Let the function f :[0,2] rightarrow R be def | vedclass

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(C) For $x \in [0, 1)$,we have $[x] = 0$,so $x - [x] = x$. Since $x^2 \le x$ for $x \in [0, 1]$,$\min \{x^2, x\} = x^2$. Thus $f(x) = e^{x^2}$.For $x

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$2e - \frac{1}{2}$

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(C) For $x \in [0, 1)$,we have $[x] = 0$,so $x - [x] = x$. Since $x^2 \le x$ for $x \in [0, 1]$,$\min \{x^2, x\} = x^2$. Thus $f(x) = e^{x^2}$.For $x \in [1, 2]$,we have $x - \log_e x$. Since $x \ge 1$,$\log_e x \ge 0$. For $x \in [1, 2]$,$1 \le x - \log_e x Now,the integral is $\int_0^2 x f(x) dx = \int_0^1 x e^{x^2} dx + \int_1^2 x e dx$.For the first part,let $u = x^2$,then $du = 2x dx$,so $\int_0^1 x e^{x^2} dx = \frac{1}{2} \int_0^1 e^u du = \frac{1}{2} [e^u]_0^1 = \frac{1}{2}(e - 1)$.For the second part,$\int_1^2 x e dx = e [\frac{x^2}{2}]_1^2 = e(\frac{4}{2} - \frac{1}{2}) = \frac{3e}{2}$.Adding them: $\frac{1}{2}(e - 1) + \frac{3e}{2} = \frac{e}{2} - \frac{1}{2} + \frac{3e}{2} = 2e - \frac{1}{2}$.

Let the function $f :[0,2] \rightarrow R$ be defined as $f(x)=\begin{cases} e^{\min \{x^2, x-[x]\}}, & x \in[0,1) \\ e^{[x-\log_e x]}, & x \in[1,2] \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then the value of the integral $\int_0^2 x f(x) dx$ is

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