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 202543 by mr W last updated on 29/Dec/23

Commented by mr W last updated on 29/Dec/23

$$unsolvedoldquestionQ201806$$

Answered by mr W last updated on 30/Dec/23

$$y=\lfloor x^2+1\rfloor$$$$forx\in [k,k+1):$$$$x=k+twith0⩽t<1$$$$x^2+1={(k+t)}^2+1=k^2+1+2kt+t^2$$$$y=\lfloor x^2+1\rfloor =k^2+m+1$$$$withm=\lfloor 2kt+t^2\rfloor$$$$0⩽m=\lfloor 2kt+t^2\rfloor <2k+1⩽2k$$$$m⩽2kt+t^2<m+1$$$$\sqrt{k^2+m}-k⩽t<\sqrt{k^2+m+1}-k$$$$\mathrm{\Delta }x=\sqrt{k^2+m+1}-\sqrt{k^2+m}$$$${\int }_k^{k+1}\lfloor x^2+1\rfloor dx$$$$=\underset{m=0}{\sum ^{2k}}y\mathrm{\Delta }x$$$$=\underset{m=0}{\sum ^{2k}}(k^2+m+1)(\sqrt{k^2+m+1}-\sqrt{k^2+m})$$$${\int }_0^n\lfloor x^2+1\rfloor dx=\underset{k=0}{\sum ^{n-1}}\underset{m=0}{\sum ^{2k}}(k^2+m+1)(\sqrt{k^2+m+1}-\sqrt{k^2+m})$$$$=n^3-\underset{k=1}{\sum ^{n^2-1}}\sqrt{k}$$$$example:$$$${\int }_{-4}^4\lfloor x^2+1\rfloor dx$$$$=2{\int }_0^4\lfloor x^2+1\rfloor dx$$$$=2(4^3-\sqrt{1}-\sqrt{2}-\sqrt{3}-...-\sqrt{15})$$$$\approx 47.061606799715✓$$$$$$$$anotherexample:$$$${\int }_{-10}^{10}\lfloor x^2+1\rfloor dx$$$$=2({10}^3-\sqrt{1}-\sqrt{2}-\sqrt{3}-...-\sqrt{99})$$$$\approx 677.074105793705$$

Commented by mr W last updated on 29/Dec/23

Terms of Service

Privacy Policy

Contact: info@tinkutara.com