Show that the sequence $f_{n}(x)=\frac{\int_{0}^{x}(1-t)^{n}dt}{\int_{0}^{1}(1-t)^{n}dt}$ converges uniformly to $1$ on $[r,1]$, where $0<r<1$.
I was trying to do this using the definition of uniform convergence.
If $f_{n}(x)$ converges uniformly to $1$, then for every $\varepsilon >0$, there exists an $N>0$ for all $x\in [r,1]$ such that if $n\geq N$, then $\left | \frac{\int_{0}^{x}(1-t)^{n}dt}{\int_{0}^{1}(1-t)^{n}dt}-1 \right |\leq \varepsilon$.
I simiplified the absolute value by finding the integrals. Is this what I am supposed to do?
$\left | \frac{\int_{0}^{x}(1-t)^{n}dt}{\int_{0}^{1}(1-t)^{n}dt}-1 \right |=\left | \frac{\frac{1}{n+1}(1-(1-x)^{n+1})-\frac{1}{n+1}}{\frac{1}{n+1}} \right |=\left | -(1-x)^{n+1} \right |$
I am not sure if I integrated this correctly. Anyway, I am stuck on whether this is the right approach or is there a better way?
Hint: $1 \ge f_n(x) \ge f_n(r)$ if $0 < r \le x \le 1$, so it suffices to show $f_n(r) \to 1$.