Let the function f:[0,2]→ℝ be defined as f(x)={emin{x2,x−[x]}, x∈[0,1)e[x−logex], x∈[1,2]where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫02xf(x)dx is

fx=ex2x∈[0,1)ex∈[1,2]∴ [x−ln⁡x]=1 for x∈[1,2]∫02xf(x)dx==∫01x.ex2dx+∫12x(e)dx=12(ex2)01+e(x22)12=12(e−1)+e(2−12)=2e−12

Let the function f:[0,2]→ℝ be defined as f(x)={emin{x2,x−[x]}, x∈[0,1)e[x−logex], x∈[1,2]where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫02xf(x)dx is

fx=ex2x∈[0,1)ex∈[1,2]∴ [x−ln⁡x]=1 for x∈[1,2]∫02xf(x)dx==∫01x.ex2dx+∫12x(e)dx=12(ex2)01+e(x22)12=12(e−1)+e(2−12)=2e−12

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Let the function $f : [0, 2] \to ℝ$ 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 $\int_{0}^{2} xf (x) dx$ is

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$2 e - 1$

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

$1 + \frac{3 e}{2}$

$(e - 1) (e^{2} + \frac{1}{2})$

answer is B.

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Detailed Solution

$f (x) = \{\begin{array}{cc} e^{x^{2}} & x \in [0, 1) \\ e & x \in [1, 2] \end{array} ∴ [x - \ln x] = 1 for x \in [1, 2]$

$\int_{0}^{2} xf (x) dx = = \int_{0}^{1} x . e^{x^{2}} dx + \int_{1}^{2} x (e) dx = \frac{1}{2} {(e^{x^{2}})}_{0}^{1} + e {(\frac{x^{2}}{2})}_{1}^{2} = \frac{1}{2} (e - 1) + e (2 - \frac{1}{2}) = 2 e - \frac{1}{2}$

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Let the function f:[0,2]→ℝ be defined as f(x)={emin{x2,x−[x]}, x∈[0,1)e[x−logex], x∈[1,2]where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫02xf(x)dx is

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Let the function $f : [0, 2] \to ℝ$ 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 $\int_{0}^{2} xf (x) dx$ is