## 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 😀

## IMS 2010 Problem 1

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

Solution

since $0, 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 😀

## 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 😀