Let [cdot]$[\cdot]$ denote the greatest integer function. If int_0^e^3left[frac1e^x - 1right]mathrmdx = alpha -log_e2$\int_{0}^{e^{3}}\left[\frac{1}{e^{x - 1}}\right]\mathrm{d}x = \alpha -\log_{e}2$, then alpha^3$\alpha^3$ is equal to ________.

### Related Formula Greatest Integer Function boundaries: [f(x)] = k quad textfor quad k le f(x) < k+1, \ k in mathbbZ$$[f(x)] = k \quad \text{for} \quad k \le f(x) < k+1, \ k \in \mathbb{Z}$$ ### Core Logic Analyze the value variations of f(x) = e^1-x$f(x) = e^{1-x}$ across the integration limits [0, e^3]$[0, e^3]$ to break down the integral into distinct piecewise continuous intervals. ### Step 1: Determine Step Function Transition Points Let y = e^1-x$y = e^{1-x}$. * At x = 0 implies y = e^1 approx 2.718$x = 0 \implies y = e^1 \approx 2.718$ * As x$x$ increases, e^1-x$e^{1-x}$ decreases monotonically. * Find x$x$ where y = 2 implies e^1-x = 2 implies 1-x = ln 2 implies x = 1 - ln 2$y = 2 \implies e^{1-x} = 2 \implies 1-x = \ln 2 \implies x = 1 - \ln 2$. * Find x$x$ where y = 1 implies e^1-x = 1 implies 1-x = 0 implies x = 1$y = 1 \implies e^{1-x} = 1 \implies 1-x = 0 \implies x = 1$. * At the final boundary x = e^3 implies y = e^1-e^3$x = e^3 \implies y = e^{1-e^3}$, which is a very small positive decimal strictly inside (0,1)$(0,1)$. ### Step 2: Split the Definite Integral Rewrite the integral based on the isolated interval blocks: I = int_0^1-ln 2 2 \, mathrmdx + int_1-ln 2^1 1 \, mathrmdx + int_1^e^3 0 \, mathrmdx$$I = \int_{0}^{1-\ln 2} 2 \, \mathrm{d}x + \int_{1-\ln 2}^{1} 1 \, \mathrm{d}x + \int_{1}^{e^3} 0 \, \mathrm{d}x$$ ### Step 3: Perform Integrations I = 2[x]_0^1-ln 2 + 1[x]_1-ln 2^1 + 0$$I = 2[x]_0^{1-\ln 2} + 1[x]_{1-\ln 2}^{1} + 0$$ I = 2(1 - ln 2 - 0) + 1(1 - (1 - ln 2)) = 2 - 2ln 2 + ln 2 = 2 - ln 2$$I = 2(1 - \ln 2 - 0) + 1(1 - (1 - \ln 2)) = 2 - 2\ln 2 + \ln 2 = 2 - \ln 2$$ ### Step 4: Solve for Alpha Cubed Compare the integrated value to alpha - ln 2$\alpha - \ln 2$: alpha - ln 2 = 2 - ln 2 implies alpha = 2$$\alpha - \ln 2 = 2 - \ln 2 \implies \alpha = 2$$ alpha^3 = 2^3 = 8$$\alpha^3 = 2^3 = 8$$ ### Pattern Recognition Always map the function values at the extreme boundary points first. Tracking the downward path from 2.71 rightarrow 2 rightarrow 1 rightarrow 0$2.71 \rightarrow 2 \rightarrow 1 \rightarrow 0$ reveals exactly where the integer thresholds are crossed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integrals