Hi my friends
can you help to find the limit of the following equation
Area = $\displaystyle \lim_{n \rightarrow \infty}$ $\displaystyle \frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})$
Hi my friends
can you help to find the limit of the following equation
Area = $\displaystyle \lim_{n \rightarrow \infty}$ $\displaystyle \frac{\pi}{n} *sin(\frac{\pi*n+\pi}{2})$
i am , doing it like this
$\displaystyle \Delta x =$$\displaystyle \frac{\pi}{n}$
$\displaystyle \lim n-->\infty \sum Sin(xi) \Delta x$
$\displaystyle \lim n-->\infty\\\\ Sin(\frac{\pi}{n} \frac{n(n+1)}{2}) \frac{\pi}{n} $
$\displaystyle \lim$ n--> $\displaystyle \infty$ $\displaystyle \frac{\pi}{n} *sin(\frac{\pi n+\pi}{2})$
I dont know if that was the right steps to find area using Riemann sum and limit
please correct me if i am wrong
It seems to me that this thread is very confused.
It seem that the OP assumes that the above limit is an approximating sum for $\displaystyle \int\limits_0^\pi {\sin (x)dx} $.
But that is totally wrong. A correct approximating sum (righthand sum) is $\displaystyle \frac{\pi }{n}\sum\limits_{k = 1}^n {\sin \left( {\frac{\pi }{n}k} \right)} $.
Thus we want this limit: $\displaystyle \lim _{n \to \infty } \frac{\pi }{n}\sum\limits_{k =1 }^n {\sin \left( {\frac{\pi }{n}k} \right)} $.