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

Question

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

Charles Anctil · Accepted Answer

#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.!

Charles Arocha · Answer

undefined To find the integral of $ F(x) = \int (x - xe^{-2x}) \, dx $ given that $ F(0) = 1 $, you can integrate the function $ x - xe^{-2x} $ with respect to $ x $ and then use the initial condition $ F(0) = 1 $ to solve for the constant of integration.

Integrating $ x - xe^{-2x} $ with respect to $ x $ yields:

$$ F(x) = \int (x - xe^{-2x}) \, dx = \frac{x^2}{2} + \frac{1}{4} e^{-2x} + C $$

Now, apply the initial condition $ F(0) = 1 $ to find the constant of integration $ C $:

$$ 1 = F(0) = \frac{0^2}{2} + \frac{1}{4} e^{-2 \cdot 0} + C $$
$$ 1 = 0 + \frac{1}{4} + C $$
$$ C = 1 - \frac{1}{4} = \frac{3}{4} $$

Thus, the function $ F(x) $ is:

$$ F(x) = \frac{x^2}{2} + \frac{1}{4} e^{-2x} + \frac{3}{4} $$