Find the area bounded by `y=max{0,lnx}` and `y lt2^x` where `1/2 ltx lt1`

Answer

To find the area bounded by the curves \( y = \max\{0, \ln x\} \) and \( y < 2^x \) for \( \frac{1}{2} < x < 1 \), we can follow these steps:

### Step 1: Understand the curves
The function \( y = \max\{0, \ln x\} \) means that for \( x \leq 1 \), \( \ln x \) is negative, so \( y = 0 \). For \( x > 1 \), \( y = \ln x \). Since we are only interested in the interval \( \frac{1}{2} < x < 1 \), we have:
- For \( \frac{1}{2} < x < 1 \), \( y = 0 \).

The function \( y = 2^x \) is an exponential function that is always positive.

### Step 2: Identify the area to be calculated
In the interval \( \frac{1}{2} < x < 1 \), the area we want to find is bounded above by the curve \( y = 2^x \) and below by \( y = 0 \). Therefore, the area \( A \) can be expressed as:
\[ A = \int_{\frac{1}{2}}^{1} (2^x - 0) \, dx = \int_{\frac{1}{2}}^{1} 2^x \, dx \]

### Step 3: Calculate the integral
To solve the integral, we use the formula for the integral of an exponential function:
\[ \int a^x \, dx = \frac{a^x}{\ln a} + C \]
For our case, \( a = 2 \):
\[ \int 2^x \, dx = \frac{2^x}{\ln 2} \]
Now we evaluate this from \( \frac{1}{2} \) to \( 1 \):
\[ A = \left[ \frac{2^x}{\ln 2} \right]_{\frac{1}{2}}^{1} \]
Calculating the upper and lower limits:
\[ A = \frac{2^1}{\ln 2} - \frac{2^{\frac{1}{2}}}{\ln 2} = \frac{2}{\ln 2} - \frac{\sqrt{2}}{\ln 2} \]
Factoring out \( \frac{1}{\ln 2} \):
\[ A = \frac{1}{\ln 2} (2 - \sqrt{2}) \]

### Final Answer
Thus, the area bounded by the curves is:
\[ A = \frac{2 - \sqrt{2}}{\ln 2} \]