Let I_{1}=int_{0}^{n}[x] d x and I_{2}=int_{0 | vedclass

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(D) We have $I_{1} = \int_{0}^{n} [x] dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k dx = \sum_{k=0}^{n-1} k(k+1-k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$.N

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$n-1$

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(D) We have $I_{1} = \int_{0}^{n} [x] dx = \sum_{k=0}^{n-1} \int_{k}^{k+1} k dx = \sum_{k=0}^{n-1} k(k+1-k) = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}$.Now, $I_{2} = \int_{0}^{n} \{x\} dx$. Since $\{x\} = x - [x]$, we have $I_{2} = \int_{0}^{n} x dx - \int_{0}^{n} [x] dx$.$I_{2} = \left[ \frac{x^2}{2} \right]_{0}^{n} - I_{1} = \frac{n^2}{2} - \frac{n(n-1)}{2} = \frac{n^2 - n^2 + n}{2} = \frac{n}{2}$.Therefore, $\frac{I_{1}}{I_{2}} = \frac{\frac{n(n-1)}{2}}{\frac{n}{2}} = n-1$.

Let $I_{1}=\int_{0}^{n}[x] d x$ and $I_{2}=\int_{0}^{n}\{x\} d x,$ where $[x]$ and $\{x\}$ are the integral and fractional parts of $x$ respectively, and $n \in N-\{1\}.$ Then, $I_{1} / I_{2}$ is equal to

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