The value of $\frac{\int_{0}^{n}{\lbrack x\rbrack dx}}{\int_{0}^{n}{\left\{ x \right\} dx}}$ is, (where $\lbrack x\rbrack\ and\ \{ x\}$ denotes the integral part and fractional part functions of $x$ and $n \in N)$

Question

Dr. Abubacker · Accepted Answer

(d)
$\because\ \int_{0}^{n}{\lbrack xbrack dx = \int_{0}^{n}{\left( x - \left\{ x ight\} ight)dx = \int_{0}^{n}{x\ dx -}\int_{0}^{n}{\left\{ x ight\} dx}}}$
$= \int_{0}^{n}{x\ dx - n}\int_{0}^{1}{\left\{ x ight\} dx}$
$= \int_{0}^{n}{x\ dx -}n\int_{0}^{1}{x\ dx} = \left( \frac{n^{2}}{2} - \frac{n}{2} ight)$
$	herefore\ \frac{\int_{0}^{n}{\lbrack xbrack dx}}{\int_{0}^{n}{\{ x\}}dx} = \frac{\frac{n^{2}}{2} - \frac{n}{2}}{\frac{n}{2}} = n - 1$

Q. The value of $\frac{\int_{0}^{n}{\lbrack x\rbrack dx}}{\int_{0}^{n}{\left\{ x \right\} dx}}$ is, (where $\lbrack x\rbrack\ and\ \{ x\}$ denotes the integral part and fractional part functions of $x$ and $n \in N)$

Explanation

Reviewed by Dr. Abubacker

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