Finding the Average Value of a Function on an Interval
Introduced| English | Meaning |
|---|---|
| average value/ˈævrɪdʒ ˈvæljuː/ | For a continuous function over an interval, its definite integral divided by the interval's length. |
The average height of a curve
- You can average a list of numbers — but how do you average a function, which has infinitely many values?
- Integration does it: the average value 平均值 of $f$ on $[a,b]$ is its total accumulation divided by the width.
- Picture the area under the curve reshaped into a rectangle of the same width — its height is the average.
- It's the calculus version of "add them up and divide by how many."
The formula
- The average value of $f$ on $[a,b]$ is
-
$$f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$$
- Compute the definite integral (the total), then divide by the interval length $b-a$.
- Units: same as $f$ itself (it's an average height, not an area).
The rectangle of height f_avg has the same area as the region under f
The equal-area rectangle
y = x²
The average value is the constant height whose rectangle encloses the same area as the curve over $[a,b]$.
The average value of $f$ on $[a,b]$ is...
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Total divided by width.
The average value has the same ____ as the function $f$, not the units of an area.
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It is an average height, not an area.
It's the "leveling out" height
- The average value is the constant height that gives the same area over $[a,b]$.
- $\displaystyle\int_a^b f_{\text{avg}}\,dx = f_{\text{avg}}(b-a)=\int_a^b f\,dx$ — the rectangle and the region match.
- So $f_{\text{avg}}$ is where you'd "level off" the curve to keep the same accumulated total.
- (The Mean Value Theorem for Integrals guarantees $f$ actually hits this average somewhere if $f$ is continuous.)
Find the average value of $f(x)=x^2$ on $[0,3]$. ($\int_0^3 x^2\,dx=9$.)
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$\tfrac{1}{3}\cdot 9=3$.
Geometrically, the average value is the height of a rectangle (width $b-a$) with the same...
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The equal-area rectangle's height is the average value.
Averages in context
- Average velocity over $[a,b]$ = $\frac{1}{b-a}\int_a^b v(t)\,dt$ = displacement ÷ time.
- Average temperature, average rate, average concentration — all the same recipe.
- Read the units to interpret: the average of a rate is still a rate.
- It answers "what single steady value would produce the same total?"
The definite integral $\int_a^b f\,dx$ by itself is the average value of $f$.
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That is the total; divide by $b-a$ for the average.
A car's displacement over $[0,2]$ h is $\int_0^2 v\,dt=120$ km. Find its average velocity (km/h).
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$\tfrac{120}{2}=60$ km/h.
The average value divides by $b-a$ — don't confuse it with the definite integral alone (that's the total, an area). $\int_a^b f\,dx$ is accumulated amount; $\frac{1}{b-a}\int_a^b f\,dx$ is average height. Forgetting the $\frac{1}{b-a}$ is the classic mistake.
Find the average value of $f(x)=x^2$ on $[0,3]$.
- $\displaystyle\int_0^3 x^2\,dx=\Big[\tfrac{x^3}{3}\Big]_0^3=9$.
- Divide by the width: $f_{\text{avg}}=\dfrac{1}{3-0}\cdot 9 = 3$.
- So a constant height of $3$ would enclose the same area over $[0,3]$.
The average value of $f$ on $[a,b]$ is $f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx$ — the total (a definite integral) divided by the interval width. It's the constant height enclosing the same area, with the same units as $f$. Never drop the $\frac{1}{b-a}$.
Write from an average value of 2
- f(x) = x on [0, 4].
- The average value is wanted.
What is the average value?
Total divided by width is the average height.
The average value of x on [0, 4] is what?
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8 / 4 = 2.