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I-0-n-x-x-dx-n-Z-




Question Number 6333 by FilupSmith last updated on 24/Jun/16
I=∫_0 ^( n) ⌊x⌋⌈x⌉dx,  n∈Z
$${I}=\int_{\mathrm{0}} ^{\:{n}} \lfloor{x}\rfloor\lceil{x}\rceil{dx},\:\:{n}\in\mathbb{Z} \\ $$
Commented by nburiburu last updated on 24/Jun/16
I=Σ_(i=0) ^n i(i+1)
$${I}=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{i}\left({i}+\mathrm{1}\right) \\ $$
Commented by prakash jain last updated on 24/Jun/16
should it be=Σ_(i=0) ^(n−1) i(i+1)
$${should}\:{it}\:{be}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{i}\left({i}+\mathrm{1}\right) \\ $$
Commented by nburiburu last updated on 27/Jun/16
I= Σ^(n−1) i^2  +Σ^(n−1) i= (((n−1)(n−1+1)(2n−2+1))/6)+(((n−1)(n−1+1))/2)    = (n−1)(n−1+1)((2n−2+4)/6)
$${I}=\:\overset{{n}−\mathrm{1}} {\sum}{i}^{\mathrm{2}} \:+\overset{{n}−\mathrm{1}} {\sum}{i}=\:\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}+\mathrm{1}\right)}{\mathrm{6}}+\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)}{\mathrm{2}}\:\: \\ $$$$=\:\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)\frac{\mathrm{2}{n}−\mathrm{2}+\mathrm{4}}{\mathrm{6}} \\ $$

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