Average Value of a Function

The average value of a continuous function over an interval is exactly what you'd expect: the integral divided by the length of the interval. Geometrically, it's the height of a rectangle with the same area and the same base as the region under the curve.

The Formula

$\bar f = \dfrac{1}{b - a}\int_{a}^{b} f(x)\,dx$

This is sometimes called the mean value of $f$ on $[a, b]$. The Mean Value Theorem for integrals guarantees that there exists at least one point $c \in (a, b)$ where $f(c) = \bar f$.

Step-by-Step Solution

Given: $f(x) = \sin x$, $a = 0$, $b = \pi$.

Find: The average value $\bar f$ on $[0, \pi]$.

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Step 1 — Compute the interval length.

$b - a = \pi - 0 = \pi$

Step 2 — Set up the formula.

$\bar f = \dfrac{1}{\pi}\int_{0}^{\pi} \sin x\,dx$

Step 3 — Find the antiderivative of $\sin x$.

$\int \sin x\,dx = -\cos x + C$

Step 4 — Apply the Fundamental Theorem.

$\int_{0}^{\pi} \sin x\,dx = \left[-\cos x\right]_{0}^{\pi}$

$= -\cos\pi - (-\cos 0)$

$= -(-1) - (-(1)) = 1 + 1 = 2$

Step 5 — Divide by the interval length.

$\bar f = \dfrac{1}{\pi}(2) = \dfrac{2}{\pi}$

Step 6 — Decimal value.

$\bar f = \dfrac{2}{\pi} \approx \dfrac{2}{3.14159} \approx 0.6366$

Step 7 — Find the points where $f$ achieves this average.

By the Mean Value Theorem, there's a $c$ where $\sin c = 2/\pi \approx 0.6366$. Solving:

$c = \arcsin(0.6366) \approx 0.6901 \;\text{or}\; c = \pi - 0.6901 \approx 2.4515$

Both are in $(0, \pi)$. The function reaches the average twice — once on the way up and once on the way down. (At the peak $c = \pi/2$, $\sin c = 1$, which is well above the average.)

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Answer: The average value of $\sin x$ on $[0, \pi]$ is

$\boxed{\;\bar f = \dfrac{2}{\pi} \approx 0.637\;}$

This is the height of the rectangle (with base $\pi$) that has the same area as the region under the sine curve on $[0, \pi]$. Note that the average is less than 1 even though the curve reaches a maximum of 1 — because much of the interval has smaller values.

Try It

• The horizontal green line marks the average value $2/\pi$.
• The two yellow dots show where the curve crosses the average — those are the $c$ values from the Mean Value Theorem.
• Toggle show rectangle to see the area under the sine curve (cyan) compared to the equivalent rectangle (green outline) — they have the same area.