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Question Number 3900 by Filup last updated on 24/Dec/15

$$\mathrm{I}\mathrm{asked}\mathrm{this}\mathrm{question}\mathrm{a}\mathrm{while}\mathrm{ago},$$$$\mathrm{but}\mathrm{I}\mathrm{forgot}\mathrm{how}\mathrm{to}\mathrm{solve}\mathrm{it}:$$$$$$$$S={\int }_0^n\lfloor x\rfloor dx$$

Answered by Yozzii last updated on 24/Dec/15

$$0⩽x<1\Rightarrow \lfloor x\rfloor =0\Rightarrow A_1=0\times (1-0)=0$$$$1⩽x<2\Rightarrow \lfloor x\rfloor =1\Rightarrow A_2=1(2-1)=1$$$$2⩽x<3\Rightarrow \lfloor x\rfloor =2\Rightarrow A_3=2(3-2)=2$$$$3⩽x<4\Rightarrow \lfloor x\rfloor =3\Rightarrow A_4=3(4-3)=3$$$$⋮$$$$n-2⩽x<n-1\Rightarrow \lfloor x\rfloor =n-2\Rightarrow A_{n-1}=n-2$$$$n-1⩽x<n\Rightarrow \lfloor x\rfloor =n-1\Rightarrow A_n=n-1$$$$S={\int }_0^n\lfloor x\rfloor dx$$$$Bythegeometricalinterpretationof$$$$theintegral,$$$$S=\underset{i=1}{\sum ^n}{A}_i=0+1+2+3+...+n-1=\frac{n(n-1)}{2}.$$$$$$$$Alternatively,fori\in \mathbb{N},$$$$S=\underset{i=1}{\sum ^n}\{{\int }_{i-1}^i\lfloor x\rfloor dx\}$$$$S=\underset{i=1}{\sum ^n}\left\{{\int }_{i-1}^i(i-1)dx\right\}(i-1⩽x<i\Rightarrow \lfloor x\rfloor =i-1)$$$$S=\underset{i=1}{\sum ^n}(i-1)\left(x{\mid }_{i-1}^i\right)$$$$S=\underset{i=1}{\sum ^n}(i-1)(i-i+1)$$$$S=\underset{i=1}{\sum ^n}(i-1)=\frac{n(n+1)}{2}-n=\frac{n(n-1)}{2}$$

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