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Question Number 2645 by Filup last updated on 24/Nov/15

$$A={\int }_{N_1}^{N_2}\lfloor x\rfloor dx$$ $$(N_1,N_2)\in \mathbb{Z},N_1<N_2$$ $$$$ $$\mathrm{Solve}\mathrm{for}A$$

Answered by Filup last updated on 24/Nov/15

$$\mathrm{let}{N}_1=N,N_2=N+k$$ $$(N_1,N_2)\in \mathbb{Z},\therefore (N,k)\in \mathbb{Z}$$ $$N_1<N_2,\therefore k>0$$ $$$$ $$N⩽x<N+1\Rightarrow \lfloor x\rfloor =N$$ $$N+1⩽x<N+2\Rightarrow \lfloor x\rfloor =N+1$$ $$⋮$$ $$N+k-1⩽x<N+k\Rightarrow \lfloor x\rfloor =N+k-1$$ $$$$ $$\mathrm{The}\mathrm{area}\mathrm{between}a\to a+1\mathrm{for}\lfloor x\rfloor$$ $$={\int }_a^{a+1}\lfloor x\rfloor =a$$ $$\because \mathrm{It}\mathrm{is}\mathrm{a}\mathrm{rectangle}\mathrm{with}l=1,h=a$$ $$$$ $$\mathrm{For}\mathrm{the}\mathrm{total}\mathrm{area}\mathrm{of}A,\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{sum}$$ $$\mathrm{all}\mathrm{the}\mathrm{rectangular}\mathrm{areas}$$ $$$$ $${\int }_N^{N+k}\lfloor x\rfloor dx=N+(N+1)+...+(N+k-1)$$ $$=N\left(k\right)+(1+2+...+(k-1))$$ $$=Nk+\underset{i=1}{\sum ^{k-1}}i$$ $$=Nk+\frac{1}{2}(k-1)(1+k-1)$$ $$=Nk+\frac{1}{2}k(k-1)$$ $$$$ $$\therefore A={\int }_N^{N+k}\lfloor x\rfloor dx=Nk+\frac{1}{2}k(k-1)$$ $$$$ $$Test:$$ $${\int }_2^5\lfloor x\rfloor dx=9\left(wolfram\right)$$ $$N=2,N+k=5\Rightarrow k=3$$ $$A=2\times 3+\frac{1}{2}3(3-1)$$ $$=6+\frac{1}{2}\left(2\right)3$$ $$=9$$ $$=True$$

Commented byYozzi last updated on 24/Nov/15

$$\underset{i=0}{\sum ^{k-1}}i=0+\underset{i=1}{\sum ^{k-1}}i=\underset{i=1}{\sum ^{k-1}}i=\frac{k-1}{2}(k-1+1)=\frac{k(k-1)}{2}\ne \frac{1}{2}{(k-1)}^2$$ $$$$

Commented byFilup last updated on 24/Nov/15

$$Thanks,\mathrm{i}\mathrm{just}\mathrm{noticed}\mathrm{the}\mathrm{mistake}\mathrm{myself}$$ $$\mathrm{haha}$$

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