Question

What is `f(x) = int xe^x-xdx` if `f(0)=-2`?

Answer 1

`f(x)=xe^x-e^x-x^2/2-1`.

Explanation:

`f(x)=int(xe^x-x)dx=intx(e^x-1)dx`.

Using the Rule of Integration by Parts (IBP) :

`"IBP :" intuv'dx=uv-intu'vdx`.

We take, `u=x, and, v'=e^x-1`.

`:. u'=1, and, v=intv'dx=int(e^x-1)dx=e^x-x`.

`:. f(x)=x(e^x-x)-int(e^x-x)dx`.

`rArr f(x)=x(e^x-x)-(e^x-x^2/2)+c, or,`

`f(x)=xe^x-e^x-x^2/2+c`.

To determine `c`, we use the condition that, `f(0)=-2`.

`:. 0e^0-e^0-0^2/2+c=-2`.

`:. c=-1`.

`rArr f(x)=xe^x-e^x-x^2/2-1`.

Answer 2

`=>f(x) = e^x(x-1)-1/2x^2 -1`

Explanation:

`f(x) = int (x e^x - x)dx`

`=>f(x) = int xe^xdx - int x dx`

`=>f(x) = int x e^x dx - 1/2 x^2 + C_2`

where `C_2` is an arbitrary constant of integration.

To solve the first integral, we will employ integration by parts. Let `u equiv x` and `dv equiv e^xdx`.

`int x e^x dx = uv - int v du`

`=>int x e^x dx = xe^x - int e^x dx + C_1`

`=>int xe^x dx = xe^x - e^x + C_1`

`=>int xe^x dx = color(blue)(e^x(x - 1) + C_1)`

where `C_1` is an arbitrary constant of integration.

We can now substitute this result back into our `f(x)`.

`f(x) = color(blue)(e^x(x - 1) + C_1) - 1/2x^2 + C_2`

`=>f(x) = e^x(x-1)-1/2x^2 + C`

where `C` is an arbitrary constant of integration (`C_1` and `C_2` were arbitrary, so they were combined into a single constant term).

Assessing at `f(0)`:

`f(0) = -2 = e^0(0-1)-1/2(0)^2+C`

`=>-2 = -1 +C`

`=>C = -1`

Hence, the final result is:

`=>color(green)(f(x) = e^x(x-1)-1/2x^2 -1)`