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Question Number 37281 by abdo.msup.com last updated on 11/Jun/18

find a better approximation for the  integrals   1) ∫_0 ^1   e^(−x^2 ) dx  2) ∫_1 ^(+∞)  e^(−x^2 ) dx .

$${find}\:{a}\:{better}\:{approximation}\:{for}\:{the} \\ $$$${integrals}\: \\ $$$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$

Commented by prof Abdo imad last updated on 18/Jun/18

we have  e^(−x^2 )    =Σ_(n=0) ^∞  (((−1)^n  x^(2n) )/(n!)) ⇒  ∫_0 ^1  e^(−x^2 ) dx = Σ_(n=0) ^∞  (((−1)^n )/(n!)) ∫_0 ^1  x^(2n) dx  =Σ_(n=0) ^∞    (((−1)^n )/(n!(2n+1))) let take 8terms we get  ∫_0 ^1  e^(−x^2 ) dx ∼ 1 −(1/3) + (1/(5.2!))  −(1/(7.3!)) +(1/(9.4!))  −(1/(11.5!))  +(1/(13.6!)) −(1/(15.7!)) +(1/(17.8!))  ⇒...  2) ∫_0 ^(+∞)  e^(−x^2 ) dx =((√π)/2) = ∫_0 ^1  e^(−x^2 ) dx +∫_1 ^(+∞)  e^(−x^2 ) dx  ⇒ ∫_1 ^(+∞)   e^(−x^2 ) dx =((√π)/2) −∫_0 ^1  e^(−x^2 ) dx and we use  approximate value from Q1...

$${we}\:{have}\:\:{e}^{−{x}^{\mathrm{2}} } \:\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} }{{n}!}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{let}\:{take}\:\mathrm{8}{terms}\:{we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:\sim\:\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}.\mathrm{2}!}\:\:−\frac{\mathrm{1}}{\mathrm{7}.\mathrm{3}!}\:+\frac{\mathrm{1}}{\mathrm{9}.\mathrm{4}!} \\ $$$$−\frac{\mathrm{1}}{\mathrm{11}.\mathrm{5}!}\:\:+\frac{\mathrm{1}}{\mathrm{13}.\mathrm{6}!}\:−\frac{\mathrm{1}}{\mathrm{15}.\mathrm{7}!}\:+\frac{\mathrm{1}}{\mathrm{17}.\mathrm{8}!}\:\:\Rightarrow... \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:+\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\Rightarrow\:\int_{\mathrm{1}} ^{+\infty} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:{and}\:{we}\:{use} \\ $$$${approximate}\:{value}\:{from}\:{Q}\mathrm{1}... \\ $$

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