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Question Number 156136 by SANOGO last updated on 08/Oct/21

$$soitE\left(x\right)lapartieentiere,p<q$$ $$alorslavaleurde$$ $${\underset{p}{\int }}^{q}E\left(x\right)dx=?$$

Answered by KONE last updated on 10/Oct/21

$${\int }_p^{q}E\left(x\right)dx={\int }_p^{p+1}E\left(x\right)dx+{\int }_{p+1^{}}^{p+2}E\left(x\right)dx+.....+{\int }_{q-2}^{q-1}E\left(x\right)dx+{\int }_{q-1}^{q}E\left(x\right)dx$$ $$=\underset{k=p}{\sum ^{q-1}}{\int }_k^{k+1}E\left(x\right)dx=\underset{k=p}{\sum ^q}k=(q-1-p+1)\frac{q+p-1}{2}$$ $$donc{\int }_p^{q}E\left(x\right)dx=(q+p-1)\frac{q-p}{2}\underset{―}{\mathcal{KAB}}$$

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