Eval limit by interpreting it as a limit of Riemann...

[confused_07]

Eval lim n-> infinity [sin(pi/n)+sin(2pi/n)+sin(3pi/n)+...+sin(npi/n) / n] by interpreting it as a limit of Riemann sums for a continuous function f defined on [0,1].

So... as n->infinity, that will remove all the 'n's from the equation. So I should be left with the integral of sin*pi + sin*2pi....etc., but where does the equation end? I know it's on [0,1], so does it end at 1?

delta x= (1 - 0) / n = 1/n

Just need help setting up the equation as I don't understand what they are asking...

 

[pka]

What you have is an approximating sum for
\(\displaystyle \L\int\limits_0^1 {\sin (\pi x)dx}\).