Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 106285 by bemath last updated on 04/Aug/20

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}?$$

Answered by bobhans last updated on 04/Aug/20

$$\mathrm{recall}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}=\sqrt{\pi }$$$$\iff {\int }_{-\mathrm{\infty }}^0{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}=\sqrt{\pi }$$$$\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{is}\mathrm{even}\mathrm{function}$$$$\mathrm{so}\underset{-\mathrm{\infty }}{\stackrel{0}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{finally}\mathrm{we}\mathrm{get}\Rightarrow 2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}=\sqrt{\pi }$$$$\therefore \underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}=\frac{\sqrt{\pi }}{2}.★$$

Answered by MAB last updated on 04/Aug/20

$$I=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-{\mathrm{x}}^2}\mathrm{dx}$$$$I^2=\left({\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx\right)\left({\int }_0^{\mathrm{\infty }}{e}^{-y^2}dy\right)$$$$I^2={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(x^2+y^2)}dxdy$$$$letx=rcos\left(\theta \right)andy=rsin\left(\theta \right)$$$$wherer⩾0and0⩽\theta ⩽\pi /2$$$$dxdy=rdrd\theta$$$$I^2={\int }_0^{\pi /2}{\int }_0^{\mathrm{\infty }}{re}^{-r^2}drd\theta$$$$I^2={\int }_0^{\pi /2}\frac{1}{2}{\left[e^{-r^2}\right]}_0^{\mathrm{\infty }}d\theta$$$$I^2=\frac{\pi }{4}$$$$henceI=\frac{\sqrt{\pi }}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com