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Question Number 10829 by Saham last updated on 26/Feb/17

lim_(x→1) ∫_( 1) ^( x) ((e^t^2 (dt))/(x^2 − 1)) (a) 1 (b) 0 (c) e/2 (d) e

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\int_{\:\mathrm{1}} ^{\:\mathrm{x}} \:\:\:\frac{\mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \:\left(\mathrm{dt}\right)}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{1}}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{1}\:\left(\mathrm{b}\right)\:\mathrm{0}\:\left(\mathrm{c}\right)\:\mathrm{e}/\mathrm{2}\:\left(\mathrm{d}\right)\:\mathrm{e} \\ $$

Answered by bahmanfeshki last updated on 27/Feb/17

f(x)=∫_1 ^x e^t^2 dt⇒f′(x)=e^x^2 ⇒lim_(x→1) (1/(x^2 −1))∫_1 ^x e^t^2 dt=lim_(x→1) (((∫_1 ^x e^t^2 dt)/(x−1)))((1/(x+1)))=f′(1)×(1/2)=(e/2)

$${f}\left({x}\right)=\int_{\mathrm{1}} ^{{x}} {e}^{{t}^{\mathrm{2}} } \:{dt}\Rightarrow{f}'\left({x}\right)={e}^{{x}^{\mathrm{2}} } \Rightarrow\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{1}} ^{{x}} {e}^{{t}^{\mathrm{2}} } \:{dt}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\int_{\mathrm{1}} ^{{x}} {e}^{{t}^{\mathrm{2}} } \:{dt}}{{x}−\mathrm{1}}\right)\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)={f}'\left(\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{2}}=\frac{{e}}{\mathrm{2}} \\ $$

Commented by FilupS last updated on 27/Feb/17

incorrect

$$\mathrm{incorrect} \\ $$

Commented by bahmanfeshki last updated on 27/Feb/17

edited

$${edited} \\ $$

Commented by Saham last updated on 27/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by FilupS last updated on 27/Feb/17

lim_(x→1) ∫_1 ^( x) (e^t^2 /(x^2 −1))dt lim_(x→1) (1/(x^2 −1))∫_1 ^( x) e^t^2 dt ∫e^x^2 dx=(1/2)(√π)erf(x) lim_(x→1) (1/(x^2 −1))×(1/2)(√π)(erf(x)−erf(1)) ((√π)/2)lim_(x→1) ((erf(x)−erf(1))/(x^2 −1))=((√π)/2)((0/0)) L′Hopital′s Law ((√π)/2)lim_(x→1) (((d/dx)(erf(x)−erf(1)))/((d/dx)(x^2 −1))) erf(1)=constant ⇒erf(x)=(2/(√π))∫e^x^2 dx ((√π)/2)lim_(x→1) ((((2/(√π))e^x^2 −0)/(2x−0))) ((√π)/2)×(2/(√π))lim_(x→1) ((e^x^2 /(2x))) (e^1 /2) ∴ lim_(x→1) ∫_1 ^( x) (e^t^2 /(x^2 −1))dt = (e/2)

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\int_{\mathrm{1}} ^{\:{x}} \frac{{e}^{{t}^{\mathrm{2}} } }{{x}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{1}} ^{\:{x}} {e}^{{t}^{\mathrm{2}} } {dt} \\ $$$$\int{e}^{{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\mathrm{erf}\left({x}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\left(\mathrm{erf}\left({x}\right)−\mathrm{erf}\left(\mathrm{1}\right)\right) \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{erf}\left({x}\right)−\mathrm{erf}\left(\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{\mathrm{0}}{\mathrm{0}}\right) \\ $$$$\mathrm{L}'\mathrm{Hopital}'\mathrm{s}\:\mathrm{Law} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\frac{{d}}{{dx}}\left(\mathrm{erf}\left({x}\right)−\mathrm{erf}\left(\mathrm{1}\right)\right)}{\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\mathrm{erf}\left(\mathrm{1}\right)=\mathrm{constant} \\ $$$$\Rightarrow\mathrm{erf}\left({x}\right)=\frac{\mathrm{2}}{\sqrt{\pi}}\int{e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{\frac{\mathrm{2}}{\sqrt{\pi}}{e}^{{x}^{\mathrm{2}} } −\mathrm{0}}{\mathrm{2}{x}−\mathrm{0}}\right) \\ $$$$\frac{\sqrt{\pi}}{\mathrm{2}}×\frac{\mathrm{2}}{\sqrt{\pi}}\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\frac{{e}^{{x}^{\mathrm{2}} } }{\mathrm{2}{x}}\right) \\ $$$$\frac{{e}^{\mathrm{1}} }{\mathrm{2}} \\ $$$$\: \\ $$$$\therefore\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\int_{\mathrm{1}} ^{\:{x}} \frac{{e}^{{t}^{\mathrm{2}} } }{{x}^{\mathrm{2}} −\mathrm{1}}{dt}\:=\:\frac{{e}}{\mathrm{2}} \\ $$

Commented by Saham last updated on 27/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by bahmanfeshki last updated on 27/Feb/17

my answer is edited...

$${my}\:{answer}\:{is}\:{edited}... \\ $$

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