Prove $F: \mathbb{R}\to\mathbb{R}$ where $F(x) = \int_a^x f(t)\, dt$ ($aProve $F: \mathbb{R}\to\mathbb{R}$ where $F(x) = \int_a^x f(t)\, dt$  ($a<x$) is surjective. 
$f$ is continuous and bounded below by $m>0$. Also $a$ belongs to $\mathbb{R}$ (reals).
 A: OK. The $a < x$ condition is wrong, and makes the theorem false, as others have pointed out. So let's get rid of it. 
Let $u \in \mathbb R$ be nonnegative.  Let $x = a + u/m$. Now estimate $F(x)$:
\begin{align}
F(x) &= \int_a^x f(t) ~dt \\
 &\ge \int_a^x m ~dt \\
 &= mx - ma \\
 &= m(a + u/m) - ma \\
 &= ma + u - ma \\
 &=  u 
\end{align}
On the other hand, $F(a) = 0$. So $F$, on the interval $[a, x]$ goes from less than $u$ to more than $u$; since $F$ is continuous (why? it's differentiable, by the Fundamental Theorem, hence continuous.) the intermediate value theorem applies, and there's a value $c \in [a, x]$ with $F(c) = u$. Since $u$ was an arbitrary nonnegative number, $F$ is surjective onto the nonnegative reals. 
A corresponding argument, with $x$ again equaling $a + u/m$, where $u/m$ is now negative, applies to negative values of $u$. Thus $F$ is surjective onto the reals. 
It also happens to be injective, because it's a strictly increasing function. 
A: I suppose it is easier to handle it using derivatives. Clearly $F'(x) = f(x) \geq m > 0$ for all $x$ so that $F(x)$ is strictly increasing. We need to show that range of $F(x)$ is whole of $\mathbb{R}$. Since $F$ is increasing it follows that either $F(x) \to L$ or $F(x) \to \infty$ as $x \to \infty$. If $F(x) \to L$ then it is obvious (by mean value theorem) that $$F(x) - F(x/2) = (x/2)F'(c) = (x/2)f(c) \geq mx/2$$ Then as $x \to \infty$ we get LHS as $L - L = 0$ and RHS as $\infty$. Hence it follows that $F(x) \to \infty$ as $x \to \infty$. Similarly it can be proved that $F(x) \to -\infty$ as $x \to -\infty$. By continuity of $F(x)$ and intermediate value theorem it follows that $F(x)$ takes all values between $-\infty$ and $\infty$ so that the range of $F$ is $\mathbb{R}$ and hence $F(x)$ is surjective.
