Let $f:\Bbb R\to\Bbb R$ be uniformly continuous and suppose $F(x)=\int_0^x f(t)dt$ goes to $+\infty$ as $x\to+\infty $. The exercise asks to show that $F$ is at most of order of growth $2$, i.e. $\lim_{x\to+\infty}\frac{F(x)}{x^a} $ is non-zero only if $a\le2$.
I could solve this only assuming the differentiability of $f$. In this case, if the order is at least $2$ we must have $$\lim_{x\to+\infty}\frac{F(x)}{x^2}=\lim_{x\to+\infty}\frac{f(x)}{2x}=\lim_{x\to+\infty}\frac{f'(x)}{2}$$ and this cannot be infinite because by the mean value theorem it would contradict the u.c. of $f$.
What if $f$ is not everywhere differentiable, or even nowhere differentiable? I think there should be an approach moredirectly linked to the definition of u.c.
Let's show that if $f$ is uniformly continuous, then $|f(x)-f(0)| \leq Cx$ for some constant $C$. I think you can take on from here, then.
So, take $\epsilon=1$. By uniform continuity, we have a $K \in \mathbb{N}$ such that if $d(y_1,y_2)\leq1/K$, then $|f(y_1)-f(y_2)|<1$.
Letting $n$ be the largest natural number such that $\frac{n}{K}< x$, we have \begin{align*} |f(x)-f(0)| &\leq \sum_{i=0}^{n-1}\left|f\left(\frac{i+1}{k}\right)-f\left(\frac{i}{K}\right)\right|+\left|f\left(x\right)-f\left(\frac{n}{K}\right)\right| \\ &\leq n+1 \\ &=K \cdot \frac{n+1}{K}\\ &\leq K \cdot x. \end{align*} Noting that $K$ depends only on the fixed $\epsilon$ and not on $x$ yields the result.