UC Calculus contest 2010 problem 4


Find \int{\frac{dx}{{{x}^{2}}\sqrt{1-{{x}^{2}}}}}?

Solution

setting x=\sin u, we get

\int{\frac{dx}{{{x}^{2}}\sqrt{1-{{x}^{2}}}}}=\int{\frac{d\sin u}{{{\sin }^{2}}u\sqrt{1-{{\sin }^{2}}u}}=\int{\frac{\cos udu}{{{\sin }^{2}}u\cos u}}}=\int{{{\csc }^{2}}udu=-\cot u+c}

therefore \int{\frac{dx}{{{x}^{2}}\sqrt{1-{{x}^{2}}}}}=-\frac{\sqrt{1-{{x}^{2}}}}{x}+c

Using the triangle below we get that -\cot u=-\frac{adjacent}{opposite}=-\frac{\sqrt{1-{{x}^{2}}}}{x}

Thank you 😀

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IMS 2010 Problem 1


Let 0<a<b. Prove that \int\limits_{a}^{b}{({{x}^{2}}+1){{e}^{-{{x}^{2}}}}dx\ge {{e}^{-{{a}^{2}}}}-{{e}^{-{{b}^{2}}}}}

Solution

since 0<a<b, we get ({{x}^{2}}+1){{e}^{-{{x}^{2}}}}\ge 2x{{e}^{-{{x}^{2}}}}(AM-GM)

then \int\limits_{a}^{b}{({{x}^{2}}+1){{e}^{-{{x}^{2}}}}\ge \int\limits_{a}^{b}{2x{{e}^{-{{x}^{2}}}}dx}}

Integral by part for \int\limits_{a}^{b}{2x{{e}^{-{{x}^{2}}}}dx}=-{{e}^{-{{x}^{2}}}}+c

therefore \int\limits_{a}^{b}{({{x}^{2}}+1){{e}^{-{{x}^{2}}}}\ge \int\limits_{a}^{b}{2x{{e}^{-{{x}^{2}}}}dx}}={{e}^{-{{a}^{2}}}}-{{e}^{-{{b}^{2}}}}

Thank you 😀

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The first welcome problem


Problem 1: let f : \left[ 0,1 \right]\to R be a continuous differentiable function, such that \int\limits_{0}^{1}{{{({{f}^{'}}(x))}^{2}}dx\le 2\int\limits_{0}^{1}{f(x)dx}} find f if f(1)=-\frac{1}{6} 😀 Thank you!!

Solution

We know that 0\le \int\limits_{0}^{1}{{{({{f}^{'}}(x)+x)}^{2}}dx=\int\limits_{0}^{1}{({{({{f}^{'}}(x))}^{2}}dx}}+2\int\limits_{0}^{1}{x{{f}^{'}}(x)dx}+\frac{1}{3}

so 0\le \int\limits_{0}^{1}{{{({{f}^{'}}(x)+x)}^{2}}dx=\int\limits_{0}^{1}{({{({{f}^{'}}(x))}^{2}}dx}}+2xf(x)|_{0}^{1}-2\int\limits_{0}^{1}{f(x)dx}+\frac{1}{3}

then 0\le \int\limits_{0}^{1}{{{({{f}^{'}}(x))}^{2}}dx}-2\int\limits_{0}^{1}{f(x)dx}

From this we get: \int\limits_{0}^{1}{{{({{f}^{'}}(x))}^{2}}dx}-2\int\limits_{0}^{1}{f(x)dx}=0

So \int\limits_{0}^{1}{{{({{f}^{'}}(x)+x)}^{2}}dx=0}

The continuity of {{f}^{'}} implies {{f}^{'}}(x)=-x. That is, f(x)=-\frac{{{x}^{2}}}{2}+a

for all x\in \left[ 0,1 \right]. From f(1)=-\frac{1}{6} we obtain a=\frac{1}{3}, that is f(x)=-\frac{{{x}^{2}}}{2}+\frac{1}{3}

Thank you 😀

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