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(D) Let $I = \int_1^n [x]\{x\} dx$.Since $[x] = k$ for $x \in [k, k+1)$,we can write the integral as:$I = \sum_{k=1}^{n-1} \int_k^{k+1} k\{x\} dx = \s

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$91$

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(D) Let $I = \int_1^n [x]\{x\} dx$.Since $[x] = k$ for $x \in [k, k+1)$,we can write the integral as:$I = \sum_{k=1}^{n-1} \int_k^{k+1} k\{x\} dx = \sum_{k=1}^{n-1} k \int_k^{k+1} (x-k) dx$.Let $u = x-k$,then $du = dx$. When $x=k, u=0$ and when $x=k+1, u=1$.So,$\int_k^{k+1} (x-k) dx = \int_0^1 u du = \left[ \frac{u^2}{2} ight]_0^1 = \frac{1}{2}$.Thus,$I = \sum_{k=1}^{n-1} k \cdot \frac{1}{2} = \frac{1}{2} \sum_{k=1}^{n-1} k = \frac{1}{2} \cdot \frac{(n-1)n}{2} = \frac{n(n-1)}{4}$.We are given $I > 2013$,so $\frac{n(n-1)}{4} > 2013$.$n(n-1) > 8052$.Since $90 	imes 89 = 8010$ and $91 	imes 90 = 8190$,the smallest integer $n$ satisfying the inequality is $91$.

For a real number $x$,let $[x]$ denote the largest integer less than or equal to $x$ and $\{x\}=x-[x]$. The smallest possible integer value of $n$ for which $\int_1^n [x]\{x\} dx$ exceeds $2013$ is

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