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Question Number 64804 by mmkkmm000m last updated on 21/Jul/19

∫_0 ^(+∞) e^(−x^2 ) dx

$$\overset{+\infty} {\int}_{\mathrm{0}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Answered by Tanmay chaudhury last updated on 21/Jul/19

x=t^(1/2)   →dx=(t^(−(1/2)) /2)dt  ∫_0 ^∞ e^(−t) ×(t^((1/2)−1) /2)dt  (1/2)∫_0 ^∞ e^(−t) ×t^((1/2)−1) dt  (1/2)×⌈((1/2))=((√π)/2)

$${x}={t}^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\rightarrow{dx}=\frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}{dt} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×\frac{{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\mathrm{2}}{dt} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

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