Find (click link for image: http://i.imgur.com/UQDs9.gif ) by evaluating an appropriate definite integral over the interval [0,1].
fix: sin should be in the numerator
\[\large\lim_{n\to\infty}\sum_{i=1}^n\frac{\sin(\frac{i\pi} n)}n\]
I don't understand why you are integrating over [0,1] and not [1,+infty] are you sure you typed the right interval?
oh wait... I think I might see what they want...
Yes, I typed it correctly. If it helps I'm in the u-substitution section
the integral over the interval [a,b] of a function f(x) is given by \[\large\lim_{n\to\infty}\sum_{i=1}^nf(a+i\Delta x)\Delta x~;~~\Delta x=\frac{b-a}n\]you have been given\[\large\lim_{n\to\infty}\sum_{i=1}^n\frac{\sin(\frac{i\pi} n)}n\]so it looks like in your case\[\Delta x=\frac1n\text{ and }f(x)=\sin(\pi x)\]
so in your case have been given\[\large\lim_{n\to\infty}\sum_{i=1}^n\frac{\sin(\frac{i\pi} n)}n=\int_0^1\sin(\pi x)dx\]which can be done with the u-sub\[u=\pi x\]
where does the \[\Delta x\] come in
\[\lim_{n\to\infty}\Delta x=\lim_{n\to\infty}\frac{b-a}n=dx\]...sort of or do you mean where is it in the summand?
I guess I'm just generally confused, sums were not my strong point
you don't actually have to to any summations here, just notice the constituent parts and identify what integral ti represents
it*
your interval is \([a,b]=[0,1]\) so we have\[\large\lim_{n\to\infty}\sum_{i=1}^nf(a+i\Delta x)\Delta x~;~~\Delta x=\frac{b-a}n\]for you\[\Delta x=\frac1n=\frac{1-0}n\]you have been given\[\large\lim_{n\to\infty}\sum_{i=1}^n\frac{\sin(\frac{i\pi} n)}n=\large\lim_{n\to\infty}\sum_{i=1}^n\sin[a+\pi i(\frac1n)]\frac1n\]since \(a=0\) this becomes\[\large\lim_{n\to\infty}\sum_{i=1}^n\sin[\pi i(\frac1n)]\frac1n=\lim_{n\to\infty}\sum_{i=1}^n\sin[\pi i\Delta x]\Delta x\]try to compare the two with the definition of the integral and you see that\[\large\lim_{n\to\infty}\sum_{i=1}^n\sin[\pi i\Delta x]\Delta x=\int_0^1\sin(\pi x)dx\]
so from there I just set u = pi*x ... ?
yep
So then ... du = pi, but then \[\int\limits_{1}^{0}\sin \pi(u)\] but sin pi is 0 ... I'm probably being an idiot with calculations right now so
Lol switch the interval
1)\[u=\pi x\implies du=\pi dx\implies dx=\frac{du}\pi\](don't forget dx, it is important!) 2) change the bounds\[u=\pi x\implies x=0\to u=0\text{ and }x=1\to u=\pi\] which leads to\[\int_0^1\sin(\pi x)dx=\int_0^\pi\sin(u)(\frac{du}\pi)=\frac1\pi\int_0^\pi\sin udu\] 3) you haven't integrated it yet. you won't get zero after integrating
Okay thanks so much
welcome!
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