Prove that $phi : C[a,b]rightarrow mathbb{R}$ given by $phi(f)=int_a^bfdx$ is uniformly continuous

Prove that $phi : C[a,b]rightarrow mathbb{R}$ given by $phi(f)=int_a^bfdx$ is uniformly continuous.

First, since $f$ is continuous $phi$ is well defined (integrals exist). Since $f$ is continuous on a compact set it is bounded, ie. $|f(x)|< M$ where $M>0$ is a constant.

Now fix $epsilon>0$. If we take $delta=frac{epsilon}{(b-a)}$ then
$$|phi(f)-phi(g)|=Bigg | int_a^b (f(x)-g(x))dx Bigg |le int_a^b |f(x)-g(x)|dx <epsilon$$
whenever $f,gin C[a,b]$ and $sup_{xin [a,b]} |f(x)-g(x)|<delta$.

Is this clear enough proof or am I missing something? thanks.