Question

What is `f(x) = int xe^x-x dx` if `f(-1) = 1`?

Answer 1

`f(x) = 2/e+ e^x(x-1)+(3-x^2)/2`

Explanation:

Using the fundamental theorem of calculus:

`f(x) = 1+ int_(-1)^x (te^t-t)dt`

as the integral is linear:

`f(x) = 1+ int_(-1)^x te^tdt- int_(-1)^x tdt`

Solve the two integrals separately:

`int_(-1)^x tdt = [t^2/2]_(-1)^x`

`int_(-1)^x tdt = x^2/2-1/2`

and:

`int_(-1)^x te^tdt = int_(-1)^x t d/dt(e^t) dt`

integrating by parts:

`int_(-1)^x te^tdt = - int_(-1)^x e^t dt + [te^t]_( -1)^x`

`int_(-1)^x te^tdt = - [e^t]_(-1)^x + xe^x+1/e`

`int_(-1)^x te^tdt = - e^x+1/e + xe^x+1/e`

`int_(-1)^x te^tdt = e^x(x-1)+2/e`

Putting together the partial results:

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

`f(x) = 2/e+ e^x(x-1)+(3-x^2)/2`