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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-

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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..
$$\mathrm{For}\:\mathrm{all}\:{n}\:\in\:\mathbb{N} \\ $$$${f}_{{n}} \left({x}\right)=\begin{cases}{\frac{{nx}}{\mathrm{2}{n}−\mathrm{1}},\:\:\:\:\:{x}\:\in\:\left[\mathrm{0},\:\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}\right]}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:,\:\:\:\:\:\:{x}\:\in\left[\frac{\mathrm{2}{n}−\mathrm{1}}{{n}},\:\mathrm{2}\right]}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{for}\:{n}\rightarrow\infty \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {f}_{{n}} \left({x}\right)\:{dx}\:\mathrm{convergences}\:\mathrm{to}.. \\ $$$$ \\ $$
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)
$$\int_{\mathrm{1}} ^{\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}} \frac{{nx}}{\mathrm{2}{n}−\mathrm{1}}{dx}+\int_{\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}} ^{\mathrm{2}} \:\mathrm{1}×{dx} \\ $$$$\frac{{n}}{\mathrm{2}{n}−\mathrm{1}}×\mid^{} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{1}} ^{\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}} \:+\mid{x}\mid_{\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}} ^{\mathrm{2}} \\ $$$$=\frac{{n}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}×\left\{\left(\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}\right)^{\mathrm{2}} −\mathrm{1}\right\}+\mathrm{2}−\left(\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}−\frac{{n}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)}+\mathrm{2}−\frac{\mathrm{2}{n}−\mathrm{1}}{{n}} \\ $$$$=\mathrm{2}−\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}{n}}−\frac{{n}}{\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$$=\mathrm{2}−\frac{\mathrm{2}−\frac{\mathrm{1}}{{n}}}{\mathrm{2}}−\frac{\frac{{n}}{{n}}}{\mathrm{2}\left(\frac{\mathrm{2}{n}}{{n}}−\frac{\mathrm{1}}{{n}}\right)} \\ $$$$=\mathrm{2}−\frac{\mathrm{2}−\frac{\mathrm{1}}{{n}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−\frac{\mathrm{1}}{{n}}\right)} \\ $$$${when}\:{n}\rightarrow\infty \\ $$$$=\mathrm{2}−\frac{\mathrm{2}−\mathrm{0}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−\mathrm{0}\right)} \\ $$$$=\mathrm{2}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$

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