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For $\displaystyle f(x)=\int_x^{x+1}\sin (\text{e}^t)\text{d}t$

Prove that : $\displaystyle \text{e}^x\left|f(x)\right|\le 2$

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up vote 4 down vote accepted

$\displaystyle f(x)=\int_x^{x+1}\sin (\text{e}^t)\text{d}t\implies \displaystyle e^xf(x)=\int_x^{x+1}e^x\sin (\text{e}^t)\text{d}t \leq \int_x^{x+1}e^t\sin (\text{e}^t)\text{d}t $ (because $x\leq t\leq x+1)$

Now substitute $z=e^t$ which gives

$\displaystyle e^xf(x)\leq\int_{e^x}^{e^{x+1}}\sin (\text{z})\text{d}z=\cos(e^{x})-\cos(e^{x+1})\implies e^x|f(x)|\leq |\cos(e^{x})|+|\cos(e^{x+1})|\leq 2$

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I can't understand last inequality $$\int_{e^x}^{e^{x+1}}|\sin z |dz\le |\cos e^x|+|\cos e^{x+1}|.$$ I think if $x$ is arbitary large, this inequalty fails. –  tetori Jan 22 '13 at 7:40
    
@tetori: i am done with editing, have a look –  Aang Jan 22 '13 at 7:52
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Seems to be correct. –  tetori Jan 22 '13 at 7:54
    
Thx, but can we put e^x straight into integral sign for the first line? –  gauss115 Jan 22 '13 at 7:58
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For that integral , $x$ is just a fixed value as it is appearing in the limits , so there is no problem in just taking the $e^x$ inside the integral –  Aang Jan 22 '13 at 8:01
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