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Question Number 152620 by Tawa11 last updated on 30/Aug/21

Commented by Tawa11 last updated on 30/Aug/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by Olaf_Thorendsen last updated on 30/Aug/21

$${\int }_0^{1}3{(1+x)}^{2}dx=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{i=0}{\sum ^n}3{(1+\frac{i}{n})}^2$$$${\left[{(1+x)}^3\right]}_0^1=\underset{n\to \mathrm{\infty }}{\lim }(\frac{3}{n}+\underset{i=1}{\sum ^n}3{(1+\frac{i}{n})}^2)$$$$\underset{n\to \mathrm{\infty }}{\lim }\underset{i=1}{\sum ^n}3{(1+\frac{i}{n})}^2={(1+1)}^3-{(1+0)}^3=7$$

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