The value of `int_0^([x]) 2^x/(2^([x])) dx` is equal to (where, [.] denotes the greatest integer function)

Options

A. `([x])/(log2)`
B. `([x])/(2 log2)`
C. `([x])/(4 log 2)`
D. None of these

Answer

To solve the integral \( \int_0^{[x]} \frac{2^x}{2^{[x]}} \, dx \), where \([x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Rewrite the Integral
The integral can be rewritten as:
\[ \int_0^{[x]} \frac{2^x}{2^{[x]}} \, dx = \int_0^{[x]} 2^{x - [x]} \, dx \]
Here, \(x - [x]\) represents the fractional part of \(x\), denoted as \(\{x\}\).

### Step 2: Identify the Limits of Integration
Let \(n = [x]\). The integral now becomes:
\[ \int_0^{n} 2^{x - n} \, dx \]
This is valid since \([x] = n\) is an integer.

### Step 3: Simplify the Integral
The expression \(2^{x - n}\) can be rewritten as:
\[ 2^{x - n} = \frac{2^x}{2^n} \]
Thus, the integral simplifies to:
\[ \int_0^{n} 2^{x - n} \, dx = \frac{1}{2^n} \int_0^{n} 2^x \, dx \]

### Step 4: Compute the Integral
Now we need to compute the integral \( \int_0^{n} 2^x \, dx \):
\[ \int 2^x \, dx = \frac{2^x}{\ln 2} + C \]
Evaluating from 0 to \(n\):
\[ \int_0^{n} 2^x \, dx = \left[ \frac{2^x}{\ln 2} \right]_0^{n} = \frac{2^n}{\ln 2} - \frac{2^0}{\ln 2} = \frac{2^n - 1}{\ln 2} \]

### Step 5: Substitute Back into the Integral
Substituting this result back into our integral gives:
\[ \int_0^{n} 2^{x - n} \, dx = \frac{1}{2^n} \cdot \frac{2^n - 1}{\ln 2} = \frac{2^n - 1}{2^n \ln 2} \]

### Step 6: Final Result
Thus, the value of the integral \( \int_0^{[x]} \frac{2^x}{2^{[x]}} \, dx \) is:
\[ \frac{2^n - 1}{2^n \ln 2} \]