Question Number 30442 by abdo imad last updated on 22/Feb/18
![prove that (1/e) ≤ ∫_0 ^1 e^(−(x−[x])^2 ) dx≤1.](https://www.tinkutara.com/question/Q30442.png)
$${prove}\:{that}\:\:\:\frac{\mathrm{1}}{{e}}\:\leqslant\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:{dx}\leqslant\mathrm{1}. \\ $$
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](https://www.tinkutara.com/question/Q30470.png)
$${You}\:{can}\:{see}\:{that}\:{if}\:{x}\in\left[\mathrm{0},\mathrm{1}\right),\:{then} \\ $$$$\left[{x}\right]=\mathrm{0}\:\Rightarrow\:{x}−\left[{x}\right]={x} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } \:{dx} \\ $$
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 .](https://www.tinkutara.com/question/Q30662.png)
$${we}\:{have}\:\left[{x}\right]\leqslant{x}<\left[{x}\right]+\mathrm{1}\:\:\Rightarrow\mathrm{0}\leqslant{x}\:−\left[{x}\right]<\mathrm{1}\Rightarrow \\ $$$$\mathrm{0}\leqslant\left({x}−\left[{x}\right]\right)^{\mathrm{2}} \:<\mathrm{1}\:\Rightarrow\:−\mathrm{1}\:<−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} \leqslant\mathrm{0}\:\Rightarrow \\ $$$$\:\:{e}^{−\mathrm{1}} <\:\:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:\leqslant\mathrm{1}\:\:\Rightarrow\:\frac{\mathrm{1}}{{e}}\:\leqslant\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } {dx}\:\leqslant\:\mathrm{1}\:. \\ $$
Answered by alex041103 last updated on 22/Feb/18
Commented by alex041103 last updated on 22/Feb/18
$${The}\:{orange}\:{function}\:{is}\:{e}^{−{x}^{\mathrm{2}} } \\ $$$${The}\:{blue}\:{area}\:{is}\:{the}\:{integral}. \\ $$$${Because}\:{of}\:{that}\:{integrals}\:{represent} \\ $$$${the}\:{area}\:{between}\:\overset{\rightarrow} {{x}}\:{and}\:{the}\:{fuction}, \\ $$$${we}\:{see}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\leqslant\:{red}\:{area}=\mathrm{1}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\leqslant\mathrm{1} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 23/Feb/18
$$\mathrm{r}\:\mathrm{u}\:\mathrm{writing}\:\mathrm{ans}.\:\mathrm{in}\:\mathrm{flight}\:,\:\mathrm{as}\:\mathrm{aeroplane} \\ $$$$\mathrm{mode}\:\mathrm{is}\:\mathrm{on}\:? \\ $$
Commented by alex041103 last updated on 24/Feb/18
$${No}.\:{I}\:{was}\:{abroad},\:{so}\:{there}\:{isn}'{t}\:{any} \\ $$$${point}\:{in}\:{using}\:{my}\:{battery}\:{for}\:{mobile} \\ $$$${service}. \\ $$
Commented by rahul 19 last updated on 24/Feb/18
$$\mathrm{hmmm},\:\mathrm{fine}. \\ $$
Answered by alex041103 last updated on 22/Feb/18
Commented by alex041103 last updated on 22/Feb/18
$${By}\:{following}\:{the}\:{same}\:{logic}\:{we}\:{can}\:{see} \\ $$$${that}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\geqslant\:{green}−{ish}\:{area}=\mathrm{1}×{e}^{−\left(\mathrm{1}\right)^{\mathrm{2}} } =\frac{\mathrm{1}}{{e}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}\geqslant\frac{\mathrm{1}}{{e}} \\ $$
Commented by abdo imad last updated on 23/Feb/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{by}\:{using}\:{graph}\:{thanks}\:{sir}… \\ $$
Answered by alex041103 last updated on 22/Feb/18
![Or as a conclusion: (1/e) ≤ ∫_0 ^1 e^(−(x−[x])^2 ) dx≤1. Questions?](https://www.tinkutara.com/question/Q30473.png)
$${Or}\:{as}\:{a}\:{conclusion}: \\ $$$$\frac{\mathrm{1}}{{e}}\:\leqslant\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−\left({x}−\left[{x}\right]\right)^{\mathrm{2}} } \:{dx}\leqslant\mathrm{1}. \\ $$$${Questions}? \\ $$