a 1, ; int_{1}^{a} [x] f'(x) dx = | vedclass

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(B) Let $n = [a]$,where $n$ is an integer such that $n \leq a < n+1$.The integral can be split as:$\int_{1}^{a} [x] f'(x) dx = \int_{1}^{2} 1 f'(x) dx

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$[a] f(a) - \{f(1) + f(2) + \dots + f([a])\}$

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(B) Let $n = [a]$,where $n$ is an integer such that $n \leq a The integral can be split as:$\int_{1}^{a} [x] f'(x) dx = \int_{1}^{2} 1 f'(x) dx + \int_{2}^{3} 2 f'(x) dx + \dots + \int_{n-1}^{n} (n-1) f'(x) dx + \int_{n}^{a} n f'(x) dx$Evaluating each integral:$= 1[f(2) - f(1)] + 2[f(3) - f(2)] + \dots + (n-1)[f(n) - f(n-1)] + n[f(a) - f(n)]$Rearranging the terms:$= -f(1) - f(2) - f(3) - \dots - f(n) + n f(a)$Since $n = [a]$,we get:$= [a] f(a) - \{f(1) + f(2) + \dots + f([a])\}$

$a > 1, \; \int_{1}^{a} [x] f'(x) dx = $

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