Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$. Calculate the limit: $\lim_{n\to+\infty}\sum_{k=1}^n\frac{\sin{\frac{k\pi}n}}{n}$ Using definite integral between the interval $[0,1]$.
It seems to me like a Riemann integral definition:
$\sum_{k=1}^n\frac{\sin{\frac{k\pi}{n}}}{n}=\frac1n(\sin{\frac{\pi}n}+...+\sin\pi)$
So $\Delta x=\frac 1n$ and $f(x)=\sin(\frac{\pi x}n)$ (Not sure about $f(x))$
How do i proceed from this point?

Remember that if $f: [a,b] \to \mathbb{R}$ is some function we can define the Riemann Sum of $f$ in the interval by choosing a partition $P = \{t_0, \dots, t_n\}$ of $[a,b]$ in a very specific manner: we simply divide the interval $[a,b]$ equally calculating it's length $b-a$ and then dividing by the number of subintervals that we want. So a partition with $n$ subintervals will have $t_i = a + i\Delta t$ where $\Delta t = (b-a)/n$ is the size of each interval. In that case the integral of $f$ on the interval $[a,b]$ will be the limit:
$$\int_a^b f(t)dt=\lim_{n\to \infty} \sum_{i=1}^n f(t_i)\Delta t$$
Look that in your case we are working on the interval $[0,1]$ so for a partition of $n$ subintervals we'll have $\Delta t = 1/n$. Now we want to compute the limit of the following sum:
$$\sum_{i=1}^n \frac{1}{n}\sin{\left(\frac{i\pi}{n}\right)}$$
Note that with this we have:
$$\sum_{i=1}^n \Delta t\sin{\left(i\Delta t \pi\right)}$$
However, since the interval starts on $0$ and since $t_i = a + i\Delta t$, we have that $t_i = i\Delta t$, so we have:
$$\sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t$$
Now on the limit this will become obviously the following integral:
$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\int_0^1 \sin(\pi t)dt$$
This integral is computed trivially by setting $u=\pi t$, $dt=du/\pi$ and so $u$ ranges from $0$ to $\pi$ and we have:
$$\lim_{n\to \infty} \sum_{i=1}^n \sin{\left(t_i \pi\right)}\Delta t=\frac{1}{\pi}\int_0^\pi \sin(u)du=\frac{2}{\pi}$$
