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if $$ \int_{2}^{5} f(x) dx = 5 $$

what is the value of: $$ \int_{2}^{5} x^2f(x)dx $$

Normally I would use integration by parts and try to reduce the $x^2$, but I can't figure out how to do that now.

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closed as unclear what you're asking by Jack D'Aurizio, Aditya Hase, Claude Leibovici, Jyrki Lahtonen, Davide Giraudo Dec 11 '14 at 16:35

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    $\begingroup$ Are you able to compute the variance of a random variable by knowing only that it is a random variable? $\endgroup$ – Jack D'Aurizio Dec 11 '14 at 15:57
  • $\begingroup$ In my opinion it is not possible to evaluate the second integral. $f$ could be $f(x)=\frac{5}{s}\delta(x-s)$ with $s$ between two and five. $\endgroup$ – Matthias Dec 11 '14 at 15:58
  • $\begingroup$ Thank you, it seems like it is not possible then. I was challenged to integrate this, so I assumed it could be evaluated. $\endgroup$ – sangi93 Dec 11 '14 at 16:01
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Hint: Consider $$f_s(x)=\frac{5}{s}\delta(x-s)$$

with $s\in[2,5]$. Then $$\int_{2}^{5} f_s(x) dx = 5$$ but

$$\int_{2}^{5} x^2f_s(x)dx=5\cdot s^2$$

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