Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove: $$ \Big|\int_1^\sqrt{3} \! \frac{\sin(x)}{e^x(x^2 +1)} \, \mathrm{d} x \Big| \leq \frac{\pi}{12e}$$

Approach: we know:

$$ \Big|\int_1^\sqrt{3} \! \frac{\sin(x)}{e^x(x^2 +1)} \, \mathrm{d} x \Big| \leq \int_1^\sqrt{3} \! \frac{|\sin(x)|}{e^x(x^2 +1)} \, \mathrm{d} x \leq \int_1^\sqrt{3} \! \frac{1}{e^x(x^2 +1)} \, \mathrm{d} x$$ Since $|\sin(x)|$ is bounded by 1 for every $x$. Can we keep estimating this integral, or should we just solve the last integral?

thanks.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Use the fact $ 1<x<\sqrt{3} \implies e<e^x<e^{\sqrt{3}}$ and $|sin(x)|\leq 1$

$$\Big|\int_1^\sqrt{3} \! \frac{\sin(x)}{e^x(x^2 +1)} \, \mathrm{d} x \Big|\leq \int_1^\sqrt{3} \! \frac{1}{e(x^2 +1)} \, \mathrm{d} x \leq \frac{\pi}{12e}$$

share|improve this answer
    
why? can you explain? –  Chasky Oct 16 '12 at 3:13
    
@LJym89: Can you see it now? –  Mhenni Benghorbal Oct 16 '12 at 3:15
    
GOT IT! thanks a lot –  Chasky Oct 16 '12 at 3:16
    
@LJym89:You are welcome. –  Mhenni Benghorbal Oct 16 '12 at 3:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.