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Question Number 126990 by physicstutes last updated on 26/Dec/20

Obtain a formula for    I_n  = ∫_0 ^n [x] dx in terms of n   where [x] is the greatest integer function of x.

$$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\: \\ $$$$\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{{n}} {\int}}\left[{x}\right]\:{dx}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\:\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}\:\mathrm{of}\:{x}. \\ $$

Answered by mr W last updated on 26/Dec/20

 I_n  = ∫_0 ^n [x] dx    =Σ_(k=0) ^(n−1)  ∫_k ^(k+1) [x] dx    =Σ_(k=0) ^(n−1) k ∫_k ^(k+1)  dx    =Σ_(k=0) ^(n−1) k=((n(n−1))/2)

$$\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{{n}} {\int}}\left[{x}\right]\:{dx} \\ $$$$\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\int_{{k}} ^{{k}+\mathrm{1}} \left[{x}\right]\:{dx} \\ $$$$\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k}\:\int_{{k}} ^{{k}+\mathrm{1}} \:{dx} \\ $$$$\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$

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