Is $ f\left(x\right) =\int^{x}_{0}\sin( e^{t^{2}})\,\mathrm{d}t$ uniformly continuous?

Question

Show that:

$$f\left(x\right) =\int^{x}_{0}\sin\left( e^{t^{2}}\right)d\,\mathrm{d}t$$

Is uniformly continuous

I have tried to integrate it and bound it by using its limits, but a little unsure about what to do from there.

Answer 1

If the derivative is bounded then it is uniformly continuous, can you see how that would apply here?

We have that the $\sin$ function is bounded by $1$, i.e. $|\sin(x)| \leq 1$. Using the fundamental theorem of calculus we obtain that $f'(x)=\sin(e^{x^2})$, and applying our knowledge of $\sin(x)$ we get $|\sin(e^{x^2})|\leq 1$. By the mean value theorem we get that

$|\sin(e^{x^2})-\sin(e^{y^2})|\leq (1)|x-y|$. Hence our function is uniformly continuous since for any $\epsilon>0$ we only need to choose $\delta<\epsilon$.

Answer 2

Note that $f$ is differentiable, with derivative $f'(x)=\sin (e^{x^2})$. Thus $|f'(x)|\le 1$. So $f$ has bounded variation and is therefore uniformly continuous.