Question

The average value of f on the interval [0, 10] is the average of the average value of f on [0, 5] and the average value of f on [5, 10]. True or False?

Answer 1

It depends what you mean by "and".

Answer 2

The average value of f on the interval [0, 10] is the average of the average value of f on [0, 5] PLUS the average value of f on [5, 10].

I think the question means PLUS where I typed above.

And means it is equal to both options simultaneously...

Answer 3

You can re-average them together.

Answer 4

so it would be true?

Answer 5

Yes.

Answer 6

The average value of the product f(x) · g(x), of two functions on an interval equals the product of the average values of f(x) and g(x) on the interval.

Answer 7

\(\frac{1}{10}\int_0^{10} f(x)\ dx = \frac{1}{2}(\frac{1}{5}\int_0^{5} f(x)\ dx + \frac{1}{5}\int_5^{10} f(x)\ dx)\)

Answer 8

Because the length of both intervals is equal \(5-0 = 5,\quad 10-5=5\).

Answer 9

what about that one?

Answer 10

i see what you mean though!

Answer 11

Do you know what a weighted average is?

Answer 12

im afraid not :/

Answer 13

A normal average would be: \[
\frac{a+ b+c}{3}
\]

Answer 14

A weighted average would be \[
\frac{aw_a+bw_b+cw_c}{w_a+w_b+w_c}
\]

Answer 15

hmmm okay so how would that apply to the product of two functions?

Answer 16

The weights \(w\) give each item being averaged a custom significance.

Answer 17

When we find the average of a continuous function, the weights we use are \(\Delta x\), which becomes the infinitesimal \(dx\).

Answer 18

So the normal formula is: \[\Large
\frac{\int_a^bf(x)\;dx}{\int_a^bdx} = \frac{F(a)-F(b)}{b-a}
\]where \(F'(x) = f(x)\).

Answer 19

If you already have calculated the average of a function for a few intervals, and you want to re-average them all, then you would take a weighted average where the weights are the interval lengths.

Answer 20

okay

Answer 21

You could average for two intervals that weren't even touching.

Answer 22

\[

\frac{(d-c)\frac{F(d)-F(c)}{d-c} +(b-a)\frac{F(b)-F(a)}{b-a}}{(d-c)+(b-a)} =
\frac{(F(d)-F(c)) +(F(b)-F(a))}{(d-c)+(b-a)}
\]

Answer 23

This is done when calculating the center of mass.

Answer 24

Let me make a correction:\[
\frac{\int_a^b f(x)\;w(x)\;dx}{\int_a^bw(x)\;dx} = \frac{G(b)-G(a)}{W(b)-W(a)}
\]where \(G'(x) = f(x)w(x)\).

Answer 25

Does this make it averages more clear to you?

Answer 26

In probability, the expected value is given by the weighted average of a random variable weighted by it's probability distribution.