Menu Close

prove-that-e-x-dx-e-x-c-




Question Number 66245 by aliesam last updated on 11/Aug/19
prove that    ∫e^x  dx = e^x  + c
provethatexdx=ex+c
Commented by Rio Michael last updated on 11/Aug/19
let  y = e^x        (dy/dx) = e^x   ∫(dy/dx) = ∫e^x dx   ⇒ y = e^x  + c
lety=exdydx=exdydx=exdxy=ex+c
Commented by mathmax by abdo last updated on 11/Aug/19
∫ e^x dx =∫ (Σ_(n=0) ^∞  (x^n /(n!)))dx =Σ_(n=0) ^∞  (1/(n!))∫ x^n dx  =Σ_(n=0) ^∞  (1/(n!))(1/(n+1))x^(n+1)  +c =Σ_(n=0) ^∞   (x^(n+1) /((n+1)!)) +c  =Σ_(n=1) ^∞  (x^n /(n!)) +c =Σ_(n=0) ^∞  (x^n /(n!)) +c−1 =e^x  +C      (C=c−1)
exdx=(n=0xnn!)dx=n=01n!xndx=n=01n!1n+1xn+1+c=n=0xn+1(n+1)!+c=n=1xnn!+c=n=0xnn!+c1=ex+C(C=c1)
Answered by mr W last updated on 12/Aug/19
since ((d(e^x +c))/dx)=((d(e^x ))/dx)+((d(c))/dx)=e^x +0=e^x   ⇒∫e^x dx=e^x +c
sinced(ex+c)dx=d(ex)dx+d(c)dx=ex+0=exexdx=ex+c

Leave a Reply

Your email address will not be published. Required fields are marked *