Evaluate $\int_0^1 cos(x) \,dx$ using Riemann Sum.
What i did
Split [0,1] into n equal sub intervals, [0,1/n],[1/n,2/n]…[(n-1)/n,1]
Δx=1/n
Sample points, $x_1*$=1/n…$x_n*$=n/n
Riemann Sum= $S_n$=cos(1/n) (1/n) +cos(2/n) (1/n) +…+cos(n/n) (1/n)
=(1/n)* $\sum_{i=0}^n cos(i * 1/n) = 1/n *(\frac{sin((n+1/2)(1/n))}{2sin(1/2n)}$-1/2)
I know the integral is the lim of the expression above, but i cant seem to evaluate out the limit and get sin (1). pls help, did i do anything wrong above?
Best Answer
$$\frac{1}{n}.\frac{\sin((n+\frac{1}{2})(\frac{1}{n}))}{2\sin(\frac{1}{2n})}=\frac{1}{2n}.\frac{\sin(1+\frac{1}{2n})}{\sin(\frac{1}{2n})}=\frac{\frac{1}{2n}}{\sin(\frac{1}{2n})}\sin\left(1+\frac{1}{2n}\right)$$
When $n$ tends to $+\infty$, the first factor tends to 1, the second to $\sin(1)$.
Moreover, by taking a first order approximation (beginning of Taylor expansion):
$$\sin\left(1+\frac{1}{2n}\right)\approx \sin(1)+\cos(1)\frac{1}{2n}$$
you know that the error is $\approx \cos(1)\frac{1}{2n}.$