Question

If $\int_{0}^{\infty} [\frac{3}{e^x}] dx = k$ then $e^k$ is equal to

• A. 2
• B. 3
• C. 8/2
• D. 6

Answer 1

9/2

Answer 2

The problem asks for the value of $e^k$, where $k = \int_{0}^{\infty} [\frac{3}{e^x}] dx$. The notation $[y]$ denotes the greatest integer less than or equal to $y$.

Let $g(x) = \frac{3}{e^x}$. As $x$ goes from $0$ to $\infty$, $g(x)$ decreases from $3$ to $0$. We need to find the intervals where $[\frac{3}{e^x}]$ takes constant integer values. \begin{enumerate} \item For $[\frac{3}{e^x}] = 2$: This occurs when $2 \le \frac{3}{e^x} < 3$. $\frac{3}{e^x} < 3 \implies e^x > 1 \implies x > 0$. $2 \le \frac{3}{e^x} \implies e^x \le \frac{3}{2} \implies x \le \ln(\frac{3}{2})$. So, for $0 < x \le \ln(\frac{3}{2})$, $[\frac{3}{e^x}] = 2$.

\item For $[\frac{3}{e^x}] = 1$: This occurs when $1 \le \frac{3}{e^x} < 2$.
$\frac{3}{e^x} < 2 \implies e^x > \frac{3}{2} \implies x > \ln(\frac{3}{2})$.
$1 \le \frac{3}{e^x} \implies e^x \le 3 \implies x \le \ln(3)$.
So, for $\ln(\frac{3}{2}) < x \le \ln(3)$, $[\frac{3}{e^x}] = 1$.

\item For $[\frac{3}{e^x}] = 0$: This occurs when $0 \le \frac{3}{e^x} < 1$.
$\frac{3}{e^x} < 1 \implies e^x > 3 \implies x > \ln(3)$.
So, for $x > \ln(3)$, $[\frac{3}{e^x}] = 0$.

\end{enumerate} Now, we split the integral: $k = \int_{0}^{\infty} [\frac{3}{e^x}] dx = \int_{0}^{\ln(3/2)} 2 dx + \int_{\ln(3/2)}^{\ln(3)} 1 dx + \int_{\ln(3)}^{\infty} 0 dx$

Evaluate each integral: $\int_{0}^{\ln(3/2)} 2 dx = 2[x]_{0}^{\ln(3/2)} = 2(\ln(\frac{3}{2}) - 0) = 2\ln(\frac{3}{2})$ $\int_{\ln(3/2)}^{\ln(3)} 1 dx = [x]_{\ln(3/2)}^{\ln(3)} = \ln(3) - \ln(\frac{3}{2}) = \ln(\frac{3}{3/2}) = \ln(2)$ $\int_{\ln(3)}^{\infty} 0 dx = 0$

Summing these gives $k$: $k = 2\ln(\frac{3}{2}) + \ln(2) = \ln((\frac{3}{2})^2) + \ln(2) = \ln(\frac{9}{4}) + \ln(2) = \ln(\frac{9}{4} \times 2) = \ln(\frac{9}{2})$

The question asks for $e^k$: $e^k = e^{\ln(9/2)} = \frac{9}{2}$.

Note: The calculated value $\frac{9}{2}$ is not present in the given options. Option (C) is $\frac{8}{2} = 4$. This suggests a potential typo in the question or options.