Average Value & the Mean Value Theorem for Integrals
The average value of \(f\) on \([a, b]\) is \(f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx\) — total accumulation divided by interval length — and the MVT for Integrals guarantees a point \(c\) where \(f(c)\) actually equals that average.
By the end you'll be able to compute a function's average value on an interval and find a point \(c\) where the function attains it.
Predict: on \([0, 3]\), \(f(x) = x^2\) runs from 0 up to 9. Is its average value closer to the low end or the high end of that range? Drag b and watch the level line to check.
Think of it as leveling off the curve: pour the area under the curve into a flat-topped tank over the same base. The level line at height \(f_{avg}\) tops a rectangle trapping exactly the same area. And because the curve is continuous, it must cross that line somewhere — the point \(c\) where \(f(c) = f_{avg}\). Hover the curve or the \(c\) marker for exact values.
f_avg = b²/3 = 3.00 · area under curve = 9.00 = rectangle area (3.00 × 3.00) · c = b/√3 ≈ 1.73
Dividing accumulated area by interval length turns an integral into an average height, and continuity guarantees the function actually reaches that height somewhere on the interval.
For continuous \(f\) on \([a, b]\), define \(f_{avg} = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx\). The motivation mirrors averaging finitely many numbers: sample \(f\) at \(n\) evenly spaced points, average them, and let \(n \to \infty\) — the Riemann sum in the numerator becomes the definite integral, and dividing by \(b-a\) rescales the accumulated area into an average height. Geometrically, \(f_{avg}\) is the height of the rectangle over \([a, b]\) whose area equals the area under \(f\) — exactly the level line in the diagram above. The Mean Value Theorem for Integrals then says: there exists at least one \(c\) in \([a, b]\) with \(f(c) = f_{avg}\). A continuous function actually attains its average value — that's the crossing point \(c\) you can hover in the diagram. Continuity is essential: a function with a jump can skip over its average height entirely.
Find the average value of \(f(x) = x^2\) on \([0, 3]\), and a point \(c\) where it is attained. First the average: \(f_{avg} = \dfrac{1}{3-0}\displaystyle\int_0^3 x^2\,dx = \dfrac{1}{3}\cdot\dfrac{x^3}{3}\Big|_0^3 = \dfrac{1}{3}(9) = \) 3. Then solve \(f(c) = 3\): \(c^2 = 3\), so \(c = \sqrt{3} \approx 1.73\), which lies in \([0, 3]\). ✓ (We take the positive root because \(c\) must live inside the interval.)
Same function, shorter interval: find the average value of \(f(x) = x^2\) on \([0, 2]\) and the point \(c\) where it's attained. Start: \(f_{avg} = \dfrac{1}{2}\displaystyle\int_0^2 x^2\,dx = \dfrac{1}{2}\cdot\dfrac{8}{3} = \dfrac{4}{3}\). Now finish it: \(c^2 = \dfrac{4}{3}\), so \(c = \) ____
Reveal the answer
\(c = \sqrt{4/3} = \dfrac{2}{\sqrt{3}} \approx 1.15\), which lies in \([0, 2]\). ✓ Check it live: set b = 2 on the slider above — the level line drops to \(f_{avg} = 4/3 \approx 1.33\) and the \(c\) marker slides to about 1.15. Notice \(c\) is not the midpoint 1.
More info — why must the curve cross its average? And how is this like the MVT for derivatives?
The average \(f_{avg}\) always lands between the minimum and maximum of \(f\) on \([a, b]\) (an average of values can't beat all of them or trail all of them). The IVT then forces continuous \(f\) to pass through every height between its min and max — including \(f_{avg}\). That's the whole proof. It's the integral-side twin of the Mean Value Theorem for derivatives from earlier in the course: there, some \(c\) has \(f'(c)\) equal to the average rate of change; here, some \(c\) has \(f(c)\) equal to the average value. Same slogan both times — a continuous quantity must hit its own average. The OpenStax section linked in Dive deeper below walks the derivation alongside the Fundamental Theorem of Calculus.
Check your understanding
What is the average value of \(f(x) = x\) on \([0, 4]\)?
For a continuous function \(f\) on \([a, b]\), what exactly does the Mean Value Theorem for Integrals guarantee?
Find the average value of \(f(x) = x^2\) on \([0, 6]\).
The Mean Value Theorem for derivatives promises a \(c\) where the instantaneous rate \(f'(c)\) equals the average RATE of change over \([a, b]\). Its integral analogue promises a \(c\) where…
Recap
- \(f_{avg} = \dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx\) — never report the bare integral as the average; dividing by \(b-a\) is what makes it a height, not an area.
- Geometrically: \(f_{avg}\) is the height of the rectangle over \([a, b]\) with the same area as the region under \(f\).
- MVT for Integrals: continuous \(f\) on \([a, b]\) attains its average — some \(c\) in \([a, b]\) has \(f(c) = f_{avg}\). Continuity is required.
- \(c\) is generally not the midpoint \((a+b)/2\) — solve \(f(c) = f_{avg}\) to find it.
Dive deeper
- OpenStax Calculus Volume 1 — 5.3 The Fundamental Theorem of Calculus Define average value and state the Mean Value Theorem for Integrals
Sources
- Average Value and the Mean Value Theorem for Integrals