Question Number 157021 by MathSh last updated on 18/Oct/21
![Find: 𝛀(n) =∫_( 1) ^( n) ([x]^2 ∙{x} + [x]∙{x}^2 )dx n∈N ; [∗]-GIF ; {x}=x-[x]](Q157021.png)
$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\left(\boldsymbol{\mathrm{n}}\right)\:=\underset{\:\mathrm{1}} {\overset{\:\boldsymbol{\mathrm{n}}} {\int}}\left(\left[\mathrm{x}\right]^{\mathrm{2}} \centerdot\left\{\mathrm{x}\right\}\:+\:\left[\mathrm{x}\right]\centerdot\left\{\mathrm{x}\right\}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{n}\in\mathbb{N}\:\:;\:\:\left[\ast\right]-\mathrm{GIF}\:\:;\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$
Answered by ghimisi last updated on 18/Oct/21
![Ω(n)=∫_1 ^n [x]{x}([x]+{x})dx=∫_1 ^n x[x](x−[x])dx= Σ_(k=1) ^(n−1) ∫_k ^(k+1) (x^2 k−xk^2 )dx=Σ_(k=1) ^(n−1) (k∙(((k+1)^3 −k^3 )/3)−k^2 ∙(((k+1)^2 −k^2 )/2))= Σ_(k=1) ^(n−1) (((3k^3 +3k^2 +k)/3)−((2k^3 +k^2 )/2))=Σ_(k=1) ^(n−1) ((3k^2 +2k)/6)= (1/6)(3∙(((n−1)n(2n−1))/6)+2∙(((n−1)n)/2))=((n(n−1)(2n+1))/(12))](Q157034.png)
$$\Omega\left({n}\right)=\underset{\mathrm{1}} {\overset{{n}} {\int}}\left[{x}\right]\left\{{x}\right\}\left(\left[{x}\right]+\left\{{x}\right\}\right){dx}=\underset{\mathrm{1}} {\overset{{n}} {\int}}{x}\left[{x}\right]\left({x}−\left[{x}\right]\right){dx}= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\underset{{k}} {\overset{{k}+\mathrm{1}} {\int}}\left({x}^{\mathrm{2}} {k}−{xk}^{\mathrm{2}} \right){dx}=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left({k}\centerdot\frac{\left({k}+\mathrm{1}\right)^{\mathrm{3}} −{k}^{\mathrm{3}} }{\mathrm{3}}−{k}^{\mathrm{2}} \centerdot\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} }{\mathrm{2}}\right)= \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{3}{k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{2}} +{k}}{\mathrm{3}}−\frac{\mathrm{2}{k}^{\mathrm{3}} +{k}^{\mathrm{2}} }{\mathrm{2}}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{3}{k}^{\mathrm{2}} +\mathrm{2}{k}}{\mathrm{6}}= \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{3}\centerdot\frac{\left({n}−\mathrm{1}\right){n}\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{6}}+\mathrm{2}\centerdot\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\right)=\frac{{n}\left({n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$
Commented by MathSh last updated on 18/Oct/21
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$