Question

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

Answer 1

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

Explanation:

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.!