Suppose I have a real positive valued continuous function $f(x)$ such that for $\Delta>0$ $$\int_0^{\Delta}f(x)\,\mathrm{d}x=1.$$ If I am not provided with any exact functional form of $f(x)$, can I still say something about $$I=\int_0^{\Delta}\left[f(x)\right]^2\,\mathrm{d}x$$ i.e., is there any simplification to $I$ given the first equation? I tried integration by parts, but that does not seem to give anything really useful.
Is there any relation between these two integrals?
1
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calculus
integration
definite-integrals
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2Cauchy Schwarz gives $1 \le I \Delta$. – 2017-01-08
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0(I deleted my answer because I failed to read "positive valued") – 2017-01-08