mvt and average values

• Jan 21st 2009, 02:11 PM
razorfever
mvt and average values
exlain how you know that, if f is continuous on [a,b], then there is a c in (a,b) such that f(c) = f avg, the average value of f on [a,b]

I have been able to show this by rearranging the mean value theorem for integrals

but the question also offers a hint:
Apply the mean-value theorem to g(x) = integral from a to x f(t) dt

can someone show me the solution using this hint explicitly
• Jan 21st 2009, 02:53 PM
skeeter
Quote:

Originally Posted by razorfever
exlain how you know that, if f is continuous on [a,b], then there is a c in (a,b) such that f(c) = f avg, the average value of f on [a,b]

I have been able to show this by rearranging the mean value theorem for integrals

but the question also offers a hint:
Apply the mean-value theorem to g(x) = integral from a to x f(t) dt

can someone show me the solution using this hint explicitly

$g(x) = \int_a^x f(t) dt$

$g(b) = \int_a^b f(t) dt$

$g(a) = 0$

by the MVT, there exists a $c$ in the open interval $(a,b)$ such that ...

$g'(c) = \frac{g(b) - g(a)}{b-a}$

$g'(c) = \frac{\int_a^b f(t) dt - 0}{b-a}$

since $g'(x) = f(x)$ by the FTC ...

$g'(c) = f(c) = \frac{\int_a^b f(t) dt}{b-a}$