Let [.] denote the greatest integer function.

Question

Let [.] denote the greatest integer function. If $\int _{0}^{{e}^{3}}[\frac{1}{{e}^{x-1}}]dx=\alpha -{\log }_{e}2$, then ${\alpha }^{3}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

To solve this, we start by evaluating the integral:

$I={\int }_{{e}^{3}}^{0}[\frac{1}{{e}^{x-1}}]\, dx$

The greatest integer function $[\cdot ]$ returns the largest integer less than or equal to the input value. Here's how we can approach the problem:

Determine the function inside the integral:

$\frac{1}{{e}^{x-1}}={e}^{1-x}$.

Identifying the intervals:

When ${e}^{1-x}\ge 2$, which simplifies to $x\le 1-\ln 2$, we have $[{e}^{1-x}]=2$.

When $1\le {e}^{1-x}When $0\le {e}\^{1-x}1$, thus $[{e}\^{1-x}]=0$ from $x=1$ to $x={e}\^{3}$.

Evaluate the integral on these intervals:

${\int }\_{1-\ln 2}\^{0}2\, dx=2(1-\ln 2)$

${\int }\_{1}\^{1-\ln 2}1\, dx=1-(1-\ln 2)=\ln 2$

${\int }\_{{e}\^{3}}\^{1}0\, dx=0$

Combine these results:

$I=2(1-\ln 2)+\ln 2+0=2-\ln 2$

Thus, we are given that:

$\alpha -\ln 2=2-\ln 2$

This implies that:

$\alpha =2$

Therefore, $

Solutions · 1