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.](Q126990.png)
$$\mathrm{Obtain}\:\mathrm{a}\:\mathrm{formula}\:\mathrm{for}\: \\ $$$$\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{{n}} {\int}}\left[{x}\right]\:{dx}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\:\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}\:\mathrm{of}\:{x}. \\ $$
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)](Q127010.png)
$$\:{I}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{{n}} {\int}}\left[{x}\right]\:{dx} \\ $$$$\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\:\int_{{k}} ^{{k}+\mathrm{1}} \left[{x}\right]\:{dx} \\ $$$$\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k}\:\int_{{k}} ^{{k}+\mathrm{1}} \:{dx} \\ $$$$\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$