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Let $I_{1}=\int_{0}^{n}[x] d x$ and $I_{2}=\int_{0}^{n}\{x\} d x,$ where $[x]$
and $\{x\}$ are integral and fractional parts of $x$ and $n \in N-\{1\} .$ Then, $I_{1} / I_{2}$ is equal to
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and $\{x\}$ are integral and fractional parts of $x$ and $n \in N-\{1\} .$ Then, $I_{1} / I_{2}$ is equal to
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Verified Answer
The correct answer is:
$n-1$
We have, $I_{1}=\int_{0}^{n}[x] d x$
$=\int_{0}^{1}[x] d x+\int_{1}^{2}[x] d x$
$\quad+\int_{2}^{3}[x] d x+\ldots+\int_{n-1}^{n}[x] d x$
$=\int_{0}^{1} 0 d x+\int_{2}^{3} 1 d x+\int_{2}^{3} 2 d x$
$\quad+\ldots+\int_{n-1}^{n}(n-1) d x$
$=0+[x]_{1}^{2}+2[x]_{2}^{3}+\ldots+(n-1)[x]_{n-1}^{n}$
$=(2-1)+2(3-2)+\ldots+$
$=\frac{n(n-1)}{2}$
$\begin{aligned} \text { Now, } I_{2} &=\int_{0}^{n}\{x\} d x=\int_{0}^{n} x-[x] d x \\ &=\int_{0}^{n} x d x-\int_{0}^{n}[x] d x \\ &=\left[\frac{x^{2}}{2}\right]_{0}^{n}-I_{1} \\ &=\frac{n^{2}}{2}-\frac{n(n-1)}{2} \\ &=\frac{n^{2}-n^{2}+n}{2} \\ &=\frac{n}{2} \\ \therefore \quad \frac{I_{1}}{I_{2}}=\frac{\frac{n(n-1)}{2}}{\frac{n}{2}} \\=&(n-1) \end{aligned}$
$=\int_{0}^{1}[x] d x+\int_{1}^{2}[x] d x$
$\quad+\int_{2}^{3}[x] d x+\ldots+\int_{n-1}^{n}[x] d x$
$=\int_{0}^{1} 0 d x+\int_{2}^{3} 1 d x+\int_{2}^{3} 2 d x$
$\quad+\ldots+\int_{n-1}^{n}(n-1) d x$
$=0+[x]_{1}^{2}+2[x]_{2}^{3}+\ldots+(n-1)[x]_{n-1}^{n}$
$=(2-1)+2(3-2)+\ldots+$
$=\frac{n(n-1)}{2}$
$\begin{aligned} \text { Now, } I_{2} &=\int_{0}^{n}\{x\} d x=\int_{0}^{n} x-[x] d x \\ &=\int_{0}^{n} x d x-\int_{0}^{n}[x] d x \\ &=\left[\frac{x^{2}}{2}\right]_{0}^{n}-I_{1} \\ &=\frac{n^{2}}{2}-\frac{n(n-1)}{2} \\ &=\frac{n^{2}-n^{2}+n}{2} \\ &=\frac{n}{2} \\ \therefore \quad \frac{I_{1}}{I_{2}}=\frac{\frac{n(n-1)}{2}}{\frac{n}{2}} \\=&(n-1) \end{aligned}$
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