For all n ∈ N f_n (x)= { ((((nx)/(2n−1)), x ∈ [0, ((2n−1)/n)])),((1 , x ∈[((2n−1)/n), 2])) :} then for n→∞ ∫_1 ^2 f


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Question Number 55638 by gunawan last updated on 01/Mar/19
$For all n ∈ N f_n (x)= { ((((nx)/(2n−1)), x ∈ [0, ((2n−1)/n)])),((1 , x ∈[((2n−1)/n), 2])) :} then for n→∞ ∫_1 ^2 f_n (x) dx convergences to..$
$For all n \in N$ $f_{n} (x) = {\begin{cases} \frac{n x}{2 n - 1}, x \in [0, \frac{2 n - 1}{n}] \\ 1, x \in [\frac{2 n - 1}{n}, 2] \end{cases}$ $then for n \to \infty$ $\int_{1}^{2} f_{n} (x) d x convergences to . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19
	$∫_1 ^((2n−1)/n) ((nx)/(2n−1))dx+∫_((2n−1)/n) ^2 1×dx (n/(2n−1))×∣^ (x^2 /2)∣_1 ^((2n−1)/n) +∣x∣_((2n−1)/n) ^2 =(n/(2(2n−1)))×{(((2n−1)/n))^2 −1}+2−(((2n−1)/n)) =(1/2)×((2n−1)/n)−(n/(2(2n−1)))+2−((2n−1)/n) =2−(((2n−1))/(2n))−(n/(2(2n−1))) =2−((2−(1/n))/2)−((n/n)/(2(((2n)/n)−(1/n)))) =2−((2−(1/n))/2)−(1/(2(2−(1/n)))) when n→∞ =2−((2−0)/2)−(1/(2(2−0))) =2−1−(1/4) 1−(1/4)=(3/4)$
	$\int_{1}^{\frac{2 n - 1}{n}} \frac{n x}{2 n - 1} d x + \int_{\frac{2 n - 1}{n}}^{2} 1 \times d x$ $\frac{n}{2 n - 1} \times ∣^{} \frac{x^{2}}{2} ∣_{1}^{\frac{2 n - 1}{n}} + ∣ x ∣_{\frac{2 n - 1}{n}}^{2}$ $= \frac{n}{2 (2 n - 1)} \times {{(\frac{2 n - 1}{n})}^{2} - 1} + 2 - (\frac{2 n - 1}{n})$ $= \frac{1}{2} \times \frac{2 n - 1}{n} - \frac{n}{2 (2 n - 1)} + 2 - \frac{2 n - 1}{n}$ $= 2 - \frac{(2 n - 1)}{2 n} - \frac{n}{2 (2 n - 1)}$ $= 2 - \frac{2 - \frac{1}{n}}{2} - \frac{\frac{n}{n}}{2 (\frac{2 n}{n} - \frac{1}{n})}$ $= 2 - \frac{2 - \frac{1}{n}}{2} - \frac{1}{2 (2 - \frac{1}{n})}$ $w h e n n \to \infty$ $= 2 - \frac{2 - 0}{2} - \frac{1}{2 (2 - 0)}$ $= 2 - 1 - \frac{1}{4}$ $1 - \frac{1}{4} = \frac{3}{4}$
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