AP Calculus AB and BC

Average Value vs MVT for Integrals

Q: Does the theorem give exactly one value of c?

It guarantees at least one. There can be several, and c is generally not the midpoint of the interval, so never assume it is.

Q: How do I find c?

Compute f_{\text{avg}} first, then solve f(c) = f_{\text{avg}} and keep only the solutions lying in the open interval (a,b).

Q: Is this the same as the Mean Value Theorem for derivatives?

No. The derivative version puts f'(c) equal to the average rate of change and needs differentiability. The integral version puts f(c) equal to the average value and needs only continuity.

Average value is a number: divide the integral of the function over an interval by the length of that interval. The Mean Value Theorem for Integrals is the guarantee that a continuous function actually takes on that number at some point c inside the interval. One computes, the other certifies.

Average value

Use when: The problem asks for the average value of $f$ on $[a,b]$ , or for an average temperature, concentration, or velocity across an interval.

MVT for integrals

Use when: The problem asks you to justify that some $c$ exists where $f(c)$ equals the average value, or that a horizontal line meets the curve.

Side by side

	Average value	MVT for integrals
What it is	A number	An existence guarantee
Statement	$f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx$	Some $c$ in $(a,b)$ has $f(c) = f_{\text{avg}}$
Hypothesis needed	Only that $f$ is integrable on $[a,b]$	$f$ continuous on $[a,b]$
When to reach for it	You are asked to compute an average	You are asked to justify that the average is attained
Common trap	Forgetting to divide by $b-a$	Claiming $c$ is unique, or assuming it is the midpoint

Average value answers how high a flat line would have to sit to enclose the same accumulated amount. Rearranging makes that literal: $f_{\text{avg}} \cdot (b-a) = \int_a^b f(x)\,dx$ , so the rectangle of height $f_{\text{avg}}$ over $[a,b]$ has exactly the area the curve has.

f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx

The theorem supplies what the formula cannot promise. A number built from an integral need not be a value the function ever takes: a function that jumps from $0$ to $2$ has average value $1$ and never equals $1$ . Continuity closes the gap, because $f_{\text{avg}}$ always lies between the minimum and maximum of $f$ , and the Intermediate Value Theorem forces a continuous function through every value in between.

Average value is not average rate

Average value of $f$ comes from an integral. Average rate of change of $f$ is $\frac{f(b)-f(a)}{b-a}$ . They meet in one place only: the average value of $f'$ on $[a,b]$ equals the average rate of change of $f$ , which is what the Fundamental Theorem says once you divide by $b-a$ . Average Value vs Average Rate of Change takes that split apart on its own page.

Frequently asked questions

Does the theorem give exactly one value of c?

It guarantees at least one. There can be several, and $c$ is generally not the midpoint of the interval, so never assume it is.

How do I find c?

Compute $f_{\text{avg}}$ first, then solve $f(c) = f_{\text{avg}}$ and keep only the solutions lying in the open interval $(a,b)$ .

Is this the same as the Mean Value Theorem for derivatives?

No. The derivative version puts $f'(c)$ equal to the average rate of change and needs differentiability. The integral version puts $f(c)$ equal to the average value and needs only continuity.

In the CED: Unit 6: Integration and Accumulation, Unit 8: Applications of Integration