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Question Number 199213 by emilagazade last updated on 29/Oct/23

Answered by witcher3 last updated on 29/Oct/23

$${\int }_{\mathrm{a}}^{\mathrm{b}}(\mathrm{x}-\left[\mathrm{x}\right]-\frac{1}{2}){\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\mathrm{dx}$$$$=\underset{\mathrm{k}=\mathrm{a}}{\sum ^{\mathrm{b}-1}}{\int }_{\mathrm{k}}^{\mathrm{k}+1}(\mathrm{x}-\left[\mathrm{x}\right]-\frac{1}{2}){\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{dx}$$$$=\underset{\mathrm{k}=\mathrm{a}}{\sum ^{\mathrm{b}-1}}{\int }_{\mathrm{k}}^{\mathrm{k}+1}(\mathrm{x}-\mathrm{k}-\frac{1}{2}){\mathrm{f}}^{\prime }\left(\mathrm{x}\right)$$$$=\underset{\mathrm{k}=\mathrm{a}[}{\sum ^{\mathrm{b}-1}}{[\mathrm{x}-\mathrm{k}-\frac{1}{2}]}_{\mathrm{k}}^{\mathrm{k}+1}\mathrm{f}\left(\mathrm{x}\right)-{\int }_{\mathrm{k}}^{\mathrm{k}+1}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$$$$=\underset{\mathrm{k}=\mathrm{a}}{\sum ^{\mathrm{b}-1}}\frac{\mathrm{f}(\mathrm{k}+1)+\mathrm{f}\left(\mathrm{k}\right)}{2}-\underset{\mathrm{a}}{\sum ^{\mathrm{b}-1}}{\int }_{\mathrm{k}}^{\mathrm{k}+1}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$$$$==\underset{\mathrm{k}=\mathrm{a}}{\sum ^{\mathrm{b}}}\mathrm{f}\left(\mathrm{k}\right)-\frac{\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}\right)}{2}-{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\iff$$$$\mathrm{\Sigma }\mathrm{f}\left(\mathrm{k}\right)={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\frac{\mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}\right)}{2}+{\int }_{\mathrm{a}}^{\mathrm{b}}(\mathrm{x}-\left[\mathrm{x}\right]-\frac{1}{2}){\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{dx}$$

Commented by emilagazade last updated on 29/Oct/23

$$nicethanks$$

Commented by witcher3 last updated on 30/Oct/23

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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