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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About Bunchhieng

I like math, programming and web design!!
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