solve-for-x-gt-0-0-x-t-2-dt-2-x-1-Q180780-reposted-

Question Number 181020 by mr W last updated on 20/Nov/22

s o l v e f o r x > 0

\int_{0}^{x} ⌊ t ⌋^{2} d t = 2 (x - 1)

(Q 180780 r e p o s t e d)

Answered by mr W last updated on 20/Nov/22

s a y x = n + f w i t h 0 ⩽ f < 1

⌊ x ⌋ = n

\int_{0}^{x} ⌊ t ⌋^{2} d t = \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} ⌊ t ⌋^{2} d t + \int_{n}^{x} ⌊ t ⌋^{2} d t

= \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} k^{2} d t + \int_{n}^{x} n^{2} d t

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

= \frac{(n - 1) n (2 n - 1)}{6} + n^{2} (x - n)

\frac{(n - 1) n (2 n - 1)}{6} + n^{2} (x - n) = 2 (x - 1)

\frac{(n - 1) n (2 n - 1)}{6} + n^{2} f = 2 (n + f - 1)

(2 n^{2} - n - 12) (n - 1) = 6 (2 - n^{2}) f

n = 1 :

f = 0 \Rightarrow x = 1

n = 2 :

(8 - 2 - 12) (2 - 1) = 6 (2 - 4) f

f = \frac{1}{2} \Rightarrow x = 2 + \frac{1}{2} = \frac{5}{2}

i . e . t h e r e a r e t w o r o o t s : x = 1, \frac{5}{2} .

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