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Question Number 30442 by abdo imad last updated on 22/Feb/18
prove that   (1/e) ≤  ∫_0 ^1  e^(−(x−[x])^2 )  dx≤1.
$${prove}\:{that}\:\:\:\frac{\mathrm{1}}{{e}}\:\leqslant\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:{dx}\leqslant\mathrm{1}. \\ $$
Commented by alex041103 last updated on 22/Feb/18
You can see that if x∈[0,1), then  [x]=0 ⇒ x−[x]=x  ⇒ ∫_0 ^1  e^(−(x−[x])^2 )  dx= ∫_0 ^1  e^(−x^2 )  dx
$${You}\:{can}\:{see}\:{that}\:{if}\:{x}\in\left[\mathrm{0},\mathrm{1}\right),\:{then} \\ $$$$\left[{x}\right]=\mathrm{0}\:\Rightarrow\:{x}−\left[{x}\right]={x} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } \:{dx} \\ $$
Commented by abdo imad last updated on 24/Feb/18
we have [x]≤x<[x]+1  ⇒0≤x −[x]<1⇒  0≤(x−[x])^2  <1 ⇒ −1 <−(x−[x])^2 ≤0 ⇒    e^(−1) <  e^(−(x−[x])^2 )  ≤1  ⇒ (1/e) ≤ ∫_0 ^1  e^(−(x−[x])^2 ) dx ≤ 1 .
$${we}\:{have}\:\left[{x}\right]\leqslant{x}<\left[{x}\right]+\mathrm{1}\:\:\Rightarrow\mathrm{0}\leqslant{x}\:−\left[{x}\right]<\mathrm{1}\Rightarrow \\ $$$$\mathrm{0}\leqslant\left({x}−\left[{x}\right]\right)^{\mathrm{2}} \:<\mathrm{1}\:\Rightarrow\:−\mathrm{1}\:<−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} \leqslant\mathrm{0}\:\Rightarrow \\ $$$$\:\:{e}^{−\mathrm{1}} <\:\:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:\leqslant\mathrm{1}\:\:\Rightarrow\:\frac{\mathrm{1}}{{e}}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } {dx}\:\leqslant\:\mathrm{1}\:. \\ $$
Answered by alex041103 last updated on 22/Feb/18
Commented by alex041103 last updated on 22/Feb/18
The orange function is e^(−x^2 )   The blue area is the integral.  Because of that integrals represent  the area between x^→  and the fuction,  we see that  ∫_0 ^1 e^(−x^2 ) dx≤ red area=1^2 =1  ⇒∫_0 ^1 e^(−x^2 ) dx≤1
$${The}\:{orange}\:{function}\:{is}\:{e}^{−{x}^{\mathrm{2}} } \\ $$$${The}\:{blue}\:{area}\:{is}\:{the}\:{integral}. \\ $$$${Because}\:{of}\:{that}\:{integrals}\:{represent} \\ $$$${the}\:{area}\:{between}\:\overset{\rightarrow} {{x}}\:{and}\:{the}\:{fuction}, \\ $$$${we}\:{see}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\leqslant\:{red}\:{area}=\mathrm{1}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\leqslant\mathrm{1} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 23/Feb/18
r u writing ans. in flight , as aeroplane  mode is on ?
$$\mathrm{r}\:\mathrm{u}\:\mathrm{writing}\:\mathrm{ans}.\:\mathrm{in}\:\mathrm{flight}\:,\:\mathrm{as}\:\mathrm{aeroplane} \\ $$$$\mathrm{mode}\:\mathrm{is}\:\mathrm{on}\:? \\ $$
Commented by alex041103 last updated on 24/Feb/18
No. I was abroad, so there isn′t any  point in using my battery for mobile  service.
$${No}.\:{I}\:{was}\:{abroad},\:{so}\:{there}\:{isn}'{t}\:{any} \\ $$$${point}\:{in}\:{using}\:{my}\:{battery}\:{for}\:{mobile} \\ $$$${service}. \\ $$
Commented by rahul 19 last updated on 24/Feb/18
hmmm, fine.
$$\mathrm{hmmm},\:\mathrm{fine}. \\ $$
Answered by alex041103 last updated on 22/Feb/18
Commented by alex041103 last updated on 22/Feb/18
By following the same logic we can see  that:  ∫_0 ^1 e^(−x^2 ) dx≥ green−ish area=1×e^(−(1)^2 ) =(1/e)  ⇒∫_0 ^1 e^(−x^2 ) dx≥(1/e)
$${By}\:{following}\:{the}\:{same}\:{logic}\:{we}\:{can}\:{see} \\ $$$${that}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\geqslant\:{green}−{ish}\:{area}=\mathrm{1}×{e}^{−\left(\mathrm{1}\right)^{\mathrm{2}} } =\frac{\mathrm{1}}{{e}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\geqslant\frac{\mathrm{1}}{{e}} \\ $$
Commented by abdo imad last updated on 23/Feb/18
your answer is correct sir by using graph thanks sir...
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{by}\:{using}\:{graph}\:{thanks}\:{sir}… \\ $$
Answered by alex041103 last updated on 22/Feb/18
Or as a conclusion:  (1/e) ≤  ∫_0 ^1  e^(−(x−[x])^2 )  dx≤1.  Questions?
$${Or}\:{as}\:{a}\:{conclusion}: \\ $$$$\frac{\mathrm{1}}{{e}}\:\leqslant\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:{dx}\leqslant\mathrm{1}. \\ $$$${Questions}? \\ $$

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