add the values, divide by how many there are. A definite integral is the continuous version of "add the values," and the width $b - a$ plays the role of "how many."
Rearranging the formula shows the geometry directly:
$$f_{\text{avg}},(b-a) = \int_a^b f(x),dx$$
The left side is the area of a rectangle with width $b - a$ and height $f_{\text{avg}}$. The right side is the area under the curve. So $f_{\text{avg}}$ is exactly the height of the rectangle that has the same base and the same area as the region under $f$. For the wider map of the subject, see the calculus overview.
How Do You Find The Average Value Of A Function?
The method is three steps, and it never changes.
Integrate. Find $\displaystyle\int_a^b f(x),dx$ using an antiderivative and the evaluation form of the Fundamental Theorem of Calculus.
Divide by the width. Multiply the integral by $\dfrac{1}{b-a}$. The width is always the larger limit minus the smaller, so it is positive.
Read the units. The answer carries the units of $f$ itself (a height, a speed, a temperature), never the units of a slope.
The result is a $y$-value, not a rate. That single fact separates this topic from average rate of change, which measures a slope. Average value answers "how tall is the curve on average," while average rate of change answers "how steeply did it climb from end to end." The two use different formulas and give different kinds of number.
What Is The Geometric Meaning Of The Average Value?
Picture the region under a positive curve as water in a container with a wavy top surface. Let the water settle. It flattens to a level line, and the area of water never changed, only its shape. That flat level is the average value $f_{\text{avg}}$.
$$\underbrace{f_{\text{avg}},(b-a)}{\text{rectangle area}} ;=; \underbrace{\int_a^b f(x),dx}{\text{area under curve}}$$
The height of that equal-area rectangle is the whole idea. Where the curve rises above the average line, it hands area down to fill the dips below it. The gold region above the line and the region below it balance exactly. This is why the average value always lands between the minimum and maximum of $f$ on the interval: a flat line above the peak or below the trough could not enclose the same area.
What Is The Mean Value Theorem For Integrals?
The rectangle picture raises a sharp question: does the curve ever actually reach its average height, or does it just balance around it? For a continuous function the answer is yes, and that promise is the Mean Value Theorem for Integrals.
If $f$ is continuous on the closed interval $[a, b]$, then there exists at least one point $c$ in $[a, b]$ such that $$f(c) = \frac{1}{b-a}\int_a^b f(x),dx = f_{\text{avg}}.$$ Equivalently, $\displaystyle\int_a^b f(x),dx = f(c),(b - a)$.
In words, a continuous curve must hit its own average value somewhere on the interval. The theorem does not say where, and it does not promise only one such $c$; it promises at least one. Continuity is the hypothesis that makes it true. A function that jumps over its average value without landing on it would have to be discontinuous.
This is the same theorem that powers Part 1 of the Fundamental Theorem of Calculus, where a thin sliver of area is replaced by a rectangle of height $f(c)$. It belongs to the same family as the derivative mean value theorem: both guarantee a point $c$ where a pointwise value matches an average behaviour across the whole interval.
How Do You Find The Value Of c? Worked Examples
Each example integrates, divides by the width to get $f_{\text{avg}}$, then solves $f(c) = f_{\text{avg}}$ for $c$ in the interval. Every integral is checked by differentiating the antiderivative back to the integrand. Decimals are given to four places.
Example 1: A parabola on a clean interval.
Find the average value of $f(x) = x^2$ on $[0, 3]$, and the $c$ that attains it.
Integrate, using antiderivative $F(x) = \dfrac{x^3}{3}$:
$$\int_0^3 x^2,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9$$
Divide by the width $b - a = 3$:
$$f_{\text{avg}} = \frac{1}{3}\cdot 9 = 3$$
Now solve $f(c) = 3$, that is $c^2 = 3$, on $[0, 3]$:
$$c = \sqrt{3} \approx 1.7321$$
Check: $F'(x) = \dfrac{d}{dx}\left(\dfrac{x^3}{3}\right) = x^2$, the integrand, so the integral is correct. And $c = \sqrt{3} \approx 1.7321$ lies in $[0, 3]$.
Final answer: $f_{\text{avg}} = 3$, attained at $c = \sqrt{3} \approx 1.7321$.
