What is F(x) = int x-xe^(-2x) dxF(x)=xxe2xdx if F(0) = 1 F(0)=1?

1 Answer
Feb 27, 2017

F(x)=x^2/2+(xe^(-2x))/2+e^(-2x)/4+3/4.F(x)=x22+xe2x2+e2x4+34.

Explanation:

We have, F(x)=int(x-xe^(-2x))dx=intxdx-intxe^(-2x)dx.F(x)=(xxe2x)dx=xdxxe2xdx.

:. 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.!