Integrability of f^2: Show f is Integrable

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Homework Statement



Let f be positive and bounded over [a,b]. If f^2 is integrable over [a,b], then show that f is as well.

The Attempt at a Solution



I'm just trying to use the fact that the upper and lower sums of f^2 over a partition P are arbitrarily close, and then somehow find an upper bound for the difference of the upper and lower sums of f based on that. I've tried separating it into cases, where the max and min of f on an interval are >=1, <1, etc. but it hasn't really led anywhere. If anyone could give me some sort of hint, i'd really appreciate it =)
 
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It's fairly easy if you know f(x)>=m for m>0 isn't it? Since f(x1)^2-f(x2)^2=(f(x1)-f(x2))*(f(x1)+f(x2)) and you know f(x1)+f(x2)>=2m. So f(x1)^2-f(x2)^2>=(f(x1)-f(x2))*2m. Now let m approach zero. Why can you ignore the part of the sums coming from f(x)<m? Remember the interval of integration is [a,b]. It's finite.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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