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

2 Answers
Apr 13, 2018

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.

Apr 13, 2018

=>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)