## anonymous 3 years ago i need some help here my friends :) $\left| \int_{1}^{\sqrt{3}} \frac{e^{-x} \sin x}{1+x^2} dx \right| \le \frac{\pi}{2 e}$

1. anonymous

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

2. anonymous

i think that will not work if im not wrong...

3. anonymous

@satellite73 i tried it but no good results.

4. anonymous

first we use theorem if a<b$\left| \int\limits _a^b f(x)dx \right|\le \int\limits_a^b \left| f(x) dx\right|$ $\left| \int\limits_1^\sqrt{3} \frac{e^{-x}\sin x}{1+x^2}dx \right|\le \int\limits_1^\sqrt{3} \left|\frac{e^{-x}\sin x}{1+x^2}dx \right|$ $\le \int\limits_1^\sqrt{3}\frac{1}{e(1+x^2)}=\frac{1}{e}\tan^{-1}x |_1^\sqrt{3}=\frac{1}{e}(\tan^{-1}\sqrt{3}-\tan^{-1}1)=\frac{1}{e}(\pi/3-\pi/4)=\frac{\pi}{12e}$ $\square$

5. anonymous

$=\frac{\pi}{12e}$

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