What is f(x) = int 3x^3-2x+xe^(x-2) dx if f(1) = 3 ?

1 Answer
Mar 27, 2017

f(x)=3x^4/4 -2x^2/2+(x-1)e^(x-2)+13/4

Explanation:

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