$\int x^k\ d\mu(x)\geq0$ for $k$ even?

Question

I am studying convex optimization, and the truncated moment problem is being discussed. I have no background in measure theory so I don't understand things taken for granted such as $$\int x^k\ d\mu(x)\geq0 \text{ for }k\text{ even}.$$ Could someone explain how this is true? I attempt to evaluate by doing: $$ \begin{align} \int x^k\ d\mu(x)&= \int x^k\mu'(x)\ dx \end{align} $$ which seems like I could pick any $\mu'(x)$ and force the integral to be $<0$. Is there perhaps some restraint on $\mu$ that can be explained?