AP Calculus AB and BC glossary

Mean Value Theorem for Integrals

Also called: MVT for integrals, Integral Mean Value Theorem

The Mean Value Theorem for Integrals says that if a function is continuous on a closed interval, then at some interior point the function equals its average value on that interval. So the definite integral equals that one height multiplied by the width of the interval.

f(c)=1b−a∫abf(x) dxfor some c in (a,b)f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx \quad \text{for some } c \text{ in } (a,b)

Continuity alone is enough, and the reason is short. The average value sits somewhere between the minimum and the maximum of ff on [a,b][a,b], and a continuous function takes every value between those two, so it has to hit the average at least once somewhere in [a,b][a,b]. That much follows from the Intermediate Value Theorem; the argument below is what sharpens it to an interior point.

It is the derivative version in disguise. Apply the Mean Value Theorem to F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt, whose derivative is ff: the conclusion F′(c)=F(b)−F(a)b−aF'(c) = \frac{F(b) - F(a)}{b-a} says exactly that f(c)f(c) equals the average value of ff. Because the Mean Value Theorem hands back a cc strictly inside, this is the argument that puts cc in the open interval (a,b)(a,b).

The mistake

Solving f(c)=f(b)−f(a)b−af(c) = \frac{f(b) - f(a)}{b-a}. That right-hand side belongs to the derivative version, f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b-a}, where the slope matches the average rate of change. Here you set the function itself, not its derivative, equal to the average value 1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx, then solve for cc.

Appears in: Unit 6: Integration and Accumulation, Unit 8: Applications of Integration