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

$$solvethisusingRiemann$$$$sumf\left(x\right)=2x;[0,4]forn=4$$

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

$${\int }_0^{4}2x\mathrm{d}x=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{4-0}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}2(0+\frac{4\mathrm{k}}{\mathrm{n}})=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{4}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{8\mathrm{k}}{\mathrm{n}}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{32}{{\mathrm{n}}^2}\cdot \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim 16}(1+\frac{1}{\mathrm{n}})=16$$$$$$

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