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Question Number 181020 by mr W last updated on 20/Nov/22

$$solveforx>0$$ $${\int }_0^x\lfloor t{\rfloor }^{2}dt=2(x-1)$$ $$$$ $$\left(Q180780reposted\right)$$

Answered by mr W last updated on 20/Nov/22

$$sayx=n+fwith0⩽f<1$$ $$\lfloor x\rfloor =n$$ $${\int }_0^x\lfloor t{\rfloor }^{2}dt=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}\lfloor t{\rfloor }^{2}dt+{\int }_n^x\lfloor t{\rfloor }^{2}dt$$ $$=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}{k}^{2}dt+{\int }_n^x{n}^{2}dt$$ $$=\underset{k=0}{\sum ^{n-1}}{k}^2+n^2(x-n)$$ $$=\frac{(n-1)n(2n-1)}{6}+n^2(x-n)$$ $$\frac{(n-1)n(2n-1)}{6}+n^2(x-n)=2(x-1)$$ $$\frac{(n-1)n(2n-1)}{6}+n^{2}f=2(n+f-1)$$ $$(2n^2-n-12)(n-1)=6(2-n^2)f$$ $$n=1:$$ $$f=0\Rightarrow x=1$$ $$n=2:$$ $$(8-2-12)(2-1)=6(2-4)f$$ $$f=\frac{1}{2}\Rightarrow x=2+\frac{1}{2}=\frac{5}{2}$$ $$i.e.therearetworoots:x=1,\frac{5}{2}.$$

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