prove-that-0-n-x-dx-n-n-1-2-and-0-n-x-dx-n-n-1-2-when-is-floor-and-is-ceil-

Question Number 88852 by M±th+et£s last updated on 13/Apr/20

p r o v e t h a t

\int_{0}^{n} ⌈ x ⌉ d x = \frac{n (n + 1)}{2} a n d \int_{0}^{n} ⌊ x ⌋ d x = \frac{n (n - 1)}{2}

w h e n ⌊ . . ⌋ i s f l o o r a n d ⌈ . . ⌉ i s c e i l

Answered by mr W last updated on 13/Apr/20

\int_{0}^{n} ⌈ x ⌉ d x = \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} ⌈ x ⌉ d x = \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} (k + 1) d x

= \underset{k = 0}{\sum^{n - 1}} (k + 1) = \underset{k = 1}{\sum^{n}} k = \frac{n (n + 1)}{2}

\int_{0}^{n} ⌊ x ⌋ d x = \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} ⌊ x ⌋ d x = \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} k d x

= \underset{k = 0}{\sum^{n - 1}} k = \frac{(n - 1) n}{2}

Commented by M±th+et£s last updated on 13/Apr/20

t h a n k y o u s i r