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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

limx11xet2(dt)x21(a)1(b)0(c)e/2(d)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(x)=1xet2dtf(x)=ex2limx11x211xet2dt=limx1(1xet2dtx1)(1x+1)=f(1)×12=e2

Commented by FilupS last updated on 27/Feb/17

incorrect

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.

Godblessyousir.

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)

limx11xet2x21dtlimx11x211xet2dtex2dx=12πerf(x)limx11x21×12π(erf(x)erf(1))π2limx1erf(x)erf(1)x21=π2(00)LHopitalsLawπ2limx1ddx(erf(x)erf(1))ddx(x21)erf(1)=constanterf(x)=2πex2dxπ2limx1(2πex202x0)π2×2πlimx1(ex22x)e12limx11xet2x21dt=e2

Commented by Saham last updated on 27/Feb/17

God bless you sir.

Godblessyousir.

Commented by bahmanfeshki last updated on 27/Feb/17

my answer is edited...

myanswerisedited...

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