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Question Number 91844 by Power last updated on 03/May/20

Commented by mr W last updated on 03/May/20

$$youareback!thesamequestionsseem$$$$tobebacktoo...$$

Commented by mr W last updated on 03/May/20

$${\int }_0^{1000}\lfloor x\rfloor dx$$$$=\underset{n=0}{\sum ^{999}}{\int }_n^{n+1}\lfloor x\rfloor dx$$$$=\underset{n=0}{\sum ^{999}}\left\{n{\int }_n^{n+1}dx\right\}$$$$=\underset{n=0}{\sum ^{999}}n$$$$=\frac{999\times 1000}{2}$$$$=499500$$

Commented by Power last updated on 03/May/20

$$\mathrm{yes}\mathrm{sir}$$

Commented by Power last updated on 03/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by john santu last updated on 03/May/20

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