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EDIT:

This question is wrong. You don't need to waste your time trying to answer it. :D

I need help showing that:

$$ \int_a^b x f(x) dx = \frac {a+b} 2\int_a^bf(x)dx$$

My attempt. $$ I = \int_a^bxf(x)dx = \int_a^b(b+a -x)f(b+a-x)dx $$ $$ = \int_a^b (b+a)f(b+a-x)dx - \int_a^bxf(b+a-x)dx $$

I doubt that probably the 2nd term is $I$ itself (though not sure). Hints?

Btw, the reason I am asking this is, if the above stated theorem is true, then how come there be a contradiction between it and this question? You see, the question is of the same form. But in that case, the integral of $f(x) = \cfrac {\cos x} {1 + \sin^2x}$ from 0 to $\pi$ is 0.

Interestingly, the above stated theorem works for $$ \int \limits_0^\pi \cfrac {x \sin x} {1 + \cos^2x} dx $$

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Parth Thakkar
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3 Answers3

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A counterexample: $a=0$, $b=1$ and $f(x)=x$. Then

$\int_0^1 x^2 dx = \frac{1}{2} \int_0^1 x dx$

iff $\frac{1}{3}=\frac{1}{4}$.

Avitus
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  • Yup. It's confirmed then that this is a wrong question. Closing it here itself! – Parth Thakkar Jun 14 '13 at 14:57
  • Ok, no problem. Btw your statement if correct for all $f$'s s.t. $f(a+b-x)=f(x)$. They can be interesting functions, as the example you describe! – Avitus Jun 14 '13 at 15:12
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My suspicion is that this integral is true only for certain $f$. Take $f(x) = \dfrac{1}{b-a}$. In Probability, we refer to this specific $f$ as the density function for the continuous uniform distribution and $\int\limits_{a}^{b}\dfrac{x}{b-a} \text{ dx} = \dfrac{a+b}{2}$ as the expected value of the continuous uniform distribution. Also, for any density function, $\int\limits_{a}^{b}f(x) \text{ dx} = 1$, so we get $\int\limits_{a}^{b}xf(x) \text{ dx} = \dfrac{a+b}{2} * 1$, which is assuming that $f(x) = \dfrac{1}{b-a}$.

EDIT: Of course, the expected value of other density functions will NOT always be $\dfrac{a+b}{2}$.

Clarinetist
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  • I don't think that this is true only for $f(x) = \frac 1 {b-a}$. As I said it works for $f(x) = \cfrac {x\sin x} {1 + \cos^2 x} dx$ also. But definitely, this is a wrong question, and it's a mere coincidence that it works for this function. Let's close it here itself! Thanks for your help though! – Parth Thakkar Jun 14 '13 at 14:57
  • Ah, that's my bad. I'll edit this answer. – Clarinetist Jun 14 '13 at 14:58
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    I think I got why it works for $f(x) = \frac {x\sin x} { 1 + \cos^2 x} $. In fact, it works for these specific limits of integration (0, $\pi$). It just so happens, that $ f(a + b - x) = f(x) $ in this case, which can be verified easily! – Parth Thakkar Jun 14 '13 at 15:01
  • If you ever do take a Probability course, you will learn that $\int\limits_{a}^{b}xf(x) \text{ d}x$ is a VERY important integral. See http://en.wikipedia.org/wiki/Expected_value#Univariate_continuous_random_variable (where $-\infty$ and $\infty$ are just simply the bounds of your variable, so it's the same thing as $a$ to $b$, as in your question). – Clarinetist Jun 14 '13 at 15:07
  • Interesting! Will surely dive into this! – Parth Thakkar Jun 14 '13 at 15:09
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If $f$ is non-negative and $\int_a^bf(x)\,dx\neq0$, then we can regard $$ \frac{f(x)}{\int_a^bf(x)\,dx} $$ as a probability distribution on the interval $[a,b]$, and the equation in the question just says that the average of this distribution is the midpoint of the interval. That's fine for symmetrical distributions (like those mentioned in answers and comments above) and for some others, but it's "usually" false.

Andreas Blass
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