Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$provethat$$$${\int }_0^n\lceil x\rceil dx=\frac{n(n+1)}{2}and{\int }_0^n\lfloor x\rfloor dx=\frac{n(n-1)}{2}$$$$when\lfloor ..\rfloor isfloorand\lceil ..\rceil isceil$$

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

$${\int }_0^n\lceil x\rceil dx=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}\lceil x\rceil dx=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}(k+1)dx$$$$=\underset{k=0}{\sum ^{n-1}}(k+1)=\underset{k=1}{\sum ^n}k=\frac{n(n+1)}{2}$$$$$$$${\int }_0^n\lfloor x\rfloor dx=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}\lfloor x\rfloor dx=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}kdx$$$$=\underset{k=0}{\sum ^{n-1}}k=\frac{(n-1)n}{2}$$

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

$$thankyousir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com