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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.

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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? –  JoeHobbit 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. –  1015 Jun 21 '13 at 15:16

3 Answers 3

up vote 8 down vote accepted

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.

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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 $f$. –  Gastón 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...

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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

$$ \int_a^b x \ f(x) \ dx $$


$$ \int_a^b f(x) \ dx $$

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

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

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