# Average value of a function

Maybe I am just bugging out, but I need a sanity check

The question:

If $\int_0^4 f(x)dx = C$, what is the average value of $f(x)$?

The question does not include a body for $f(x)$ or a table of values/graph for the output of f(x), so can this question be answered?

I know that: $$\int_a^b f(x)\,dx = F(b)-F(a),$$ will evaluate the integral across a given interval, but the question is asking what the average value of f(x) is, not its integral.

If I take the derivative of the above integral, I can get the function itself, but what do I need to do about the interval $[0, 4]$?

If I take the derivative of the integral, will I be left with $f(x) = 0$ (since the constant will disappear).

Thanks for the help.

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You write "the question is asking what the average value of $f(x)$ is", but the question as you quoted it doesn't ask for the average value. It would make a lot more sense if it did, so perhaps you've misquoted it? – joriki Oct 28 '11 at 4:18
Is the question asking for the average value of $f(x)$ on that interval? If so the definition of average value of $f(x)$ on an interval $[a,b]$ is $1/(b-a) \int_a^b f(x)\,dx$ so in your case it would just be $C/(4-0)=0.25C$. – Bill Cook Oct 28 '11 at 4:19
For $f(x)\ge 0$, the area under the curve from $0$ to $4$ is, as you know, $\int_a^b f(x)\,dx$. Imagine that area is flattened out to form a rectangle of base the interval from $a$ to $b$. Then the average value of $f(x)$ is the height of that rectangle. Therefore the average value is $\frac{1}{b-a}\int_a^b f(x)\,dx$. the same formula also works when $f(x)$ is not necessarily $\ge 0$, but the geometry is less clear. – André Nicolas Oct 28 '11 at 5:09

Remember that if $f(x)$ is continuous on $[a,b]$, then the average value of $f(x)$ on $[a,b]$ is defined to be $$\text{average value of }f\text{ on }[a,b] = \frac{1}{b-a}\int_a^b f(x)\,dx.$$

So if you need to know the average value of $f(x)$ on $[0,4]$ (as opposed to "the value of $f(x)$"), then just use that formula.

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I somehow missed this formula. Is this an identity? Can you direct me to a proof of this? Thanks. – Dylan Oct 28 '11 at 4:37
Also, I am not sure why the numerator is 1. Or is it just a constant? – Dylan Oct 28 '11 at 4:40
@Dylan: It's a definition. That's what "the average value of $f(x)$" is defined to be. Yes, the numerator of the fraction is 1. Intuition: you are "adding up" all the values of $f(x)$ with the integral, and then you are dividing by the length of the interval to get the "average". – Arturo Magidin Oct 28 '11 at 4:42
If I understand you correctly, the one is a scaling factor for the sum of whatever all the values of f(x), which will be some constant? – Dylan Oct 28 '11 at 4:55
@Dylan: Indeed, it does not make sense. But "does not make sense" does not seem to be a bar for the occasional student to say something, so better to make sure. – Arturo Magidin Oct 28 '11 at 5:13

Consider that average speed over the course of a journey is given by distance travelled divided by time elapsed; if the journey took place from time $t=a$ to time $t=b$, and if s(t) is your odometer reading at time $t$, then $${\rm average\ speed}={s(b)-s(a)\over b-a}$$ Now your speed $v(t)$ is the rate of change of your position, so $v(t)=s'(t)$, making $s(t)$ an antiderivative of $v(t)$. Thus, the numerator in that displayed equation is just $\int_a^bv(t)\,dt$. And so we get $${\rm average\ of\ }v(t)={1\over b-a}\int_a^bv(t)\,dt$$ (without writing down any Riemann sums or other things that tend to scare 1st-year Calculus students) (and, yes, I am aware that I am not being careful to distinguish speed from velocity; sue me).

Well, speed is just a function; what works for $v(t)$ should work for any function (since $v(t)$ could be any function). And that (I maintain) is how to get to the formula Arturo gives in his answer.

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A good proof of how the average is derived can be found in here:

Average value of a function

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The $\int_0^4f(x).dx$ gives you the area, which is the same thing as the sum of all the values of f from 0 to 4, multiplied by the dx (the horizontal distance between the values):

$\int_0^4f(x).dx$ = F(4)-F(0) = $[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx$

divide this number by dx multiplied by the number of f's and you get the mean value.

$\frac{[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx}{(ndx)}$ where n is the number of f's (or points)

Remember, dx is just the (horizontal) length of the area you're integrating, divided by the number of points (as the number of points goes to infinity). So, ndx=n(horizontal length)/n is just the horizontal length of the area you're integrating over.

Simplifying the previous thing, you get $[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx/(ndx) = [F(4)-F(0)]/\text{(horizontal length)}= [F(4)-F(0)]/(4-0) = C/4$

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And what, precisely, do you mean by .00000...2? There are ways to make infinitesimals rigorous, but this isn't one of them. I'd encourage you to delete this answer. – Gerry Myerson Oct 28 '11 at 6:18
A common sense representation used in answering a basic question. And no, I'm not going to delete the best answer to this question. – john smith Oct 28 '11 at 14:05