# 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.

• You should specify the relationship that you want / are looking at. Are any of the standard analysis inequalities sufficient? For example, $|\int_a^b xf(x) \, dx |^2 \leq \int_a^b |f(x)|^2 \, dx \times \frac{ b^3 - a^3 } {3}$ by Cauchy Schwarz. – Calvin Lin Jun 11 '13 at 0:42
• Could you elaborate on the nature of the problem you are facing? Also, do you know what a and b are? – User3910 Jun 11 '13 at 0:45
• @CalvinLin Ah... Sorry. I'm looking to compute the numeric value of the second integral, so equalities are what I'm looking for. (Forgot that, just because I know what I'm looking for, not everyone else knows! ;) Thanks for the reminder.) – apnorton Jun 11 '13 at 0:45
• @JoeHobbit See Edit 2. :) – apnorton Jun 11 '13 at 0:49
• Not in general. For example, $\int_a^bxe^{-x^2}dx$ is trivial by variable change, while $\int_a^be^{-x^2}dx$ is non elementary. – Julien Jun 21 '13 at 15:16

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.

• Can the non-existence of an equality be proven, or is it a conjecture? (Just curious... I always like to play with things that haven't necessarily been proven before. ;)) – apnorton Jun 11 '13 at 0:56
• @anorton, equality cant be proven since does not hold for some $f$. – Gaston Burrull Jun 11 '13 at 1:06
• Ah! I see now... – apnorton Jun 11 '13 at 1:11

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...

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

$$\int_{-\infty}^{\infty} f(x) \ dx = 1$$