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
![∫_0 ^1 3(1+x)^2 dx = lim_(n→∞) (1/n)Σ_(i=0) ^n 3(1+(i/n))^2 [(1+x)^3 ]_0 ^1 = lim_(n→∞) ((3/n)+Σ_(i=1) ^n 3(1+(i/n))^2 ) lim_(n→∞) Σ_(i=1) ^n 3(1+(i/n))^2 = (1+1)^3 −(1+0)^3 = 7](Q152621.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{3}\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \:{dx}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{3}\left(\mathrm{1}+\frac{{i}}{{n}}\right)^{\mathrm{2}} \\ $$$$\left[\left(\mathrm{1}+{x}\right)^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{3}}{{n}}+\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}\left(\mathrm{1}+\frac{{i}}{{n}}\right)^{\mathrm{2}} \right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{3}\left(\mathrm{1}+\frac{{i}}{{n}}\right)^{\mathrm{2}} \:=\:\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{3}} \:=\:\mathrm{7} \\ $$