함수의 평균값

This seems like it might be true: if $f$ is getting closer and closer to zero, as $x$ grows towards infinity, then over time the later values will become a larger and larger portion of the total interval we're looking at, and so the average should look more and more like $f(x)$ when $x$ is large–that is, like 0. That's some intuition to start us out, but it isn't a rigorous argument. To be sure we haven't overlooked something, let's use the definition of an average to express $A(x)$.

$$\begin{aligned} A(x)&=\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t\\ \lim_{x \to \infty}A(x)&=\lim_{x \to \infty}\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \\ \text{If $\int_0^\infty f(t) \,\mathrm{d}t$ converges, then this limit is 0, and the statement is true. So, suppose it does not converge. Since $f(x)$ is positive, that means $\lim\limits_{x \to \infty} \int_0^x f(t)\,\mathrm{d}(t) = \infty$, so we can use l'H\^{o}pital's rule. To differentiate the numerator, we use the Fundamental Theorem of Calculus part 1.} \\ \lim_{x \to \infty}A(x)&=\lim_{x \to \infty}\frac{\int_0^x f(t)\,\mathrm{d}t}{x} = \lim_{x \to \infty}\frac{f(t)}{1}=0 \end{aligned}$$

So, the statement is true whether $\int_0^xf(t)\,\mathrm{d}t$ converges or not.