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mukushla
Group Title
i need some help here my friends :)
\[\left \int_{1}^{\sqrt{3}} \frac{e^{x} \sin x}{1+x^2} dx \right \le \frac{\pi}{2 e}\]
 one year ago
 one year ago
mukushla Group Title
i need some help here my friends :) \[\left \int_{1}^{\sqrt{3}} \frac{e^{x} \sin x}{1+x^2} dx \right \le \frac{\pi}{2 e}\]
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
yikes did you try the simple method of locating the max on the interval? i am just asking, i didn't go it, or not sure this will work
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
i think that will not work if im not wrong...
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 i tried it but no good results.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
first we use theorem if a<b\[\left \int\limits _a^b f(x)dx \right\le \int\limits_a^b \left f(x) dx\right\] \[\left \int\limits_1^\sqrt{3} \frac{e^{x}\sin x}{1+x^2}dx \right\le \int\limits_1^\sqrt{3} \left\frac{e^{x}\sin x}{1+x^2}dx \right\] \[\le \int\limits_1^\sqrt{3}\frac{1}{e(1+x^2)}=\frac{1}{e}\tan^{1}x _1^\sqrt{3}=\frac{1}{e}(\tan^{1}\sqrt{3}\tan^{1}1)=\frac{1}{e}(\pi/3\pi/4)=\frac{\pi}{12e}\] \[\square\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.1
\[=\frac{\pi}{12e}\]
 one year ago
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