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Sum of Coefficients of All Terms

Prove that ...

Question

Prove that $\int_{0}^{x} [t] d t = \frac{[x] ([x] - 1)}{2} + [x] (x - [x])$ , where $[\cdot]$ denotes the greatest integer function.

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Solution

Given : $\int_{0}^{x} [t] d t = \frac{[x] ([x] - 1)}{2} + [x] (x - [x])$
$I = \int_{0}^{x} [t] d t I = \int_{x = 0}^{1} [t] d t + \int_{x = 1}^{2} [t] d t + \int_{x = 2}^{3} [t] d t . . . . . . . . . + \int_{x = [x]}^{x} [t] d t I = 0 + \int_{1}^{2} 1 d t + 2 \int_{2}^{3} d t . . . . . . . . . + [x] \int_{x = [x]}^{x} d t I = 1 (2 - 1) + 2 (3 - 2) + . . . . . . . . .1 . ([x] - 1) + [x] (x - [x]) I = 1 + 2 + 3 + . . . . . . . . . ([x] - 1) + [x] (x - [x]) I = \sum_{i = 1}^{\infty - 1} n + [x] (x - [x]) I = \frac{[x] ([x] - 1)}{2} + [x] (x - [x])$
Hence proved that $\int_{0}^{x} [t] d t = \frac{[x] ([x] - 1)}{2} + [x] (x - [x])$

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