If $f(x) \rightarrow c > 0$ when $x \rightarrow \infty$, is it true that its anti-derivative $F(x) \rightarrow \infty$ when $x \rightarrow \infty$? If $f$ is a continous function and $f(x) \rightarrow c > 0$ when $x \rightarrow \infty$, is it true that the function's anti-derivative $F(x) \rightarrow \infty$ when $x \rightarrow \infty$?
Intuitively, I would say that this is true because I believe that there should be an infinitely large area constrained by the graph of $f(x)$ (which is above the positive $x-axis$) and the positive $x$-axis as $x \rightarrow \infty$. However, I cannot prove this. Is my intuition correct and might a more rigorous argument look like?
 A: Its true and a rigourous argument is just across the road. Pick $M$ such that for $x>M$, we have the bound $$f(x) > c/2>0.$$ It is enough to show that $\lim_{y\to\infty} \int_M^y f  = \infty$. And this is true, because...
A: You know that there is some $X \in \mathbb{R}$ such that for all $x > X$, $f(x) \in [\frac{1}{2}c, \frac{3}{2}c]$, so $f(x) > \frac{c}{2} > 0$, since $f(x)$ comes arbitrarily close to $c$ if $x$ gets big enough. Hence for all $x > X$:
$$
\int_0^x f(s) ds = \int_0^X f(s) ds + \int_X^x f(s) ds > \int_0^X f(s) ds + (x-X)\frac{c}{2}.
$$
As $x \to \infty$, the second term blows up, so $F(x) \to \infty$ as $x\to \infty$.
A: Yes, it's true. Your intuition is correct.Here's a formal way to show it.
Because $f$ converges to $c$, there exists $u\in\mathbb R$ such that if $x\geq u$, then $f(x)\geq \frac c 2$.
Then $$F(x)=F(u)+\int_u^x f(t)dt \geq F(u) +\int_u^x \frac{c}{2}dt=F(u) +\frac c 2 (x-u)$$
And the latter $\rightarrow+\infty$.
A: Yes, that is true. One way to see it is using the FTC: you can say that $f(x)>c/2$ for large enough $x$, say $x>a$) (by the definition of limit) and then
$$
F(x)=F(a)+\int\limits_a^xF'(x)\,dx=F(a)+\int\limits_a^x f(x)\,dx
$$
$$
\ge F(a)+\int\limits_a^x \frac{c}{2}\,dx \ge F(a)+\frac{c}{2}(x-a)
$$
so $F$ grows at least linearly at infinity.
