Question Number 38100 by maxmathsup by imad last updated on 21/Jun/18
![let A_n = ∫_0 ^n (x−[x])^2 dx 1) calculate A_n 2) find lim_(n→+∞) A_n](https://www.tinkutara.com/question/Q38100.png)
$${let}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \left({x}−\left[{x}\right]\right)^{\mathrm{2}} {dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Commented by prof Abdo imad last updated on 22/Jun/18
![A_n = ∫_0 ^n ( x^2 −2x[x] +[x]^2 )dx = ∫_0 ^n x^2 dx −2 ∫_0 ^n x[x]dx +∫_0 ^n [x]^2 dx but ∫_0 ^n x[x]dx=Σ_(k=0) ^(n−1) ∫_k ^(k+1) kx dx =Σ_(k=0) ^(n−1) k ( (((k+1)^2 )/2) −(k^2 /2)) =Σ_(k=0) ^(n−1) k((k^2 +2k +1−k^2 )/2) =Σ_(k=0) ^(n−1) k( k +(1/2))=Σ_(k=0) ^(n−1) k^2 +(1/2)Σ_(k=0) ^(n−1) k (((n−1)(n−1+1)(2(n−1)+1))/6) +(1/2)(((n−1)n)/2) =((n(n−1)(2n−1))/6) +((n(n−1))/4) =((n(n−1))/2){ ((2n−1)/3) +(1/2)} =((n(n−1))/2){ ((4n−2 +3)/6)} =((n(n−1)(4n+1))/(12)) also ∫_0 ^n x^2 dx =[(x^3 /3)]_0 ^n = (n^3 /3) ∫_0 ^n [x]^2 dx =Σ_(k=0) ^(n−1) ∫_k ^(k+1) k^2 dx =Σ_(k=0) ^(n−1) k^2 =((n(n−1)(2n−1))/6) ⇒ A_n = (n^3 /3) −2 ((n(n−1)(4n+1))/(12)) +((n(n−1)(4n+1))/6) A_n =(n^3 /3) 2) lim_(n→+∞) A_n =+∞](https://www.tinkutara.com/question/Q38172.png)
$${A}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \left(\:{x}^{\mathrm{2}} \:−\mathrm{2}{x}\left[{x}\right]\:+\left[{x}\right]^{\mathrm{2}} \right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{{n}} \:{x}^{\mathrm{2}} {dx}\:\:−\mathrm{2}\:\int_{\mathrm{0}} ^{{n}} \:{x}\left[{x}\right]{dx}\:\:+\int_{\mathrm{0}} ^{{n}} \:\left[{x}\right]^{\mathrm{2}} {dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{{n}} \:{x}\left[{x}\right]{dx}=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:{kx}\:{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\:\left(\:\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}\:−\frac{{k}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\frac{{k}^{\mathrm{2}} \:+\mathrm{2}{k}\:+\mathrm{1}−{k}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}\left(\:{k}\:+\frac{\mathrm{1}}{\mathrm{2}}\right)=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}^{\mathrm{2}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k} \\ $$$$\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}\left({n}−\mathrm{1}\right)+\mathrm{1}\right)}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}}\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}} \\ $$$$ \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}\:+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{4}} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\left\{\:\:\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\left\{\:\frac{\mathrm{4}{n}−\mathrm{2}\:+\mathrm{3}}{\mathrm{6}}\right\} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{1}\right)}{\mathrm{12}}\:{also} \\ $$$$\int_{\mathrm{0}} ^{{n}} \:{x}^{\mathrm{2}} {dx}\:=\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{{n}} \:=\:\frac{{n}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{{n}} \:\left[{x}\right]^{\mathrm{2}} {dx}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{k}^{\mathrm{2}} \:{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{k}^{\mathrm{2}} \:=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}\:\Rightarrow \\ $$$${A}_{{n}} \:=\:\frac{{n}^{\mathrm{3}} }{\mathrm{3}}\:\:\:−\mathrm{2}\:\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{1}\right)}{\mathrm{12}}\:\:+\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$${A}_{{n}} \:=\frac{{n}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =+\infty \\ $$