Just as you can average a set of numbers, you can average a continuous function over an interval. The average value of a function tells you the constant value that would give the same total area over the interval. This is also known as the Mean Value Theorem for Integrals.
🎯 In this section you will learn
- The formula for the average value of a function: $f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) dx$
- How to interpret the average value geometrically
- The Mean Value Theorem for Integrals (guarantees there is a $c$ where $f(c) = f_{\text{avg}}$)
- Applications in physics, economics, and other fields
📌 The Average Value Formula
$$ f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) \, dx $$
The average value of $f$ on $[a, b]$
💡 Geometric Interpretation
The average value is the height of a rectangle on $[a, b]$ whose area equals $\int_a^b f(x) dx$. In other words, the area under the curve equals the area of a rectangle of height $f_{\text{avg}}$ and width $b-a$.
Find the average value of $f(x) = 2x$ on $[0, 4]$.
①
Set up the formula
$f_{\text{avg}} = \frac{1}{4-0} \int_0^4 2x dx$
②
Evaluate the integral
$\int_0^4 2x dx = [x^2]_0^4 = 16$
③
Divide by length
$f_{\text{avg}} = \frac{16}{4} = 4$
Find the average value of $f(x) = x^2$ on $[0, 3]$.
- $f_{\text{avg}} = \frac{1}{3-0} \int_0^3 x^2 dx = \frac{1}{3} \left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3$
Find the average value of $f(x) = \sin x$ on $[0, \pi]$.
- $f_{\text{avg}} = \frac{1}{\pi-0} \int_0^{\pi} \sin x dx = \frac{1}{\pi} [-\cos x]_0^{\pi} = \frac{1}{\pi} (-\cos\pi - (-\cos 0)) = \frac{1}{\pi} (-(-1) + 1) = \frac{1}{\pi} (1 + 1) = \frac{2}{\pi} \approx 0.6366$
📌 Mean Value Theorem for Integrals
$$ \text{If } f \text{ is continuous on } [a, b], \text{ then there exists } c \in [a, b] \text{ such that } f(c) = \frac{1}{b-a} \int_a^b f(x) dx $$
The function attains its average value at some point $c$
💡 Intuition
If you drive a car and your average speed over a trip is 60 mph, the Mean Value Theorem guarantees that at some moment during the trip, your speed was exactly 60 mph (assuming continuous speed).
For $f(x) = x^2$ on $[0, 3]$, the average value is $3$. Find the value(s) of $c$ where $f(c) = 3$.
- $c^2 = 3$ ⇒ $c = \sqrt{3} \approx 1.732$ (since $c$ must be in $[0, 3]$)
The temperature during a 24-hour day is given by $T(t) = 70 + 10\sin\left(\frac{\pi t}{12}\right)$ where $t$ is in hours from midnight. Find the average temperature.
①
Set up the formula
$T_{\text{avg}} = \frac{1}{24-0} \int_0^{24} \left[70 + 10\sin\left(\frac{\pi t}{12}\right)\right] dt$
②
Integrate
$\int_0^{24} 70 dt = 70 \cdot 24 = 1680$
$\int_0^{24} 10\sin\left(\frac{\pi t}{12}\right) dt = 10 \left[-\frac{12}{\pi}\cos\left(\frac{\pi t}{12}\right)\right]_0^{24} = -\frac{120}{\pi}(\cos(2\pi) - \cos 0) = -\frac{120}{\pi}(1 - 1) = 0$
③
Divide by 24
$T_{\text{avg}} = \frac{1680}{24} = 70$°
⚠️ Common Mistakes
- Forgetting the factor $\frac{1}{b-a}$: The average value is not just the integral; you must divide by the length of the interval.
- Confusing average value with average rate of change: Average value is $\frac{1}{b-a}\int f(x)dx$; average rate of change is $\frac{f(b)-f(a)}{b-a}$.
- Thinking the average value always occurs at the midpoint: It occurs at some $c$, but not necessarily the midpoint.
🔍 Key Takeaways
- Average value: $f_{\text{avg}} = \frac{1}{b-a} \int_a^b f(x) dx$.
- Geometrically: the height of a rectangle with the same area as the region under the curve.
- Mean Value Theorem for Integrals: There exists $c \in [a, b]$ such that $f(c) = f_{\text{avg}}$.
- Applications include average temperature, average current, average population, etc.
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