The value of `:.int_(0)^([x]) (2^(x))/(2^([x]))dx` is

Options

A. log 2
B. `([x])/(log2)`
C. `(1)/(2)([x])/(log 2)`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{x} \frac{2^t}{2^{\lfloor t \rfloor}} \, dt \), we will break it down step by step.

### Step 1: Rewrite the Integral
We can rewrite the integrand:
\[ I = \int_{0}^{x} 2^{t - \lfloor t \rfloor} \, dt \]
Here, \( t - \lfloor t \rfloor \) is the fractional part of \( t \), denoted as \( \{t\} \).

### Step 2: Identify the Behavior of the Fractional Part
The fractional part \( \{t\} \) varies between 0 and 1 as \( t \) goes from any integer \( n \) to \( n+1 \). Therefore, we can break the integral into intervals:
\[ I = \sum_{n=0}^{\lfloor x \rfloor - 1} \int_{n}^{n+1} 2^{t - n} \, dt + \int_{\lfloor x \rfloor}^{x} 2^{t - \lfloor x \rfloor} \, dt \]

### Step 3: Evaluate the Integral Over Each Interval
For \( n = 0, 1, \ldots, \lfloor x \rfloor - 1 \):
\[ \int_{n}^{n+1} 2^{t - n} \, dt = \int_{0}^{1} 2^{u} \, du \quad \text{(where \( u = t - n \))} \]
This integral evaluates to:
\[ \int_{0}^{1} 2^{u} \, du = \left[ \frac{2^{u}}{\ln 2} \right]_{0}^{1} = \frac{2^{1} - 2^{0}}{\ln 2} = \frac{2 - 1}{\ln 2} = \frac{1}{\ln 2} \]

### Step 4: Sum Over All Intervals
There are \( \lfloor x \rfloor \) such intervals, so:
\[ \sum_{n=0}^{\lfloor x \rfloor - 1} \int_{n}^{n+1} 2^{t - n} \, dt = \lfloor x \rfloor \cdot \frac{1}{\ln 2} \]

### Step 5: Evaluate the Remaining Integral
Now we evaluate the last part:
\[ \int_{\lfloor x \rfloor}^{x} 2^{t - \lfloor x \rfloor} \, dt = \int_{0}^{x - \lfloor x \rfloor} 2^{u} \, du \quad \text{(where \( u = t - \lfloor x \rfloor \))} \]
This integral evaluates to:
\[ \int_{0}^{x - \lfloor x \rfloor} 2^{u} \, du = \left[ \frac{2^{u}}{\ln 2} \right]_{0}^{x - \lfloor x \rfloor} = \frac{2^{x - \lfloor x \rfloor} - 1}{\ln 2} \]

### Step 6: Combine the Results
Thus, the total integral \( I \) becomes:
\[ I = \lfloor x \rfloor \cdot \frac{1}{\ln 2} + \frac{2^{x - \lfloor x \rfloor} - 1}{\ln 2} \]
Combining these, we have:
\[ I = \frac{\lfloor x \rfloor + (2^{x - \lfloor x \rfloor} - 1)}{\ln 2} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{\lfloor x \rfloor + 2^{x - \lfloor x \rfloor} - 1}{\ln 2} \]