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mukushla

  • 3 years ago

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

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  1. anonymous
    • 3 years ago
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    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. mukushla
    • 3 years ago
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    i think that will not work if im not wrong...

  3. mukushla
    • 3 years ago
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    @satellite73 i tried it but no good results.

  4. Jonask
    • 3 years ago
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    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. Jonask
    • 3 years ago
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    \[=\frac{\pi}{12e}\]

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