Average Value of a Function Calculator

Q: What is the average value of a function?

f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx. It is the height of the rectangle on [a, b] that has exactly the same area as the region under the curve.

Q: Is there always a point where f equals its average?

Yes, if f is continuous: that is the Mean Value Theorem for Integrals. The calculator marks every such point c.

Find the average height of a curve over an interval. The graph draws the rectangle with the same area, so you can see the average instead of just computing it.

Function f(x)

Reads as

\sin\left(x\right)

Try:

Average value

f_{avg} = 0.63662

Violet = area under f. Amber rectangle = same area, height = average.

Step by step

Formula
$f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx$
Integrate
Antiderivative: $F(x) = -\cos\left(x\right)$
$\int_{0}^{\pi} \sin\left(x\right)\,dx = 2$
Divide by the interval length
$f_{avg} = \frac{2}{\pi} = 0.63662$
Where f equals its average
Mean Value Theorem for Integrals.
$c = 0.690107,\ c = 2.45149$

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The rectangle that tells you the average

Flatten the region under the curve into a rectangle over the same interval, like levelling sand in a box. Its height is the average value. Where the curve pokes above the rectangle, it fills the gaps where the curve dips below. Formally $f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx$ , and the integral comes from the integral calculator.

Questions students ask

What is the average value of a function?

$f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx$ . It is the height of the rectangle on $[a, b]$ that has exactly the same area as the region under the curve.

Why divide by b − a?

For a list of numbers you add them and divide by how many. For a function you "add" with an integral and divide by the length of the interval, the continuous version of "how many".

Is there always a point where f equals its average?

Yes, if $f$ is continuous: that is the Mean Value Theorem for Integrals. The calculator marks every such point $c$ .

Example: average of sin x on [0, π]?

$\frac{1}{\pi}\int_0^\pi \sin x\,dx = \frac{2}{\pi} \approx 0.6366$ . Try it in the calculator.