Question ID: #605
Let $[t]$ denote the greatest integer function. If $\int_{0}^{e^3} \left[\frac{1}{e^{x-1}}\right] dx = \alpha – \log_{e}2$, then $\alpha^{3}$ is equal to:
Solution:
Let the integrand be $f(x) = \frac{1}{e^{x-1}} = e^{1-x}$.
We need to find the points where $f(x)$ becomes an integer within the limit $0$ to $e^3$.
Check for integer values:
$$ e^{1-x} = 2 \Rightarrow 1-x = \ln 2 \Rightarrow x = 1 – \ln 2 $$
$$ e^{1-x} = 1 \Rightarrow 1-x = \ln 1 \Rightarrow x = 1 $$
Now, split the integral into three intervals based on these critical points:
1. **Interval $0 \le x < 1 - \ln 2$:** $$ 1-x > \ln 2 \Rightarrow e^{1-x} > 2 $$
At $x=0$, $e^{1-0} = e \approx 2.71$. Thus, $[e^{1-x}] = 2$.
2. **Interval $1 – \ln 2 \le x < 1$:** $$ 0 < 1-x \le \ln 2 \Rightarrow 1 < e^{1-x} \le 2 $$ Thus, $[e^{1-x}] = 1$.
3. **Interval $1 \le x \le e^3$:**
$$ 1-x \le 0 \Rightarrow e^{1-x} \le 1 $$
For $x > 1$, $e^{1-x} < 1$. Thus, $[e^{1-x}] = 0$.
Now, write the integral as a sum:
$$ I = \int_{0}^{1-\ln 2} 2 \, dx + \int_{1-\ln 2}^{1} 1 \, dx + \int_{1}^{e^3} 0 \, dx $$
Evaluate each part:
$$ I = 2[x]_{0}^{1-\ln 2} + 1[x]_{1-\ln 2}^{1} + 0 $$
$$ I = 2(1 – \ln 2 – 0) + (1 – (1 – \ln 2)) $$
$$ I = 2 – 2\ln 2 + 1 – 1 + \ln 2 $$
$$ I = 2 – \ln 2 $$
Equating this to the given value $\alpha – \ln 2$:
$$ 2 – \ln 2 = \alpha – \ln 2 $$
$$ \Rightarrow \alpha = 2 $$
Calculate $\alpha^3$:
$$ \alpha^3 = 2^3 = 8 $$
Ans. (8)
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