Question

Find the value(s) of guaranteed by the
Mean Value Theorem for Integrals for the function over the
given interval. @NCount

Answer 1

\[f(x)=\frac{ 9 }{ x ^{3} }\] [1,3]

Answer 2

\[f(c) = \frac{1}{3-1} \int\limits_{1}^{3} \frac{9}{x^2} dx\]

solve for c

Answer 3

Thanks dude!

Answer 4

this is the mean value theorem for INTEGRAL

Answer 5

Whoops! Thanks sourwing. I saw MVT and started running without reading that bit.

Answer 6

so how did you get the \[\frac{ 1 }{ 1-3 }\]

Answer 7

Here is the general formula sourwing pulled that from:
\[f(c)=\frac{ 1 }{ b-a }\int\limits_{b}^{a}f(x) dx\]

Answer 8

EDIT: the integral should be from a to b, NOT b to a.

Answer 9

Thank you thank you

Answer 10

Sure thing!