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.!