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(B) The fractional part function $\{x\}$ is defined as $\{x\} = x - \lfloor x \rfloor$.For the interval $[0, 2)$, we have:$\{x\} = x$ for $0 \leq x <

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(B) The fractional part function $\{x\}$ is defined as $\{x\} = x - \lfloor x \rfloor$.For the interval $[0, 2)$, we have:$\{x\} = x$ for $0 \leq x $\{x\} = x - 1$ for $1 \leq x Therefore, the integral can be split as:$\int_0^2 \{x\} dx = \int_0^1 x dx + \int_1^2 (x - 1) dx$Evaluating the first part: $\int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$Evaluating the second part: $\int_1^2 (x - 1) dx$. Let $u = x - 1$, then $du = dx$. When $x=1, u=0$; when $x=2, u=1$.$\int_0^1 u du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}$Adding both parts: $\frac{1}{2} + \frac{1}{2} = 1$.

Let $\{x\}$ denote the fractional part of a real number $x$. Then $\int_0^2 \{x\} dx = $

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