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Question-91844




Question Number 91844 by Power last updated on 03/May/20
Commented by mr W last updated on 03/May/20
you are back! the same questions seem  to be back too...
$${you}\:{are}\:{back}!\:{the}\:{same}\:{questions}\:{seem} \\ $$$${to}\:{be}\:{back}\:{too}… \\ $$
Commented by mr W last updated on 03/May/20
∫_0 ^(1000) ⌊x⌋dx  =Σ_(n=0) ^(999) ∫_n ^(n+1) ⌊x⌋dx  =Σ_(n=0) ^(999) {n∫_n ^(n+1) dx}  =Σ_(n=0) ^(999) n  =((999×1000)/2)  =499500
$$\int_{\mathrm{0}} ^{\mathrm{1000}} \lfloor{x}\rfloor{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\int_{{n}} ^{{n}+\mathrm{1}} \lfloor{x}\rfloor{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}\left\{{n}\int_{{n}} ^{{n}+\mathrm{1}} {dx}\right\} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\mathrm{999}} {\sum}}{n} \\ $$$$=\frac{\mathrm{999}×\mathrm{1000}}{\mathrm{2}} \\ $$$$=\mathrm{499500} \\ $$
Commented by Power last updated on 03/May/20
yes sir
$$\mathrm{yes}\:\mathrm{sir} \\ $$
Commented by Power last updated on 03/May/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by john santu last updated on 03/May/20
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