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Question Number 180786 by Vynho last updated on 17/Nov/22

$$f\left(t\right)={\int }_0^{t}x-\lfloor x\rfloor dx$$

Answered by mr W last updated on 17/Nov/22

$$letn=\lfloor t\rfloor$$$$f\left(t\right)=\underset{k=0}{\sum ^{n-1}}{\int }_k^{k+1}(x-\lfloor x\rfloor )dx+{\int }_n^t(x-\lfloor x\rfloor )dx$$$$f\left(t\right)=\underset{k=0}{\sum ^{n-1}}{\int }_0^1(k+\xi -k)d(k+\xi )+{\int }_0^{t-n}(n+\xi -n)d(n+\xi )$$$$f\left(t\right)=\underset{k=0}{\sum ^{n-1}}{\int }_0^1\xi d\xi +{\int }_0^{t-n}\xi d\xi$$$$f\left(t\right)=\underset{k=0}{\sum ^{n-1}}\left(\frac{1}{2}\right)+\frac{{(t-n)}^2}{2}$$$$f\left(t\right)=\frac{n}{2}+\frac{{(t-n)}^2}{2}$$$$f\left(t\right)=\frac{\lfloor t\rfloor +{(t-\lfloor t\rfloor )}^2}{2}$$

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