Definite integral of unknown function [closed]

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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Are you able to compute the variance of a random variable by knowing only that it is a random variable? – Jack D'Aurizio Dec 11 '14 at 15:57
• 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. – Matthias Dec 11 '14 at 15:58
• Thank you, it seems like it is not possible then. I was challenged to integrate this, so I assumed it could be evaluated. – sangi93 Dec 11 '14 at 16:01

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