Example 2: A trig function with two values of c.
Find the average value of $f(x) = \sin x$ on $[0, \pi]$, and every $c$ that attains it.
Integrate, using antiderivative $F(x) = -\cos x$:
$$\int_0^{\pi} \sin x,dx = \big[-\cos x\big]_0^{\pi} = (-\cos\pi) - (-\cos 0) = (1) - (-1) = 2$$
Divide by the width $b - a = \pi$:
$$f_{\text{avg}} = \frac{2}{\pi} \approx 0.6366$$
Solve $\sin c = \dfrac{2}{\pi}$ on $[0, \pi]$. The sine curve reaches this height on the way up and again on the way down, so there are two solutions:
$$c = \arcsin\left(\frac{2}{\pi}\right) \approx 0.6901 \qquad \text{and} \qquad c = \pi - 0.6901 \approx 2.4515$$
Check: $F'(x) = \dfrac{d}{dx}(-\cos x) = \sin x$, so the integral is correct. Both values lie in $[0, \pi]$, which shows the theorem's "at least one $c$" can be more than one.
Final answer: $f_{\text{avg}} = \dfrac{2}{\pi} \approx 0.6366$, attained at $c \approx 0.6901$ and $c \approx 2.4515$.
Example 3: Average velocity from a velocity function.
A particle moves with velocity $v(t) = 3t^2$ (metres per second) for $t$ in $[0, 4]$ seconds. Find its average velocity, and the instant $c$ when it travels at that speed.
Average velocity is the average value of $v$:
$$v_{\text{avg}} = \frac{1}{4-0}\int_0^4 3t^2,dt = \frac{1}{4}\big[t^3\big]_0^4 = \frac{1}{4}(64) = 16$$
This matches the physical definition of average velocity as total displacement over elapsed time: the displacement is $\int_0^4 3t^2,dt = 64$ metres, over $4$ seconds, giving $16$ m/s. Now find $c$ from $v(c) = 16$:
$$3c^2 = 16 ;\Rightarrow; c^2 = \frac{16}{3} ;\Rightarrow; c = \frac{4}{\sqrt{3}} \approx 2.3094 \text{ s}$$
Check: $\dfrac{d}{dt}(t^3) = 3t^2 = v(t)$, so the integral is correct, and $c \approx 2.3094$ lies in $[0, 4]$.
Final answer: average velocity $= 16$ m/s, reached at $c = \dfrac{4}{\sqrt{3}} \approx 2.3094$ s.
Example 4: Average temperature over a day.
Suppose the temperature in degrees Celsius over a six-hour window is modelled by $T(t) = 20 + 6t - t^2$ for $t$ in $[0, 6]$ hours. Find the average temperature and the times $c$ when it is attained.
Integrate, using antiderivative $F(t) = 20t + 3t^2 - \dfrac{t^3}{3}$:
$$\int_0^6 \left(20 + 6t - t^2\right)dt = \left[20t + 3t^2 - \frac{t^3}{3}\right]_0^6 = 120 + 108 - 72 = 156$$
Divide by the width $b - a = 6$:
$$T_{\text{avg}} = \frac{156}{6} = 26 ;^\circ\text{C}$$
Solve $T(c) = 26$: $20 + 6c - c^2 = 26$, which rearranges to $c^2 - 6c + 6 = 0$, so
$$c = \frac{6 \pm \sqrt{36 - 24}}{2} = 3 \pm \sqrt{3} ;\Rightarrow; c \approx 1.2679 \text{ h} \quad \text{or} \quad c \approx 4.7321 \text{ h}$$
Check: $F'(t) = 20 + 6t - t^2 = T(t)$, so the integral is correct, and both times lie in $[0, 6]$.
Final answer: average temperature $= 26,^\circ$C, reached at $c \approx 1.2679$ h and $c \approx 4.7321$ h.
What Is The Average Value Of Common Functions?
This table collects a few standard results, each computed with the same three steps and verified by differentiating the antiderivative back. Use it to sanity-check your own work.
Table: Average value and the attained point $c$ for common functions.
