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If the $(\int_a ^b f(x))/(a-b)$ is the arithmetic average of all the values of $f(x)$ between $a$ and $b$, what is the expression representing the geometric average of all these values?

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I know GNU Supporter already answered, but I wanted to explain why its the answer.

We'll first discuss arithmetic calculus (discrete and continuous). Let:

$_{h}\Delta_x[f(x)] = f(x+h) - f(x)$

Be the discrete forward derivative in $h$-step form. The usual operator has $h=1$. Likewise:

$\sum_x[f(x)]$

Is the discrete integral: a summation. You might be wondering about the strange notation. This represents an "indefinite" sum, just as you have indefinite integrals. The following notation is more familiar:

$\sum_{x=a}^{b-1}[f(x)]$

Which is the definite forward sum. And of course you know the continuous versions of them:

$\lim_{h\to 0}[{_{h}\Delta_x[f(x)] \over h}] = {d \over dx}[f(x)] = {df \over dx}$

$\lim_{h\to 0}[h\sum_{k=1}^n[f(x_k)]]=\int_a^b[f(x)dx]$

Where $n = {(b-a) \over h}$. I know this seems repetitive, but now we construct geometric functions that mirror them. Let:

$_{h}⩍_x[f(x)] = {f(x+h) \over f(x)}$

Be the discrete forward geometric derivative (that's a mouthful). There isn't a standard notation as far as I'm aware, so I chose a rounded delta. Again, usually $h=1$. Then you have the discrete geometric integral:

$\prod_{x}[f(x)]$

Which is just the capital pi product function. Finally, the continuous versions are:

$\lim_{h\to 0}[\sqrt[h]{_{h}⩍_x[f(x)]}] = \sqrt[dx]{q}[f(x)]=\sqrt[dx]{qf}$

$\lim_{h\to 0}[\prod_{k=1}^n[f(x_k)^h]] = \mathcal{P}_a^b[f(x)^{dx}]$

Maybe now you're scratching your head? All we've done is substituted quotients for differences and products for sums in our previous definitions (and roots for quotients and powers for products). Then the $q$ under the root symbol stands for quotient, just as $d$ in the derivative stands for difference. Similarly the integral symbol is an elongated $S$ for sum, so the geometric integral symbol is a fancy $P$ for product. Great, but how do we evaluate these? As it turns out, there's a neat relationship between the arithmetic and geometric operators:

$\sqrt[dx]{q}[f(x)] = \exp({d \over dx}[\ln(f(x))])$

$\mathcal{P}_a^b[f(x)^{dx}] = \exp(\int[\ln(f(x))dx])$

This is amazing, because we don't need to write a new rulebook for evaluating the operators: we just use their counterparts! And that's why there's $\exp()$ and $\ln()$ functions in the geometric average. (As a side note, the second relationship classifies the integral as "Type II". A "Type I" would simply not have $\ln()$ inside the integral. But I'm not going to go into that). I Hope it bettered your understanding.

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I think there is a typo in the first answer, namely the "$n$". Let $f\: [a, b] \to {\Bbb R}$ be continuous and strictly positive on $[a, b]$. Then the geometric mean of $f$ on $[a, b]$ is $$ \exp\left(\frac{1}{b-a}\int_a^b \ln(f(x)) \,\mathrm{d}x \right) $$ where I use $\ln$ rather than $\log$, just out of habit.

The explanation given by Andrew above may be correct, but is overly complicated. Here is a simpler approach just using basic calculus concepts. If we partition $[a, b]$ into $n$ equal subintervals as one does routinely in calculus, and for each $i$, choose an arbitrary point $x_i^{\ast}$ in the $i$th subinterval, then the geometric mean at the $n$ sample points will be $$ a_n = \bigl(f(x_1^{\ast})f(x_2^{\ast})\cdots f(x_n^{\ast})\bigr)^{1/n}. $$ Of course we want to find $\lim_{n \to \infty}a_n$, which will be the geometric mean of $f$ on $[a, b]$. Now, using rules for logarithms, and with $\Delta x = \dfrac{b - a}{n}$ as usual, we have $$ \ln(a_n) = \dfrac{1}{n}\sum_{i = 1}^{n} \ln(f(x_i^{\ast})) = \dfrac{1}{b - a}\sum_{i = 1}^{n} \ln(f(x_i^{\ast}))\Delta x, $$ and so $$\lim_{n \to \infty}\ln(a_n) = \dfrac{1}{b - a}\int_{a}^{b}\ln(f(x))\, dx. $$ Exponentiating, we obtain that $$ \lim_{n \to \infty}a_n = \exp\left(\frac{1}{b-a}\int_a^b \ln(f(x))\,\mathrm{d}x \right), $$ which is therefore the geometric mean of $f$ on $[a, b]$.

It's also worth noting that by Jensen's Inequality, the geometric mean is less than or equal to the arithmetic mean, that is, with the same hypotheses, $$ \exp\left(\frac{1}{b-a}\int_a^b \ln(f(x))\,\mathrm{d}x \right) \leq \frac{1}{b - a}\int_{a}^{b}f(x)\, dx. $$ As references, see Rudin's Real & Complex Analysis, as well as any standard calculus text.

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