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How do I express a limit as a definite integral $\int_0^1 f(x) dx$ and then evaluate the limit $\lim_{n \to \infty} \frac{1}{n} ( \cos \frac{\pi}{2n} + \cos \frac{2 \pi}{2n} + ... + \cos \frac{k \pi}{2n} + .... + \cos \frac{(n-1) \pi}{2n} + \cos \frac{n\pi}{2n})$

What am I supposed to do with the limit as I have not encountered this type of problem relating definite integrals and limits before. Is this part of a discontinuous integrad?

I saw the answers and it states that

$\lim_{n \to \infty} \sum_1^n \frac{1}{n} \sin \frac{k\pi}{2n} = \lim_{n \to \infty} \sum_1^n \frac{1}{n} \cos (\frac{k}{n} \frac{\pi}{2})$

How did they even arrive to these steps using which method?

user307640
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  • Do you know to define the definite integral $\int_a^b f(x) , dx$ as a limit? – Lee Mosher Nov 20 '22 at 15:31
  • I notice that the post linked which already answers the question is lacking some detail. Do you understand the solution? – Ogglie Ostrich Nov 21 '22 at 04:41
  • @OgglieOstrich not really. I am not sure how to convert it – user307640 Nov 21 '22 at 06:25
  • We define integration as:$$\int_{a}^{b}f(x)dx=\lim_{\Delta x \to 0}\sum_{n=0}^Nf(x_n)\Delta x,: \Delta x = \frac{b-a}{N}, : x_n= a + n \Delta x$$ $$\implies \int_{a}^{b}f(x)dx=\lim_{N \to \infty}\sum_{n=0}^Nf(x_n)\Delta x$$ So: $$\int_{0}^{1}f(x)dx=\lim_{N \to \infty}\sum_{n=0}^Nf(n\cdot \frac{1}{N})\frac{1}{N}$$ Now try substituting $\sin(\frac{\pi}{2}x)$ and likewise for cos, then compare the values of the two integrals and relate the sums. Since your sum is from 0 to N instead of from 1 to N you might have to do a little extra reasoning. – Ogglie Ostrich Nov 23 '22 at 13:49

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