Uniform Convergence of integrals

This is tautological. By definition, $\int_a^b f_n$ tends to $\int_a^bf$ means the $\epsilon$ condition you are trying to prove. So if you already know the convergence, there is nothing to do. But actually, I suggest you prove that $\int_a^b f_n$ tends to $\int_a^bf$. It will be a good exercise to clarify these notions.

Hint: the integral is linear and $$ |\int_a^bg(t)dt|\leq \int_a^b|g(t)|dt\leq (b-a)\sup_{t\in[a,b]}|g(t)|=(b-a)\|g\|_\infty. $$

It doesn't make sense to speak of uniform convergence in this context. The integrals $$I_n := \int_a^b f_n(x) \, dx \qquad \qquad I:=\int_a^b f(x) \, dx$$ are real numbers which do not depend on another parameter (since $a,b$ are fixed). $$I_n = \int_a^b f_n(x) \, dx \to I=\int_a^b f(x) \, dx \qquad (n \to \infty)$$ is (by the definition of convergence of sequences) equivalent to the following statement: $$\forall \varepsilon>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: |I_n - I| \leq \varepsilon \\ \Leftrightarrow \forall \varepsilon>0 \, \exists N \in \mathbb{N} \, \forall n \geq N: \left|\int_a^b f_n(x) \, dx - \int_a^b f(x) \, dx \right| \leq \varepsilon$$