Let $f: [0,1]\to \mathbb R$ be continuous. Let us define $F_n: [0,1]\to \mathbb R$ by $$ F_n(x)=\sum_{k=1}^{n} \frac{x}{n}f\left(\frac{kx}{n} \right). $$ Then
a. Find $F=\lim_{n\to \infty}F_n$.
b. Show uniform convergence.
I think that the answer for a. is $\int_0^1f(x)dx$. I came up with this by showing that for each $x \in [0,1]$ $F_n(x)$ is a Riemann sum with the partition $ P=\{0 ,\frac{x}{n}, \frac{2x}{n} , \frac{3x}{n},\dots,1 \}$ and by knowing $f$ is continuous we get the integral above.
Correct me if I'm wrong.
My problem is with b. I tried showing by definition and by Cauchy definition without any success.
You're close: the limit is $\int_0^{\color{red}{x}} f(t)\, dt$. When $x = 0$, $F_n(x) = 0$, and for each $x > 0$, $F_n(x)$ is a Riemann sum for $f$ on the interval $[0,x]$.
To see that $F_n$ converges uniformly to $F$, fix $\epsilon > 0$, and using uniform continuity of $f$ on $[0,1]$, choose $\eta > 0$ such that for all $s,t\in [0,1]$, $\lvert s - t\rvert < \eta$ implies $\lvert f(s) - f(t)\rvert < \frac{\epsilon}{2}$. Let $N\in \Bbb N$ such that $\dfrac{1}{N} < \eta$, and write
$$\sum_{k = 1}^n \frac{x}{n}f \bigl(\frac{kx}{n}\bigr) - \int_0^x f(t)\, dt = \sum_{k = 1}^n \int_{(k-1)x/n}^{kx/n} \left[f\bigl(\frac{kx}{n}\bigr) - f(t)\right]\, dt$$
If $n > N$ and $k \in \{1,2,\ldots, n\}$, $\lvert f(\frac{kx}{n}) - f(t)\rvert < \frac{\epsilon}{2}$ on $[\frac{(k-1)x}{n}, \frac{kx}{n}]$. So the latter sum is bounded by
$$\sum_{k = 1}^n \frac{\epsilon}{2} \frac{x}{n} = \frac{\epsilon}{2} x < \epsilon$$
Since $\epsilon$ was arbitrary, the result follows.