find lim_(n→+∞) ∫_0 ^n sin(((πx)/n))dx .


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Question Number 48174 by Abdo msup. last updated on 20/Nov/18

$f i n d {l i m}_{n \to + \infty} \int_{0}^{n} s i n (\frac{π x}{n}) d x .$
Commented by Abdo msup. last updated on 20/Nov/18
$let A_n =∫_0 ^n sin(((πx)/n))dx ⇒ A_n =[−(n/π)cos(((πx)/n))]_0 ^n =−(n/π){cos(π)−1} =((2n)/π) ⇒ lim_(n→+∞) A_n =+∞$
$l e t A_{n} = \int_{0}^{n} s i n (\frac{π x}{n}) d x \Rightarrow$ $A_{n} = {[- \frac{n}{π} c o s (\frac{π x}{n})]}_{0}^{n} = - \frac{n}{π} {c o s (π) - 1}$ $= \frac{2 n}{π} \Rightarrow {l i m}_{n \to + \infty} A_{n} = + \infty$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Nov/18

	$\int_{0}^{n} s i n (\frac{π x}{n}) d x$ $= {[\frac{- c o s (\frac{π x}{n})}{\frac{π}{n}}]}_{0}^{n} = \frac{- n}{π} [c o s π - c o s 0]$ $= \frac{- n}{π} [- 2] = \frac{2 n}{π}$ $w h e n n \to \infty l i m i t a l s o \to \infty$
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