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