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Question Number 3900 by Filup last updated on 24/Dec/15
I asked this question a while ago, but I forgot how to solve it: S=∫_0 ^( n) ⌊x⌋dx
$$\mathrm{I}\:\mathrm{asked}\:\mathrm{this}\:\mathrm{question}\:\mathrm{a}\:\mathrm{while}\:\mathrm{ago}, \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{forgot}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}: \\ $$$$ \\ $$$${S}=\int_{\mathrm{0}} ^{\:{n}} \lfloor{x}\rfloor{dx} \\ $$
Answered by Yozzii last updated on 24/Dec/15
0≤x<1⇒⌊x⌋=0⇒A_1 =0×(1−0)=0 1≤x<2⇒⌊x⌋=1⇒A_2 =1(2−1)=1 2≤x<3⇒⌊x⌋=2⇒A_3 =2(3−2)=2 3≤x<4⇒⌊x⌋=3⇒A_4 =3(4−3)=3 ⋮ n−2≤x<n−1⇒⌊x⌋=n−2⇒A_(n−1) =n−2 n−1≤x<n⇒⌊x⌋=n−1⇒A_n =n−1 S=∫_0 ^n ⌊x⌋dx By the geometrical interpretation of the integral, S=Σ_(i=1) ^n A_i =0+1+2+3+...+n−1=((n(n−1))/2). Alternatively, for i∈N, S=Σ_(i=1) ^n {∫_(i−1) ^i ⌊x⌋dx} S=Σ_(i=1) ^n {∫_(i−1) ^i (i−1)dx} (i−1≤x<i⇒⌊x⌋=i−1) S=Σ_(i=1) ^n (i−1)(x∣_(i−1) ^i ) S=Σ_(i=1) ^n (i−1)(i−i+1) S=Σ_(i=1) ^n (i−1)=((n(n+1))/2)−n=((n(n−1))/2)
$$\mathrm{0}\leqslant{x}<\mathrm{1}\Rightarrow\lfloor{x}\rfloor=\mathrm{0}\Rightarrow{A}_{\mathrm{1}} =\mathrm{0}×\left(\mathrm{1}−\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{1}\leqslant{x}<\mathrm{2}\Rightarrow\lfloor{x}\rfloor=\mathrm{1}\Rightarrow{A}_{\mathrm{2}} =\mathrm{1}\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{2}\leqslant{x}<\mathrm{3}\Rightarrow\lfloor{x}\rfloor=\mathrm{2}\Rightarrow{A}_{\mathrm{3}} =\mathrm{2}\left(\mathrm{3}−\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{3}\leqslant{x}<\mathrm{4}\Rightarrow\lfloor{x}\rfloor=\mathrm{3}\Rightarrow{A}_{\mathrm{4}} =\mathrm{3}\left(\mathrm{4}−\mathrm{3}\right)=\mathrm{3} \\ $$$$\vdots \\ $$$${n}−\mathrm{2}\leqslant{x}<{n}−\mathrm{1}\Rightarrow\lfloor{x}\rfloor={n}−\mathrm{2}\Rightarrow{A}_{{n}−\mathrm{1}} ={n}−\mathrm{2} \\ $$$${n}−\mathrm{1}\leqslant{x}<{n}\Rightarrow\lfloor{x}\rfloor={n}−\mathrm{1}\Rightarrow{A}_{{n}} ={n}−\mathrm{1} \\ $$$${S}=\int_{\mathrm{0}} ^{{n}} \lfloor{x}\rfloor{dx} \\ $$$${By}\:{the}\:{geometrical}\:{interpretation}\:{of} \\ $$$${the}\:{integral}, \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{A}_{{i}} =\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}−\mathrm{1}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}. \\ $$$$ \\ $$$${Alternatively},\:{for}\:{i}\in\mathbb{N}, \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{\int_{{i}−\mathrm{1}} ^{{i}} \lfloor{x}\rfloor{dx}\right\} \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{\int_{{i}−\mathrm{1}} ^{{i}} \left({i}−\mathrm{1}\right){dx}\right\}\:\:\:\:\:\:\left({i}−\mathrm{1}\leqslant{x}<{i}\Rightarrow\lfloor{x}\rfloor={i}−\mathrm{1}\right) \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}−\mathrm{1}\right)\left({x}\mid_{{i}−\mathrm{1}} ^{{i}} \right) \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}−\mathrm{1}\right)\left({i}−{i}+\mathrm{1}\right) \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({i}−\mathrm{1}\right)=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$

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