मान लीजिए f(x) = min {[x-1], [x-2], ldots, [x | vedclass

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(B) दिया गया है $f(x) = \min \{[x-1], [x-2], \ldots, [x-10]\}$.चूंकि $[x-n] = [x]-n$,इसलिए $f(x) = \min \{[x]-1, [x]-2, \ldots, [x]-10\} = [x]-10$.जब

Vedclass · Accepted Answer

$385$

Vedclass · Answer

(B) दिया गया है $f(x) = \min \{[x-1], [x-2], \ldots, [x-10]\}$.चूंकि $[x-n] = [x]-n$,इसलिए $f(x) = \min \{[x]-1, [x]-2, \ldots, [x]-10\} = [x]-10$.जब $x \in [n, n+1)$,तब $[x] = n$,इसलिए $f(x) = n-10$ जहाँ $n = 0, 1, \ldots, 9$.$\int_{0}^{10} f(x) \, dx = \sum_{n=0}^{9} \int_{n}^{n+1} (n-10) \, dx = \sum_{n=0}^{9} (n-10) = (-10) + (-9) + \ldots + (-1) = -\frac{10 	imes 11}{2} = -55$.$\int_{0}^{10} (f(x))^2 \, dx = \sum_{n=0}^{9} \int_{n}^{n+1} (n-10)^2 \, dx = \sum_{n=0}^{9} (n-10)^2 = (-10)^2 + (-9)^2 + \ldots + (-1)^2 = 10^2 + 9^2 + \ldots + 1^2 = \frac{10 	imes 11 	imes 21}{6} = 385$.$\int_{0}^{10} |f(x)| \, dx = \sum_{n=0}^{9} \int_{n}^{n+1} |n-10| \, dx = \sum_{n=0}^{9} |n-10| = 10 + 9 + \ldots + 1 = \frac{10 	imes 11}{2} = 55$.अतः,योग $-55 + 385 + 55 = 385$ है।

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