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#
