Answer
$\displaystyle \int f(x)dx=-\frac{1}{4}e^{-2x}(2x+1)+C$
Rreasonable, because $f(x)$ behaves as $F'(x)$ should.
Work Step by Step
By parts, we take $\displaystyle \left[\begin{array}{ll}
u=x & dv=e^{-2x}dx\\
& \\
du=dx & v=-\frac{1}{2}e^{-2x}
\end{array}\right],\quad\int udv=uv-\int vdu$
$I=-\displaystyle \frac{1}{2}xe^{-2x}-\int-\frac{1}{2}e^{-2x}dx$
$=-\displaystyle \frac{1}{2}xe^{-2x}+\int\frac{1}{2}e^{-2x}dx$
$=-\displaystyle \frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C$
$=-\displaystyle \frac{1}{4}e^{-2x}(2x+1)+C$
Graphing (F is blue, f is red, C=0)
we recognize the characteristics of $f(x)=F'(x)$:
As F decreases (from the left to the y-axis, f (the derivative) is negative.
When F reaches its minimum, f is zero, changing from negative to positive.
As F rises, f is positive.
So, the answer is reasonable.
