Definite integral of function multiplied by its dependent variable It is well known that $$ \int_a^b f(x) dx = (b-a)\bar{f} $$
where $\bar{f}$ is the average value of $f(x)$ in $[a,b]$
I was wondering if there is a similar approach for $$ \int_a^b xf(x) dx $$
involving the average value. I am pretty sure that i have seen something about this in the past but I can't find anything right now since I don't even know how to search for it
Thank you all in advance
 A: The first formula you wrote is usually taken to be the definition of "average value."
The second integral is closely related to a similar quantity known as the "first moment of $f$" (you've written down "the first moment around $0$").
If you think of $f(x)$ as the amount of mass at position $x$ from the center of a lever-arm (which is centered at $x = 0$), then the integral represents the torque about the pivot of the lever.
Because of the appearance of the integral, a natural way to evaluate it is to use integration by parts (at least when $f$ has an obvious antiderivative).
Anyhow, if you search for "moment integrals", you'll find a great deal more information.
A: No, there is no connection. Specifically, there is no function of $\overline{f}$, $a$, and $b$ that reliably produces the given integral. To see this, we just need to find two functions $f$ and $g$, over a common interval $[a, b]$, with the same average value $\overline{f} = \overline{g}$.
Let $f(x) = 1$ and $g(x) = 2x$, over the interval $[0, 1]$. Then $\overline{f} = \overline{g} = 1$, but
$$\int_0^1 xf(x)\, \mathrm{d}x = \int_0^1 x\, \mathrm{d}x = \frac{1}{2}$$
and
$$\int_0^1 xg(x) \, \mathrm{d}x = \int_0^1 2x^2 \, \mathrm{d}x = \frac{2}{3}.$$

Generally speaking, the two integrals have very little to do with each other. Even the difficulty of computing each integral tends to be incomparable. The classic example is $\int e^{x^2} \, \mathrm{d}x$, which cannot be expressed with elementary functions. But, if we multiply by $x$, then we get the very tractable integral
$$\int xe^{x^2} \, \mathrm{d}x = \frac{1}{2}e^{x^2} + C.$$
But, multiplying $xe^{x^2}$ by $x$ again, we get an integral as intractable as the first:
$$\int x^2e^{x^2} \, \mathrm{d}x = \int x\left(xe^{x^2}\right) \, \mathrm{d}x = x\cdot \frac{1}{2}e^{x^2} - \frac{1}{2}\int e^{x^2} \, \mathrm{d}x,$$
hence we can compute $\int x^2e^{x^2} \, \mathrm{d}x$ if and only if we can compute $\int e^{x^2} \, \mathrm{d}x$ (which we can't, easily).
