Let's first evaluate the indefinite integral:
f(x)=int(3x^3-2x+xe^(x-2))dx
=int(3x^3)dx -int(2x)dx+int(xe^(x-2))dx
=3x^4/4 -2x^2/2+C+int(xe^(x-2))dx where C is the constant of integration.
We can evaluate int(xe^(x-2))dx using integration by parts as follows:
int(xe^(x-2))dx = e^-2int(xe^(x))dx
Evaluating int(xe^(x))dx:
int(xe^(x-2))dx = xe^x - int(e^x)dx
= xe^x - e^x +C
Therefore, int(xe^(x-2))dx = e^-2(xe^x - e^x)+C
=(x-1)e^(x-2)+C
Therefore, the original integral is equal to:
f(x)=3x^4/4 -2x^2/2+(x-1)e^(x-2)+C
Plugging in x=1, we get:
f(1)=3/4 -1+0+C =-1/4+C
But f(1)=3
So, 3=-1/4+C => C=3+1/4=13/4
Therefore,
f(x)=3x^4/4 -2x^2/2+(x-1)e^(x-2)+13/4