Question

Mean Value Theorum with Integration

Answer 1

I know:
\[\frac{ 1 }{ (b-a) }\int\limits_{}^{}f(x) = f(x)_{avg}\]

Answer 2

But I dont know how to find C? Is it the same as usual - set the derivative equal to what I got in the first equation?

Answer 3

\[\frac{ 1 }{ (b-a) }\int\limits\limits_{}^{}f(x) = f(x)_{avg}=c\]c is your average.

Answer 4

So whatever this integral stuff comes out to, that is your c.

Answer 5

Err maybe I'm misreading it :) sec sec

Answer 6

That seems sort of redundant for them to ask me the same thing twice, no?

Answer 7

Dont worry, if you are, youre not the only one lol

Answer 8

ya ya maybe I'm mistaken.
I think they mean f(c) is your average.
And they want the value c.

Answer 9

So plug the average back into the function?

Answer 10

\[\large\rm \frac{1}{3-(-3)}\int\limits_{-3}^3 |4x|dx\]Any trouble with integrating this?
Requires splitting it into a couple integrals, or geometry if you want to avoid integration.

Answer 11

I havent worked it yet honestly, just a sec. Im going to try and break it up into a piecewise function

Answer 12

\[\frac{ 1 }{ 6 }\int\limits_{0}^{3}(4x) + \frac{ 1 }{ 6 }\int\limits_{-3}^{0}-(4x)\]

Answer 13

That look right?

Answer 14

Mmm yes

Answer 15

ok, solving now

Answer 16

Alright so the average here is 6, but what do I need to do with that to get C?

Answer 17

\(\large\rm f(x)_{ave}=6\)
Ok great.
So we want to know exactly where (this means x location) the function takes on this average value.
\(\large\rm f(x)_{ave}=|4x|\)

Answer 18

I think I do just set it equal to the original function

Answer 19

yup so +/- (2/3)

Answer 20

Perfect, I know how to do these now! Thanks @zepdrix

Answer 21

Noiiceeee, good job :)

Answer 22

3/2, I think that was a typo

Answer 23

yea 3/2 I mean