solve-this-using-Riemann-sum-f-x-2x-0-4-for-n-4-

Question Number 104359 by bemath last updated on 21/Jul/20

$${solve}\:{this}\:{using}\:{Riemann} \\ $$$${sum}\:{f}\left({x}\right)=\mathrm{2}{x}\:;\:\left[\mathrm{0},\mathrm{4}\right]\:{for}\:{n}=\mathrm{4} \\ $$

Answered by Ar Brandon last updated on 21/Jul/20

∫_0 ^4 2xdx=lim_(n→∞) ((4−0)/n)Σ_(k=1) ^n 2(0+((4k)/n))=lim_(n→∞) (4/n)Σ_(k=1) ^n ((8k)/n) =lim_(n→∞) ((32)/n^2 )∙((n(n+1))/2)=lim_(n→∞) 16(1+(1/n))=16

$$\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{2}{x}\mathrm{d}{x}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4}−\mathrm{0}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{2}\left(\mathrm{0}+\frac{\mathrm{4k}}{\mathrm{n}}\right)=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{8k}}{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{32}}{\mathrm{n}^{\mathrm{2}} }\centerdot\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}16}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)=\mathrm{16} \\ $$$$ \\ $$

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