Let's first evaluate the indefinite integral:
f(x)=int(3x^3-2x+xe^(x-2))dxf(x)=∫(3x3−2x+xex−2)dx
=int(3x^3)dx -int(2x)dx+int(xe^(x-2))dx=∫(3x3)dx−∫(2x)dx+∫(xex−2)dx
=3x^4/4 -2x^2/2+C+int(xe^(x-2))dx=3x44−2x22+C+∫(xex−2)dx where CC is the constant of integration.
We can evaluate int(xe^(x-2))dx∫(xex−2)dx using integration by parts as follows:
int(xe^(x-2))dx = e^-2int(xe^(x))dx∫(xex−2)dx=e−2∫(xex)dx
Evaluating int(xe^(x))dx∫(xex)dx:
int(xe^(x-2))dx = xe^x - int(e^x)dx∫(xex−2)dx=xex−∫(ex)dx
= xe^x - e^x +C=xex−ex+C
Therefore, int(xe^(x-2))dx = e^-2(xe^x - e^x)+C∫(xex−2)dx=e−2(xex−ex)+C
=(x-1)e^(x-2)+C=(x−1)ex−2+C
Therefore, the original integral is equal to:
f(x)=3x^4/4 -2x^2/2+(x-1)e^(x-2)+Cf(x)=3x44−2x22+(x−1)ex−2+C
Plugging in x=1x=1, we get:
f(1)=3/4 -1+0+C =-1/4+Cf(1)=34−1+0+C=−14+C
But f(1)=3f(1)=3
So, 3=-1/4+C => C=3+1/4=13/43=−14+C⇒C=3+14=134
Therefore,
f(x)=3x^4/4 -2x^2/2+(x-1)e^(x-2)+13/4f(x)=3x44−2x22+(x−1)ex−2+134
