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Question Number 88852 by M±th+et£s last updated on 13/Apr/20

prove that ∫_0 ^n ⌈x⌉dx= ((n(n+1))/2) and ∫_0 ^n ⌊x⌋dx=((n(n−1))/2) when ⌊..⌋ is floor and ⌈..⌉ is ceil

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{{n}} \lceil{x}\rceil{dx}=\:\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:{and}\:\int_{\mathrm{0}} ^{{n}} \lfloor{x}\rfloor{dx}=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${when}\:\lfloor..\rfloor\:{is}\:{floor}\:{and}\:\lceil..\rceil\:{is}\:{ceil} \\ $$

Answered by mr W last updated on 13/Apr/20

∫_0 ^n ⌈x⌉dx=Σ_(k=0) ^(n−1) ∫_k ^(k+1) ⌈x⌉dx=Σ_(k=0) ^(n−1) ∫_k ^(k+1) (k+1)dx =Σ_(k=0) ^(n−1) (k+1)=Σ_(k=1) ^n k=((n(n+1))/2) ∫_0 ^n ⌊x⌋dx=Σ_(k=0) ^(n−1) ∫_k ^(k+1) ⌊x⌋dx=Σ_(k=0) ^(n−1) ∫_k ^(k+1) kdx =Σ_(k=0) ^(n−1) k=(((n−1)n)/2)

$$\int_{\mathrm{0}} ^{{n}} \lceil{x}\rceil{dx}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lceil{x}\rceil{dx}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \left({k}+\mathrm{1}\right){dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left({k}+\mathrm{1}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{{n}} \lfloor{x}\rfloor{dx}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \lfloor{x}\rfloor{dx}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} {kdx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{k}=\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}} \\ $$

Commented by M±th+et£s last updated on 13/Apr/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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