To show $\lim_{x \to \infty} \int_{x}^{x+1}f(t)dt=0$ if $\lim_{x\to\infty} f(x)=0$

$f:\mathbb{R}\to \mathbb{R}$ is Riemann integrable on any bounded interval and $\lim_{x\to\infty} f(x)=0$.

Define $g(x)=\int_{x}^{x+1}f(t)dt$, we need to show $\lim_{x\to\infty} g(x)=0$.

Please give me hint, I want to try myself. Thank you, I was trying to apply fundamental theorem of calculus, like taking the derivative of $g$ which is $f(x+1)-f(x)$ but then don't know what to do.