I-asked-this-question-a-while-ago-but-I-forgot-how-to-solve-it-S-0-n-x-dx-

Question Number 3900 by Filup last updated on 24/Dec/15

I asked this question a while ago,

but I forgot how to solve it :

S = \int_{0}^{n} ⌊ x ⌋ d x

Answered by Yozzii last updated on 24/Dec/15

0 ⩽ x < 1 \Rightarrow ⌊ x ⌋ = 0 \Rightarrow A_{1} = 0 \times (1 - 0) = 0

1 ⩽ x < 2 \Rightarrow ⌊ x ⌋ = 1 \Rightarrow A_{2} = 1 (2 - 1) = 1

2 ⩽ x < 3 \Rightarrow ⌊ x ⌋ = 2 \Rightarrow A_{3} = 2 (3 - 2) = 2

3 ⩽ x < 4 \Rightarrow ⌊ x ⌋ = 3 \Rightarrow A_{4} = 3 (4 - 3) = 3

⋮

n - 2 ⩽ x < n - 1 \Rightarrow ⌊ x ⌋ = n - 2 \Rightarrow A_{n - 1} = n - 2

n - 1 ⩽ x < n \Rightarrow ⌊ x ⌋ = n - 1 \Rightarrow A_{n} = n - 1

S = \int_{0}^{n} ⌊ x ⌋ d x

B y t h e g e o m e t r i c a l i n t e r p r e t a t i o n o f

t h e i n t e g r a l,

S = \underset{i = 1}{\sum^{n}} A_{i} = 0 + 1 + 2 + 3 + \dots + n - 1 = \frac{n (n - 1)}{2} .

A l t e r n a t i v e l y, f o r i \in N,

S = \underset{i = 1}{\sum^{n}} {\int_{i - 1}^{i} ⌊ x ⌋ d x}

S = \underset{i = 1}{\sum^{n}} {\int_{i - 1}^{i} (i - 1) d x} (i - 1 ⩽ x < i \Rightarrow ⌊ x ⌋ = i - 1)

S = \underset{i = 1}{\sum^{n}} (i - 1) (x ∣_{i - 1}^{i})

S = \underset{i = 1}{\sum^{n}} (i - 1) (i - i + 1)

S = \underset{i = 1}{\sum^{n}} (i - 1) = \frac{n (n + 1)}{2} - n = \frac{n (n - 1)}{2}

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