Let [.] denote the greatest integer function. | vedclass

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(A) Let $f(x) = \frac{1}{e^{x-1}} = e^{1-x}$. We need to evaluate the integral $I = \int_0^{e^3} [f(x)] dx$.The function $f(x) = e^{1-x}$ is a strictl

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$8$

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(A) Let $f(x) = \frac{1}{e^{x-1}} = e^{1-x}$. We need to evaluate the integral $I = \int_0^{e^3} [f(x)] dx$.The function $f(x) = e^{1-x}$ is a strictly decreasing function.At $x=0$,$f(0) = e^1 \approx 2.718$.At $x=1$,$f(1) = e^0 = 1$.At $x=1+\ln 2$,$f(1+\ln 2) = e^{1-(1+\ln 2)} = e^{-\ln 2} = \frac{1}{2} = 0.5$.Since $f(x)$ is decreasing,we find the intervals where $[f(x)]$ is constant:For $x \in [0, 1-\ln 2)$,$f(x) \in (2, e]$,so $[f(x)] = 2$.For $x \in [1-\ln 2, 1)$,$f(x) \in [1, 2)$,so $[f(x)] = 1$.For $x \in [1, e^3]$,$f(x) \in (0, 1]$,so $[f(x)] = 0$.Thus,$I = \int_0^{1-\ln 2} 2 dx + \int_{1-\ln 2}^1 1 dx + \int_1^{e^3} 0 dx$.$I = 2(1-\ln 2 - 0) + 1(1 - (1-\ln 2)) + 0$.$I = 2 - 2\ln 2 + \ln 2 = 2 - \ln 2$.Given $I = \alpha - \ln 2$,we have $\alpha - \ln 2 = 2 - \ln 2$,which implies $\alpha = 2$.Therefore,$\alpha^3 = 2^3 = 8$.

Let $[.]$ denote the greatest integer function. If $\int_0^{e^3}\left[\frac{1}{e^{x-1}}\right] d x=\alpha-\log _e 2$,then $\alpha^3$ is equal to . . . . . . .

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