A-N-1-N-2-x-dx-N-1-N-2-Z-N-1-lt-N-2-Solve-for-A-

Question Number 2645 by Filup last updated on 24/Nov/15

A = \int_{N_{1}}^{N_{2}} ⌊ x ⌋ d x

(N_{1}, N_{2}) \in Z, N_{1} < N_{2}

Solve for A

Answered by Filup last updated on 24/Nov/15

let N_{1} = N, N_{2} = N + k

(N_{1}, N_{2}) \in Z, ∴ (N, k) \in Z

N_{1} < N_{2}, ∴ k > 0

N ⩽ x < N + 1 \Rightarrow ⌊ x ⌋ = N

N + 1 ⩽ x < N + 2 \Rightarrow ⌊ x ⌋ = N + 1

⋮

N + k - 1 ⩽ x < N + k \Rightarrow ⌊ x ⌋ = N + k - 1

The area between a \to a + 1 for ⌊ x ⌋

= \int_{a}^{a + 1} ⌊ x ⌋ = a

∵ It is a rectangle with l = 1, h = a

For the total area of A, we need to sum

all the rectangular areas

\int_{N}^{N + k} ⌊ x ⌋ d x = N + (N + 1) + \dots + (N + k - 1)

= N (k) + (1 + 2 + \dots + (k - 1))

= N k + \underset{i = 1}{\sum^{k - 1}} i

= N k + \frac{1}{2} (k - 1) (1 + k - 1)

= N k + \frac{1}{2} k (k - 1)

∴ A = \int_{N}^{N + k} ⌊ x ⌋ d x = N k + \frac{1}{2} k (k - 1)

T e s t :

\int_{2}^{5} ⌊ x ⌋ d x = 9 (w o l f r a m)

N = 2, N + k = 5 \Rightarrow k = 3

A = 2 \times 3 + \frac{1}{2} 3 (3 - 1)

= 6 + \frac{1}{2} (2) 3

= 9

= T r u e

Commented by Yozzi last updated on 24/Nov/15

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

Commented by Filup last updated on 24/Nov/15

T h a n k s, i just noticed the mistake myself

haha

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