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e-2x-2-5x-3-dx-help-me-




Question Number 99541 by student work last updated on 21/Jun/20
∫_(−∞) ^∞ e^(−2x^2 −5x−3) dx=?   help me
e2x25x3dx=?helpme
Answered by smridha last updated on 21/Jun/20
e^(−3) ∫_(−∞) ^(+∞) e^(−2x^2 +(−5)x) dx  =e^(−3) .(√(𝛑/2)).e^((25)/8) =(√(𝛑/2)).e^(1/8)   [using this ∫_(−∞) ^(+∞) e^(−𝛂x^2 +𝛃x) dx=(√(𝛑/𝛂)).e^(𝛃^2 /(4𝛂))
e3+e2x2+(5)xdx=e3.π2.e258=π2.e18[usingthis+eαx2+βxdx=πα.eβ24α
Commented by student work last updated on 21/Jun/20
thanks sir good
thankssirgood
Commented by smridha last updated on 21/Jun/20
welcome
Answered by aleks041103 last updated on 21/Jun/20
I=∫_(−∞) ^∞ e^(−2x^2 −5x−3) dx=  =∫_(−∞) ^∞ e^(−(2x^2 +5x+3)) dx  We complete the square:  2x^2 +5x+3=  =((√2)x)^2 +2((√2)x)(5/(2(√2)))+((5/(2(√2))))^2 −((5/(2(√2))))^2 +3=  =((√2)x+(5/(2(√2))))^2 +3−((25)/8)=  =((√2)x+(5/(2(√2))))^2 −(1/8)  Let  u=(√2)x+(5/(2(√2))) ⇒du=(√2) dx ⇒ dx=(du/( (√2)))  Also   x→+∞ ⇒ u→+∞  x→−∞ ⇒ u→−∞  Then  I=∫_(−∞) ^(+∞) e^(−(u^2 −1/8)) (du/( (√2)))=(e^(1/8) /( (√2)))∫_(−∞) ^(+∞) e^(−u^2 ) du  But ∫_(−∞) ^(+∞) e^(−u^2 ) du=(√π) is the gaussian  integral. Then for I we have:  I=e^(1/8) (√(π/2))  Or:  I=∫_(−∞) ^∞ e^(−2x^2 −5x−3) dx=e^(1/8) (√(π/2))
I=e2x25x3dx==e(2x2+5x+3)dxWecompletethesquare:2x2+5x+3==(2x)2+2(2x)522+(522)2(522)2+3==(2x+522)2+3258==(2x+522)218Letu=2x+522du=2dxdx=du2Alsox+u+xuThenI=+e(u21/8)du2=e1/82+eu2duBut+eu2du=πisthegaussianintegral.ThenforIwehave:I=e1/8π2Or:I=e2x25x3dx=e1/8π2
Commented by student work last updated on 21/Jun/20
thanks sir
thankssir
Answered by mathmax by abdo last updated on 21/Jun/20
I =∫_(−∞) ^(+∞)  e^(−2x^2 −5x−3) dx ⇒ I =∫_(−∞) ^(+∞)  e^(−2{ x^2 +(5/2)x+(3/2)})  dx  =∫_(−∞) ^(+∞)  e^(−2{ x^2  +2.(5/4)x +((25)/(16))+(3/2)−((25)/(16))}) dx =∫_(−∞) ^(+∞ ) e^(−2{(x+(5/4))^2  −(1/(16))}) dx  (x+(5/4))(√2)=u  =e^(1/8)  ∫_(−∞) ^(+∞)   e^(−u^2 )  (du/( (√2))) =(e^(1/8) /( (√2)))×(√π) ⇒I =^8 (√e)×(√(π/2))
I=+e2x25x3dxI=+e2{x2+52x+32}dx=+e2{x2+2.54x+2516+322516}dx=+e2{(x+54)2116}dx(x+54)2=u=e18+eu2du2=e182×πI=8e×π2

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