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For each continuous function $f: [0,1] \rightarrow R,$ let $I\left(f\right) = \int_0^1 x^2 f\left(x\right)\: \textrm{d}x.$ and $J\left(f\right) = \int_0^1 x \left(f\left(x\right)\right)^2 \: \textrm{d}x.$ Find the maximum value $I\left(f\right) - J\left(f\right)$ over all such functions f.

So the problem is:

Step 1. For what values of $x$ does $\frac{\textrm{d}x}{\textrm{d}y} \left[ \int_0^1 x^2f\left(x\right)\: \textrm{d}x - \int_0^1 xf\left(x\right)^2\:\textrm{d}x\right] = 0 $

Step 2. For what values of $x$ is this negative $\frac{\textrm{d}^2x}{\textrm{d}y^2} \left[ \int_0^1 x^2f\left(x\right)\: \textrm{d}x - \int_0^1 xf\left(x\right)^2\:\textrm{d}x\right] = 0 $

Not sure exactly how to do that.

Page 281 Problem #80 in Calculus 9$^{th}$ edition by Larson No, its not homework its way to difficult for class, but I like math and the last problem is the most fun and I learn the most from.

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    Info on formatting: http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117 also http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference also http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto2012-11-17

1 Answers 1

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Step1. For the stationary $f$, $f(0)=0$. Otherwise, we can re-define $f$ on a right neighbourhood of zero and make the RHS strictly increase.

Step2. Assuming $x\in(0,1]$, take $g(x)=\frac{f(x)}{x}$. Then we have: $RHS=\int_{(0,1]}\left(x^2 f-x f^2\right)\,dx = \int_{(0,1]}x^3 g (1-g)\,dx, $ but $g(1-g)\leq\frac{1}{4}$ by the AM-GM inequality, so: $ RHS\leq \int_{(0,1]}\frac{x^3}{4}\,dx = \frac{1}{16}, $ with equality reached only when $g$ is constantly $\frac{1}{2}$, or $f(x)=\frac{x}{2}$.

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    Could you name a good book to learn this in? I have to study this more.2012-11-17