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(B) Let $I = \int_{1}^{n} [x][\sqrt{x}] \, dx$. We evaluate the integral by splitting it into intervals $[k, k+1)$ where $[x]$ is constant. For $x \in

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$9$

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(B) Let $I = \int_{1}^{n} [x][\sqrt{x}] \, dx$. We evaluate the integral by splitting it into intervals $[k, k+1)$ where $[x]$ is constant. For $x \in [k, k+1)$,$[x] = k$. Thus,$I = \sum_{k=1}^{n-1} \int_{k}^{k+1} k [\sqrt{x}] \, dx$. Calculating the values for each interval: For $k=1, x \in [1, 2), [\sqrt{x}] = 1 \implies \int_{1}^{2} 1 \cdot 1 \, dx = 1$. For $k=2, x \in [2, 3), [\sqrt{x}] = 1 \implies \int_{2}^{3} 2 \cdot 1 \, dx = 2$. For $k=3, x \in [3, 4), [\sqrt{x}] = 1 \implies \int_{3}^{4} 3 \cdot 1 \, dx = 3$. For $k=4, x \in [4, 5), [\sqrt{x}] = 2 \implies \int_{4}^{5} 4 \cdot 2 \, dx = 8$. For $k=5, x \in [5, 6), [\sqrt{x}] = 2 \implies \int_{5}^{6} 5 \cdot 2 \, dx = 10$. For $k=6, x \in [6, 7), [\sqrt{x}] = 2 \implies \int_{6}^{7} 6 \cdot 2 \, dx = 12$. For $k=7, x \in [7, 8), [\sqrt{x}] = 2 \implies \int_{7}^{8} 7 \cdot 2 \, dx = 14$. For $k=8, x \in [8, 9), [\sqrt{x}] = 2 \implies \int_{8}^{9} 8 \cdot 2 \, dx = 16$. Summing these values: $1 + 2 + 3 + 8 + 10 + 12 + 14 + 16 = 66$. Since $66 > 60$,the smallest integer $n$ is $9$.

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$. The smallest positive integer $n$ for which the integral $\int_{1}^{n} [x][\sqrt{x}] \, dx$ exceeds $60$ is

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