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(B) The integral is $\int_0^3 \frac{e^x + e^{-x}}{[x]!} dx$. We split the interval $[0, 3)$ into $[0, 1), [1, 2), [2, 3)$.For $x \in [0, 1)$,$[x] = 0$

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$\frac{1}{2} (e^2 + e^3 - e^{-2} - e^{-3})$

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(B) The integral is $\int_0^3 \frac{e^x + e^{-x}}{[x]!} dx$. We split the interval $[0, 3)$ into $[0, 1), [1, 2), [2, 3)$.For $x \in [0, 1)$,$[x] = 0$,so $[x]! = 0! = 1$. The integral is $\int_0^1 (e^x + e^{-x}) dx = [e^x - e^{-x}]_0^1 = (e - e^{-1}) - (1 - 1) = e - e^{-1}$.For $x \in [1, 2)$,$[x] = 1$,so $[x]! = 1! = 1$. The integral is $\int_1^2 (e^x + e^{-x}) dx = [e^x - e^{-x}]_1^2 = (e^2 - e^{-2}) - (e - e^{-1}) = e^2 - e^{-2} - e + e^{-1}$.For $x \in [2, 3)$,$[x] = 2$,so $[x]! = 2! = 2$. The integral is $\int_2^3 \frac{e^x + e^{-x}}{2} dx = \frac{1}{2} [e^x - e^{-x}]_2^3 = \frac{1}{2} ((e^3 - e^{-3}) - (e^2 - e^{-2})) = \frac{1}{2}e^3 - \frac{1}{2}e^{-3} - \frac{1}{2}e^2 + \frac{1}{2}e^{-2}$.Summing these: $(e - e^{-1}) + (e^2 - e^{-2} - e + e^{-1}) + (\frac{1}{2}e^3 - \frac{1}{2}e^{-3} - \frac{1}{2}e^2 + \frac{1}{2}e^{-2}) = \frac{1}{2}e^3 + \frac{1}{2}e^2 - \frac{1}{2}e^{-2} - \frac{1}{2}e^{-3} = \frac{1}{2}(e^3 + e^2 - e^{-2} - e^{-3})$.

Let $[\cdot]$ denote the greatest integer function. Then the value of $\int_0^3 \left( \frac{e^x + e^{-x}}{[x]!} \right) dx$ is :

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