Let the function f:[0,2] ightarrow \mathbb{R} be defined as f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x] ight\},} & x \in[0,1) \ e^{\left[x-\log _{e} x ight]}, & 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\limits_{0

Question

Let the function f:[0,2] ightarrow \mathbb{R} be defined as f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]ight\},} & x \in[0,1) \ e^{\left[x-\log _{e} xight]}, & 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\limits_{0}^{2} x f(x) d x is :

Accepted Answer

Answer: B. 2 e-\frac{1}{2} \begin{aligned} \operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}ight\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\ {\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}ight] } & =1 ; \mathrm{x} \in[1,2) \end{aligned} 	herefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}ight. \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}ight|_1 ^2 \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned}

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