Question

Find (click link for image: http://i.imgur.com/UQDs9.gif ) by evaluating an appropriate definite integral over the interval [0,1].

Answer 1

fix: sin should be in the numerator

Answer 2

\[\large\lim_{n\to\infty}\sum_{i=1}^n\frac{\sin(\frac{i\pi} n)}n\]

Answer 3

I don't understand why you are integrating over [0,1] and not [1,+infty]
are you sure you typed the right interval?

Answer 4

oh wait... I think I might see what they want...

Answer 5

Yes, I typed it correctly. If it helps I'm in the u-substitution section

Answer 6

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\[\Delta x=\frac1n\text{ and }f(x)=\sin(\pi x)\]

Answer 7

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\[u=\pi x\]

Answer 8

where does the \[\Delta x\] come in

Answer 9

\[\lim_{n\to\infty}\Delta x=\lim_{n\to\infty}\frac{b-a}n=dx\]...sort of
or do you mean where is it in the summand?

Answer 10

I guess I'm just generally confused, sums were not my strong point

Answer 11

you don't actually have to to any summations here, just notice the constituent parts and identify what integral ti represents

Answer 12

it*

Answer 13

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\[\large\lim_{n\to\infty}\sum_{i=1}^n\sin[\pi i\Delta x]\Delta x=\int_0^1\sin(\pi x)dx\]

Answer 14

so from there I just set u = pi*x ... ?

Answer 15

yep

Answer 16

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

Answer 17

Lol switch the interval

Answer 18

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

Answer 19

Okay thanks so much

Answer 20

welcome!