Function $f(x)$ | Interval $[a,b]$ | $\displaystyle\int_a^b f,dx$ | $f_{\text{avg}}$ | Attained at $c$ |
|---|---|---|---|---|
$x^2$ | $[0, 3]$ | $9$ | $3$ | $\sqrt{3} \approx 1.7321$ |
$\sin x$ | $[0, \pi]$ | $2$ | $\dfrac{2}{\pi} \approx 0.6366$ | $0.6901,; 2.4515$ |
$6x$ | $[2, 5]$ | $63$ | $21$ | $3.5$ (the midpoint) |
$e^x$ | $[0, 1]$ | $e - 1$ | $e - 1 \approx 1.7183$ | $\ln(e-1) \approx 0.5413$ |
$\dfrac{1}{x}$ | $[1, e]$ | $1$ | $\dfrac{1}{e-1} \approx 0.5820$ | $e - 1 \approx 1.7183$ |
Notice the linear row: for any straight line, the average value is the value at the midpoint of the interval, and $c$ is that midpoint. That neat coincidence holds only for linear functions, which is exactly why it becomes a trap on curved ones.
Why Does The Average Value Formula Work?
The formula is not an arbitrary recipe. It is the continuous limit of the ordinary average you already know.
It starts as a plain mean. Sample the function at $n$ evenly spaced points and average the heights: $\dfrac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$. This is the mean of a finite list.
More samples sharpen it. As $n$ grows, the sum of heights times the spacing $\Delta x$ becomes the area under the curve, a Riemann sum tightening into $\int_a^b f(x),dx$. Dividing by $n$ becomes dividing by the width $b - a$.
The rectangle is the payoff. The limit gives $f_{\text{avg}} = \dfrac{1}{b-a}\int_a^b f(x),dx$, the height that trades every rise for a matching dip so the flat rectangle and the curved region enclose equal area.
Seen this way, the average value of a function is the same idea as averaging test scores, stretched from a handful of numbers to the infinitely many heights of a smooth curve. The integral does the adding; the width does the dividing.
Who Shaped The Idea Of The Average Value Of A Function?
Averaging a changing quantity is an old instinct, but pinning it to a rigorous integral came with the nineteenth-century drive to put calculus on solid ground.
Two figures anchor the timeline:
Augustin-Louis Cauchy (1789–1857, France) gave the definite integral its limit-of-sums definition, the foundation the average-value formula rests on.
Bernhard Riemann (1826–1866, Germany) widened that definition to a broad class of functions, so the average value could be computed even for curves with corners or jumps, provided they stay integrable.
Where Is The Average Value Of A Function Used In The Real World?
Any time a quantity changes continuously and you want one representative number, the average value of a function is the tool.
Motion and travel: average velocity is displacement over elapsed time, which is the average value of the velocity function. See velocity and acceleration for the full kinematics picture.
Weather and climate: the mean temperature over a day, month, or year is the average value of a temperature-versus-time curve, not just the midpoint of the high and low.
Electrical engineering: the root-mean-square voltage of an alternating current is an average value of a squared signal, and it is what sets the effective power delivered to a device.
Economics: average cost per unit over a production run, or average revenue over a season, is the average value of the relevant rate function across the period.
Medicine and biology: the average concentration of a drug in the bloodstream over a dosing interval is an average value that guides how often a dose should be given.
One formula turns a whole varying curve into a single honest number, which is why every field that measures something over time reaches for it.
What Are The Most Common Mistakes With The Average Value Of A Function?
These four errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and course common-error handouts such as the University of Kentucky MA113 sheet.
Forgetting to divide by the width.
Where it slips in:
A student computes $\int_a^b f(x),dx$, writes that number as the answer, and never multiplies by $\dfrac{1}{b-a}$.
Don't do this:
Do not report the integral as the average value. The integral is an area, not a height.
The correct way:
Always finish with the division: $f_{\text{avg}} = \dfrac{1}{b-a}\displaystyle\int_a^b f(x),dx$. For $f(x)=x^2$ on $[0,3]$, the integral is $9$, but the average value is $\dfrac{9}{3} = 3$.
Confusing average value with average rate of change.
Where it slips in:
Seeing the word "average," a student reaches for $\dfrac{f(b) - f(a)}{b - a}$, the slope formula, instead of the integral.
