If $f$ and $f'$ are integrable, then $f'$ has integral $0$ I have the problem

Prove that if $f$  and $f'$ are Lebesgue integrable over $\mathbb{R}$, then $\int_\mathbb{R}f' = 0$, where $f'$ is defined everywhere.

Honestly, not sure where to start. I had the idea to perhaps use the fundamental theorem of calculus $f(x) = \int_0^x f' + f(0)$ and take limits as $x$ goes to $0$ and trying to work from there. However, I'm not sure that the theorem holds in this case. Any help getting started is appreciated!

$$f(x) = f(0) + \int_0^x f'$$
Taking the appropriate limits...
$$0=f(\infty) = f(0) + \int_0^\infty f'$$
$$0=f(-\infty) = f(0) + \int_0^{-\infty} f'=f(0) - \int_{-\infty}^0 f'$$
$$\therefore \int_{-\infty}^\infty f'= \int_{-\infty}^0 f' + \int_0^\infty f'=f(0)-f(0)=0$$
Where we needed $f$ to be Lebesgue integrable to know it tended to $0$ on the tails, and $f'$ to be Lebesgue integrable to rearrange those integrals.
