Question

MVT for integrals

Answer 1

For an even function \
\[\int\limits_{-4}^{4}f(x)=2\int\limits_{0}^{4}f(x)\]

Where the average:
\[\frac{ 1 }{ b-a }\int\limits_{a}^{b}f(x)\]

But the question states this is an odd function, and I dont remember learning how to deal with that.

Answer 2

Why do you have | .... |, they aren't absolute values, are they?

Answer 3

This is an example of an odd function

Answer 4

For this kind of function \[\int\limits_{-a}^{a}f(x)dx = 0\]

Answer 5

If I can ignore these || bars in the question in your attachment:


For an odd function f(x) where \(\displaystyle\color{red}{\int_{4}^{0}f(x)dx=3}\),
determine the average value of f(x) over [-4,4].

Answer 6

Well, if \(\displaystyle\color{red}{\int_{4}^{0}f(x)dx=3}\)

then, \(\displaystyle\color{red}{\int_{0}^{4}f(x)dx=-3}\)

Answer 7

And your function is symmetric about the origin, since it is odd.

Answer 8

so, due to this symmetry, whatever the area you have over [-4,0], you will have exactly the same area by magnitude but with an opposite sign that will cancel it out over the interval [0,4].

Answer 9

This is why odd functions satisfy

\(\displaystyle\color{red}{\int_{-a}^{a}f(x)dx=0}\)'

as lord pointed out.

Answer 10

@SolomonZelman no, its a glitch in the program

Answer 11

Ah that makes perfect sense, I was thinking about it too much. Thank you both

Answer 12

Yes, so you don't even need to know that \({\tiny\\[0.3em]}\)
\(\displaystyle\color{red}{\int_{4}^{0}f(x)dx=3}\)

As soon as you know that the function is odd, then
\(\displaystyle\color{blue}{\int_{-b}^{b}f(x)dx=0}\)

because the function is symmetric about the origin,
and thus, as I said, areas on both sides cancel each other
out, and also the average value is 0.

Answer 13

(If overall area under a curve over specified interval is 0, then obviously the average values is 0 as well... so:))