Math · Calculus I · Worked example

Find the average value of a function

Find the average value of f(x) = x² on [0, 3], and find where f takes that value.

f (x) = x^{2}

Use the average value formula

The average value is the integral divided by the length of the interval: the height of the rectangle on [0, 3] with the same area as the region under the curve.

f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x

Integrate

An antiderivative of x² is x³/3.

\begin{array}{l} f (x) = x^{2} \\ \int_{0}^{3} x^{2} d x = \frac{3^{3}}{3} - \frac{0^{3}}{3} = 9 \end{array}

Divide by the width

The interval has length 3 − 0 = 3.

\frac{9}{3 - 0} = 3

Find where f equals its average

x² = 3 gives x = √3 ≈ 1.73, which is in [0, 3]. The Mean Value Theorem for integrals guarantees such a point for every function that is continuous on the interval.

\sqrt{3} \approx 1.73

Result

The average value is 3, and f(√3) = 3, where √3 ≈ 1.73.

Your turn

Find the average value of f(x) = 2x + 1 on [1, 4].

Show the answer and explanation

6.

∫₁⁴ (2x + 1) dx = (16 + 4) − (1 + 1) = 18, and 18/3 = 6. For a linear function the average is the value at the midpoint: f(2.5) = 6.

\begin{array}{l} f (x) = 2 x + 1 \\ \int_{1}^{4} (2 x + 1) d x = (4^{2} + 4) - (1^{2} + 1) = 18 \\ \frac{18}{4 - 1} = 6 \end{array}

\begin{array}{c} f (x) = 2 x + 1 \\ \int_{1}^{4} (2 x + 1) d x = 18 \\ \frac{18}{4 - 1} = 6 \end{array}

Keep exploring

In Graph, the line y = 3 crosses y = x² at x = √3 ≈ 1.73. The 3-by-3 rectangle under the line has the same area, 9, as the region under the parabola.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

See the average height in Graph Check the average in Math Open worked example on a board Fundamental Theorem in Math Reference

Your existing work stays on this device. Examples open as editable copies.