Relationship between $\int_a^b f(x)\,dx$ and $\int_a^bxf(x)\,dx$ I was working a problem, and came to a point where it would help greatly if there was a relationship between the following two expressions:
1) The numeric value of $\int_a^b f(x)\,dx$, and
2) The numeric value of $\int_a^bxf(x)\,dx$.
That is, if I know the numeric value for the first integral, but nothing about $f(x)$, is it possible to determine the numeric value for the second?
I've tried experimenting with integration by parts, but that always seems to need the indefinite integral of $F(x)$.
EDIT: In response to Calvin Lin's comment: I'm looking to compute the second integral, so equalities are the type of relationship that I'm looking for.
EDIT 2: In response to JoeHobbit's comment: The particular problem that I'm trying to solve is really this one here.  However, this sprung off some other thoughts, not necessarily tied to specific problems.
 A: It's impossible to determine the numerical value of $\int_a^bxf(x)dx$ exactly from $\int_a^bf(x)dx$ (for example, $\int_0^1 xdx = \int_0^1(1-x)dx = \frac{1}{2}$, but $\int_0^1x\cdot xdx = \frac{1}{3} \ne \int_0^1x(1-x)dx = \frac{1}{6}$. However, you still know some things. As you mentioned, integration by parts tells you that $\int_a^bxf(x)dx = bF(b) - aF(a) + \int_a^bF(x)dx$, where $F(x)$ is an antiderivative of $f(x)$. If you could find bounds to the antiderivative, you could tell a bit about this integral. Another inequality was given in the comments, but you can't get an equality.
A: I'm not sure if this is the answer you're looking for, but there is a relationship in probability theory.
If $f(x)$ is a probability density function, then we can define the average of $x$ on the interval $[a,b]$, to be
$$
\int_a^b x \ f(x) \ dx
$$
where
$$
\int_a^b f(x) \ dx
$$
represents the probability (if $f$ is normalized on the interval) that $x \in [a,b]$.
Note: If $f$ is a normalized probability density function, then
$$
\int_{-\infty}^{\infty} f(x) \ dx = 1
$$
A: For estimation if $f(x)$ is non-decreasing and non-negative, Steffensen's inequality came into my mind:
$$
\int_{b - k}^{b} f(x) \, dx \leq \int_{a}^{b} f(x) g(x) \, dx \leq \int_{a}^{a + k} f(x) \, dx,$$
where $g: [a,b]\to [0,1]$ is integrable, and $k = \int^a_b g(x) \,dx$. So we can let $g(x) = (x-a)/(b-a)$, then $k = (b-a)/2$:
$$
\int_{(b+a)/2}^{b} f(x) \, dx \leq \int_{a}^{b} f(x) \frac{x-a}{b-a} \, dx \leq \int_{a}^{(b+a)/2} f(x) \, dx,$$
which gives us the bound:
$$
b\int_{m}^{b} f(x)\, dx  + a\int_{a}^{m} f(x) \, dx \leq \int_{a}^{b} xf(x)  \, dx \leq b\int_{a}^{m} f(x) \, dx + a\int_{m}^{b} f(x)\, dx,$$
where $m$ is the midpoint of $[a,b]$. Above bound is sharper than Cauchy-Schwarz by looking at the error term of Steffensen. For the curious, if $f(x)$ is a polynomial, above bound is very very sharp. The bound also implies, if $f(x)$ is not changing very fast on the interval, then
$$
\frac{a+b}{2}\int^b_a f(x) \,dx 
$$
is a very good approximation to the integral (Midpoint rule in disguise...).
For general $f(x)$, I don't know if there is any good method on estimating the first moment based on the zeroth moment...
