The area (in sq. units) of the region bounded by the curves `y=2^(x)` and `y=|x+1|`, in the first quadrant is :

Options

A. `(3)/(2)-(1)/(log_(e )2)`
B. `log_(e )2+(3)/(2)`
C. `(3)/(2)`
D. `(1)/(2)`

Answer

To find the area bounded by the curves \( y = 2^x \) and \( y = |x + 1| \) in the first quadrant, we will follow these steps:

### Step 1: Identify the curves and their intersection points
The first curve is \( y = 2^x \), which is an exponential function. The second curve is \( y = |x + 1| \), which is a linear function that has a vertex at \( (-1, 0) \).

To find the points of intersection, we set the two equations equal to each other:
\[ 2^x = x + 1 \]

### Step 2: Find intersection points
We can check for intersection points by substituting values:
- For \( x = 0 \):
\[ 2^0 = 1 \quad \text{and} \quad 0 + 1 = 1 \quad \Rightarrow \quad (0, 1) \text{ is an intersection point.} \]
- For \( x = 1 \):
\[ 2^1 = 2 \quad \text{and} \quad 1 + 1 = 2 \quad \Rightarrow \quad (1, 2) \text{ is another intersection point.} \]

Thus, the curves intersect at the points \( (0, 1) \) and \( (1, 2) \).

### Step 3: Set up the integral for the area
In the first quadrant, from \( x = 0 \) to \( x = 1 \), the area between the curves can be found by integrating the upper curve minus the lower curve:
\[ \text{Area} = \int_{0}^{1} \left( (x + 1) - 2^x \right) \, dx \]

### Step 4: Calculate the integral
First, we compute the integral:
\[ \int_{0}^{1} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{0}^{1} = \left( \frac{1^2}{2} + 1 \right) - \left( 0 + 0 \right) = \frac{1}{2} + 1 = \frac{3}{2} \]

Next, we compute the integral of \( 2^x \):
\[ \int_{0}^{1} 2^x \, dx = \left[ \frac{2^x}{\ln 2} \right]_{0}^{1} = \left( \frac{2^1}{\ln 2} - \frac{2^0}{\ln 2} \right) = \left( \frac{2}{\ln 2} - \frac{1}{\ln 2} \right) = \frac{1}{\ln 2} \]

### Step 5: Combine the results
Now, we can find the area:
\[ \text{Area} = \frac{3}{2} - \frac{1}{\ln 2} \]

### Final Answer
Thus, the area of the region bounded by the curves in the first quadrant is:
\[ \text{Area} = \frac{3}{2} - \frac{1}{\ln 2} \]