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Question Number 199213 by emilagazade last updated on 29/Oct/23
Answered by witcher3 last updated on 29/Oct/23
∫ab(x−[x]−12)f′(x)=dx=∑b−1k=a∫kk+1(x−[x]−12)f′(x)dx=∑b−1k=a∫kk+1(x−k−12)f′(x)=∑b−1k=a[[x−k−12]kk+1f(x)−∫kk+1f(x)dx=∑b−1k=af(k+1)+f(k)2−∑b−1a∫kk+1f(x)dx==∑bk=af(k)−f(a)+f(b)2−∫abf(x)dx⇔Σf(k)=∫abf(x)dx+f(a)+f(b)2+∫ab(x−[x]−12)f′(x)dx
Commented by emilagazade last updated on 29/Oct/23
nicethanks
Commented by witcher3 last updated on 30/Oct/23
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