What is $F(x) = int x-xe^(-2x) dx$ if $F(0) = 1$ ?

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Answer 1 · 2017-02-27T13:23:44

$F(x)=x^2/2+(xe^(-2x))/2+e^(-2x)/4+3/4.$

We have, $F(x)=int(x-xe^(-2x))dx=intxdx-intxe^(-2x)dx.$

$:. F(x)=x^2/2-I, where, I=intxe^(-2x)dx.$

To evaluate $I,$ we need the following Rule of Integration by Parts (IBP) :

$"IBP : "intuvdx=uintvdx-int{((du)/dx)(intvdx)}dx.$

With, $u=x, &, v=e^(-2x) rArr (du)/dx=1" & "intvdx=e^(-2x)/-2;$

$I=(-x/2)e^(-2x)-int{(1)(e^(-2x)/-2)}dx,$

$=-(xe^(-2x))/2+1/2(e^(-2x)/-2),$

$:. F(x)=x^2/2+(xe^(-2x))/2+e^(-2x)/4+C.$

To determine $C,$ we use the cond. $: F(0)=1,$

$rArr 0+0+1/4+C=1 rArr C=1-1/4=3/4.$

Finally, we get,,

$F(x)=x^2/2+(xe^(-2x))/2+e^(-2x)/4+3/4.$

Enjoy Maths.!

What is F(x) = int x-xe^(-2x) dxF(x) = int x-xe^(-2x) dx if F(0) = 1 F(0) = 1 ?