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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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    This is a problem in calculus of variations. I don't think steps 1 and 2 are accurate.2012-11-17
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    Check out Euler Lagrange condition for more details.2012-11-17
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    @GautamShenoy I have googled **Euler Lagrange condition** and I vagle understand what I am reading, could you please explain more. Or atleast point me to a PDF textbook that will explain. I am in first year calculus.2012-11-17
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    If you are in first-year Calculus, then Calculus of Variations is a bit out of reach, I'm afraid. Did this question come up in first-year Calculus? or did you find it somewhere else?2012-11-17
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    @GerryMyerson Its in my text book, edited the post.2012-11-17
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    Good. Now, what's the name of the chapter that has this problem?2012-11-17
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    @GerryMyerson Integration2012-11-17
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    MathJax/LaTeX tip: You *really* should not overuse `\left` and `\right` like that - especially when writing actual LaTeX and not MathJax because there are issues with whether you get opening/closing fences (`\mathopen/\mathclose`) or not.2012-11-17
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    @kahen is there a primer for using MathJax for here?2012-11-17
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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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    What is AM-GM inequality?2012-11-17
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    To Gerry Myerson: typo corrected. @MaoYiyi: AM-GM is the inequality between the arithmetic and the geometric mean - http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means.2012-11-17
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    What is the justification for replacing $g(1-g)$ with $1/4$?2012-11-17
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    Again: AM-GM. If two real numbers sum to one, their product is at most $1/4$, and it is exactly one fourth only when the two numbers are equal to $1/2$.2012-11-17
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    Is that because of the Riemann sum. sorry that I am not getting this.2012-11-17
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    Assume that $g(x)$ is different from $1/2$ for some $x$. Since $g$ is continous, $g(1-g)$ is strictly less than $1/4$ in an open neighbourhood of $x$, then $\int_{0}^{1}x^3 g(1-g)dx$ is stricly less than $1/16$.2012-11-17
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    Could you name a good book to learn this in? I have to study this more.2012-11-17