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(D) We need to evaluate the integral $I = \int_{0}^{3} \{x\}^{[x]} \, dx$. Since the function involves $[x]$,we split the integral at integer points:

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$11/6$

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(D) We need to evaluate the integral $I = \int_{0}^{3} \{x\}^{[x]} \, dx$. Since the function involves $[x]$,we split the integral at integer points: $I = \int_{0}^{1} \{x\}^{[x]} \, dx + \int_{1}^{2} \{x\}^{[x]} \, dx + \int_{2}^{3} \{x\}^{[x]} \, dx$. For $x \in [0, 1)$,$[x] = 0$,so $\{x\}^{[x]} = \{x\}^{0} = 1$ (for $x 
eq 0$). Thus,$\int_{0}^{1} 1 \, dx = 1$. For $x \in [1, 2)$,$[x] = 1$,so $\{x\}^{[x]} = \{x\}^{1} = x - 1$. Thus,$\int_{1}^{2} (x - 1) \, dx = [\frac{(x-1)^2}{2}]_{1}^{2} = \frac{1}{2} - 0 = \frac{1}{2}$. For $x \in [2, 3)$,$[x] = 2$,so $\{x\}^{[x]} = \{x\}^{2} = (x - 2)^2$. Thus,$\int_{2}^{3} (x - 2)^2 \, dx = [\frac{(x-2)^3}{3}]_{2}^{3} = \frac{1}{3} - 0 = \frac{1}{3}$. Adding these values: $I = 1 + \frac{1}{2} + \frac{1}{3} = \frac{6+3+2}{6} = \frac{11}{6}$.

Let $y = \{x\}^{[x]}$ where $\{x\}$ denotes the fractional part of $x$ and $[x]$ denotes the greatest integer $\le x$. Then $\int_{0}^{3} y \, dx = $

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