OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

fix: sin should be in the numerator

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (turingtest):

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$u=\pi x$

6 years ago
OpenStudy (anonymous):

where does the $\Delta x$ come in

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (turingtest):

it*

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (turingtest):

yep

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

Lol switch the interval

6 years ago
OpenStudy (turingtest):

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

6 years ago
OpenStudy (anonymous):

Okay thanks so much

6 years ago
OpenStudy (turingtest):

welcome!

6 years ago