The minimum value of the function f(x) = int | vedclass

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(A) For $x \leq 0$,$f(x) = \int \limits_0^2 e^{t-x} dt = e^{-x}(e^2-1)$.For $0 < x < 2$,$f(x) = \int \limits_0^x e^{x-t} dt + \int \limits_x^2 e^{t-x}

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$2(e-1)$

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(A) For $x \leq 0$,$f(x) = \int \limits_0^2 e^{t-x} dt = e^{-x}(e^2-1)$.For $0 For $x \geq 2$,$f(x) = \int \limits_0^2 e^{x-t} dt = e^{x-2}(e^2-1)$.For $x \leq 0$,$f(x)$ is decreasing and for $x \geq 2$,$f(x)$ is increasing.Therefore,the minimum value of $f(x)$ lies in the interval $x \in (0, 2)$.In the interval $(0, 2)$,$f(x) = e^x + e^{2-x} - 2$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$:$e^x + e^{2-x} \geq 2 \sqrt{e^x \cdot e^{2-x}} = 2 \sqrt{e^2} = 2e$.Thus,the minimum value of $f(x)$ is $2e - 2 = 2(e-1)$.

The minimum value of the function $f(x) = \int \limits_0^2 e^{|x-t|} dt$ is

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