Summing up the floor

Calculus Level 4

$\large I = \int_{0}^{x} \lfloor t \rfloor \ dt$

Find $I$ in terms of $x$ , where $x$ is a real positive number

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

None of these choices

\dfrac{1}{2}x(x-1)

\dfrac{1}{2}x(x+1)

\dfrac{1}{2} \lfloor x \rfloor \left( x+\{ x \} - 1 \right)

\dfrac{1}{2} \lfloor x \rfloor \left( x+\{ x \} +1 \right)

1 solution

Sabhrant Sachan
Jul 9, 2016

$I=\displaystyle \int_{0}^{x} \lfloor t \rfloor \cdot dt = \displaystyle \int_{0}^{1} 0\cdot dt+\displaystyle \int_{1}^{2} 1\cdot dt+\cdots+\displaystyle \int_{\lfloor x \rfloor -1}^{\lfloor x \rfloor } \lfloor x \rfloor-1\cdot dt+\displaystyle \int_{\lfloor x \rfloor }^{x}\lfloor x \rfloor \cdot dt \\ I = 0+1+2+3+\cdots +\lfloor x \rfloor-1+\lfloor x \rfloor(x-\lfloor x \rfloor) \\ I = \dfrac{\lfloor x \rfloor(\lfloor x \rfloor-1)}{2}+\lfloor x \rfloor \cdot \{ x \} \\ I = \dfrac{\lfloor x \rfloor}{2} \left( \lfloor x \rfloor-1+2\{ x \} \right) \\ \boxed{I = \dfrac{\lfloor x \rfloor}{2} \left( x+\{ x\}-1 \right) }$

Summing up the floor

1 solution

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