If [.] denotes the greatest integer function, then `int_(0)^(oo) [2e^(-x)]dx=`

Options

A. 0
B. `e^(2)`
C. `2//e`
D. log 2

Answer

To solve the integral \( \int_{0}^{\infty} [2e^{-x}] \, dx \) where \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Understand the function \( y = 2e^{-x} \)
The function \( y = 2e^{-x} \) is a decreasing function. As \( x \) increases, \( e^{-x} \) decreases, which means \( y \) will also decrease.

### Step 2: Determine the behavior of \( y \)
- At \( x = 0 \):
\[ y = 2e^{0} = 2 \]
- As \( x \to \infty \):
\[ y \to 0 \]

### Step 3: Identify the intervals for the greatest integer function
- The greatest integer function \([y]\) will take different values based on the value of \( y \):
- For \( 0 \leq x < \log 2 \), \( 1 < y < 2 \) so \([y] = 1\).
- For \( x \geq \log 2 \), \( y < 1 \) so \([y] = 0\).

### Step 4: Break the integral into two parts
We can break the integral into two parts based on the value of \( x \):
\[ \int_{0}^{\infty} [2e^{-x}] \, dx = \int_{0}^{\log 2} [2e^{-x}] \, dx + \int_{\log 2}^{\infty} [2e^{-x}] \, dx \]

### Step 5: Evaluate the first integral
For \( x \in [0, \log 2] \):
\[ \int_{0}^{\log 2} [2e^{-x}] \, dx = \int_{0}^{\log 2} 1 \, dx = x \bigg|_{0}^{\log 2} = \log 2 - 0 = \log 2 \]

### Step 6: Evaluate the second integral
For \( x \in [\log 2, \infty) \):
\[ \int_{\log 2}^{\infty} [2e^{-x}] \, dx = \int_{\log 2}^{\infty} 0 \, dx = 0 \]

### Step 7: Combine the results
Adding the two parts together:
\[ \int_{0}^{\infty} [2e^{-x}] \, dx = \log 2 + 0 = \log 2 \]

### Final Answer:
\[ \int_{0}^{\infty} [2e^{-x}] \, dx = \log 2 \]

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