one function out of two with integrals I'm looking to have a single function $f(x)$ that is drawn from and satisfies two base functions
$$\int_{-\infty}^\infty (f(x)) \, dx = 1$$
$$\int_{-\infty}^\infty x(f(x)) \, dx = 0$$
I'm sure there's some shortcut to it, an obvious function given the outstanding $x$ in function two that makes it zero and the other $1$. I just don't see it. I'm sure someone here will get it instantly but I'm deaf and dumb right now.

Any probability density $f$ that is symmetric about $0$ works provided the integral of $|x|f(x)\,dx$ is finite.  For example
$$
f(x) = \begin{cases} \frac 1 2 & \text{if } -1<x<1, \\[10pt] 0 & \text{otherwise}. \end{cases}
$$
This is the uniform distribution on the interval $[-1,1]$.
Or you can use the standard normal distribution.
