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If $f(x)$ is a continuous function on the interval $[x, x+h]$, find $$\lim_{h\to 0} f(x)_{avg}$$

I suspect I'm using the limit definition of the derivative, and to obtain the average value I've integrated over $[{x, x+ h}]$: $$\frac{\int_x^{x+h}f(x + h) - \int_x^{x+h}f(x )}{(x+h) - x} $$

What is the next step, or what have I done incorrectly? I'm not sure what the goal is.

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    If $f(x)$ is continuous, then $\lim f(g(x))=f(\lim g(x))$ – 2017-01-31
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    The average of $f$ on $[x,x+h]$ is $\frac{1}{h} \int_x^{x+h} f(t) dt\,$. Not sure I follow where your difference of two integrals came from. – 2017-01-31
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    @dxiv Probably the limit definition of a derivative of an integral, and OP is likely trying to derive FTOC I imagine. – 2017-01-31
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    @SimplyBeautifulA yes I was trying to derive the FTC I. thanks for the help – 2017-01-31
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    @SimplyBeautifulArt Guess it's about FTC indeed, still point remains that that's not the right expression `to obtain the average value`. – 2017-01-31
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    @dxiv isn't the average value $\frac{1}{(b-a)}$ multiplied by the integral of $f(x)$? – 2017-01-31
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    @J.T. Right, as I also wrote in a previous comment. But you are essentially taking the difference between average values on $[x+h,x+2h]$ and $[x,x+h]$. It's that difference which I am not sure what's supposed to mean or do. – 2017-01-31
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    ah i see what you mean. I agree it is not correct, the bounds on my second integral were wrong. I was trying to get to a point to use the FTC I with the limit definition of a derivative . – 2017-01-31
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    `I was trying to get to a point to use the FTC` Use it, or prove it? +1 to @SimplyBeautifulArt for a nicely written answer, but (not that it matters) I am still not quite clear on what the question was. – 2017-01-31
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    D: I still do not think what I did was as good as it could've been, so I made it a community post to promote outside editing. – 2017-01-31
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    @dxiv I was just as confused as you are. The question was part of a review problem set, and did not give me any more information than in my post. The question, exactly, was If $f(x)$ is a continuous function on the interval $[x, x+h]$, find $\lim_{h\to 0}$ of the average value of the function. I assume it was more of a food for thought thing, and it pulled me in. – 2017-01-31

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I will note you did the first step wrong. Note that you should have

$$g(x)=\int_a^xf(t)\ dt$$

Differentiating:

$$g'(x)=\lim_{h\to0}\frac{\int_a^{x+h}f(t)\ dt-\int_a^xf(t)\ dt}h=\lim_{h\to0}\frac1h\int_x^{x+h}f(t)\ dt$$

If we suppose the following statement, perhaps as axiom, that $\int_x^{x+h}f(t)\ dt=\int_x^{x+h}f_{avg}(x,x+h)\ dt$ where $f_{avg}(a,b)$ is the average value of $f$ for $a\le x\le b$, then,

$$g'(x)=\lim_{h\to0}\frac1h\int_x^{x+h}f_{avg}(x,x+h)\ dt=\lim_{h\to0}\frac1hf_{avg}(x,x+h)\int_x^{x+h}1\ dt\\=\lim_{h\to0}\frac1hf_{avg}(x,x+h)\cdot h$$

This integral is a trivial one that may be geometrically solved for, and from here, we see the $h$'s cancel, leaving us with

$$g'(x)=\lim_{h\to0}f_{avg}(x,x+h)\stackrel?=f_{avg}(x,x)=f(x)$$

which I suppose is what you want. Somewhere along the line though, we need continuity, which I suppose could come through with a limit property:

if $f(x)$ is continuous, then $\lim f(g(x))=f(\lim g(x))$.