Don't do this:
Do not use the endpoints alone. Average rate of change is a slope; average value is an average height, and they are different numbers.
The correct way:
Average value integrates the whole curve: $\dfrac{1}{b-a}\displaystyle\int_a^b f(x),dx$. Average rate of change uses only the two endpoints. Match the formula to the question before computing.
Using the wrong interval width.
Where it slips in:
A student divides by $a - b$ instead of $b - a$, or by a single limit, flipping the sign of the answer.
Don't do this:
Do not subtract the limits in the wrong order. The width of $[a, b]$ is $b - a$, always the larger minus the smaller.
The correct way:
Write the width first and confirm it is positive. On $[0, 3]$ the width is $3 - 0 = 3$, never $0 - 3 = -3$.
Expecting a single, midpoint value of c.
Where it slips in:
A student assumes the point $c$ from the Mean Value Theorem for Integrals is unique, or that it sits at the midpoint of the interval as it does for a straight line.
Don't do this:
Do not assume one $c$, and do not place it at the midpoint for a curved function. The midpoint rule holds only for linear $f$.
The correct way:
Solve $f(c) = f_{\text{avg}}$ honestly and keep every solution in $[a, b]$. For $\sin x$ on $[0, \pi]$ there are two: $c \approx 0.6901$ and $c \approx 2.4515$.
Practice Problems On The Average Value Of A Function
Find the average value on the given interval, then the point $c$ where it is attained. Answers are verified by differentiating the antiderivative back.
$f(x) = 4x^3$ on $[0, 2]$.
(Answer: $\int_0^2 4x^3,dx = [x^4]0^2 = 16$, so $f{\text{avg}} = \tfrac{16}{2} = 8$; $4c^3 = 8 \Rightarrow c = \sqrt[3]{2} \approx 1.2599$.)$f(x) = 6x$ on $[2, 5]$.
(Answer: $\int_2^5 6x,dx = [3x^2]2^5 = 75 - 12 = 63$, so $f{\text{avg}} = \tfrac{63}{3} = 21$; $6c = 21 \Rightarrow c = 3.5$, the midpoint.)$f(x) = \cos x$ on $[0, \tfrac{\pi}{2}]$.
(Answer: $\int_0^{\pi/2}\cos x,dx = [\sin x]0^{\pi/2} = 1$, so $f{\text{avg}} = \tfrac{2}{\pi} \approx 0.6366$; $\cos c = \tfrac{2}{\pi} \Rightarrow c \approx 0.8807$.)$f(x) = e^x$ on $[0, 1]$.
(Answer: $\int_0^1 e^x,dx = e - 1 \approx 1.7183$, so $f_{\text{avg}} = e - 1 \approx 1.7183$; $e^c = e - 1 \Rightarrow c = \ln(e-1) \approx 0.5413$.)$f(x) = \dfrac{1}{x}$ on $[1, e]$.
(Answer: $\int_1^e \tfrac{1}{x},dx = [\ln x]1^e = 1$, so $f{\text{avg}} = \tfrac{1}{e-1} \approx 0.5820$; $\tfrac{1}{c} = \tfrac{1}{e-1} \Rightarrow c = e - 1 \approx 1.7183$.)$f(x) = x^2$ on $[1, 4]$.
(Answer: $\int_1^4 x^2,dx = \left[\tfrac{x^3}{3}\right]1^4 = \tfrac{64}{3} - \tfrac{1}{3} = 21$, so $f{\text{avg}} = \tfrac{21}{3} = 7$; $c^2 = 7 \Rightarrow c = \sqrt{7} \approx 2.6458$.)
Where Should You Go Next After The Average Value Of A Function?
The average value sits at the crossroads of integration and the mean value idea, and several natural doors open from here.
Mean Value Theorem. The derivative twin of the theorem you just met, guaranteeing a point where the slope equals the average slope.
Applications of integration. Area, volume, and average value are all the same integral read for different purposes.
Properties of definite integrals. The additivity and bound rules that make average-value computations quicker and safer.
If your child is meeting the average value of a function for the first time, a live Bhanzu trainer teaches it from the rectangle-of-equal-area picture up, so the formula and the geometry arrive together, in the Bhanzu math program.
Was this article helpful?
Your feedback helps us write better content