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(B) Let $I = \int_0^n \cos(2 \pi [x] \{x\}) dx$.Since $[x] = k$ for $x \in [k, k+1)$ where $k$ is an integer,we can split the integral as:$I = \sum_{k

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(B) Let $I = \int_0^n \cos(2 \pi [x] \{x\}) dx$.Since $[x] = k$ for $x \in [k, k+1)$ where $k$ is an integer,we can split the integral as:$I = \sum_{k=0}^{n-1} \int_k^{k+1} \cos(2 \pi k (x-k)) dx$.For $k=0$,the integral is $\int_0^1 \cos(0) dx = \int_0^1 1 dx = 1$.For $k \geq 1$,let $u = x-k$,then $du = dx$. The integral becomes $\int_0^1 \cos(2 \pi k u) du$.This is $\left[ \frac{\sin(2 \pi k u)}{2 \pi k} \right]_0^1 = \frac{\sin(2 \pi k) - \sin(0)}{2 \pi k} = \frac{0-0}{2 \pi k} = 0$.Thus,$I = 1 + 0 + 0 + \dots + 0 = 1$.

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$ and $\{x\} = x - [x]$. Let $n$ be a positive integer. Then,$\int_0^n \cos(2 \pi [x] \{x\}) dx$ is equal to

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