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 of the region bounded by the curves \( y = 2^x \) and \( y = |x + 1| \) in the first quadrant, we can follow these steps:

### Step 1: Identify the curves
We have two curves:
1. \( y = 2^x \)
2. \( y = |x + 1| \)

In the first quadrant, \( |x + 1| \) simplifies to \( x + 1 \) since \( x \) is greater than \(-1\).

### Step 2: Find the points of intersection
To find the area between the curves, we need to determine where they intersect. We set the equations equal to each other:
\[ 2^x = x + 1 \]
We can check for intersections by substituting values.

- For \( x = 0 \):
\[ 2^0 = 1 \quad \text{and} \quad 0 + 1 = 1 \quad \Rightarrow \quad (0, 1) \]

- For \( x = 1 \):
\[ 2^1 = 2 \quad \text{and} \quad 1 + 1 = 2 \quad \Rightarrow \quad (1, 2) \]

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

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be expressed as:
\[ A = \int_0^1 (x + 1) \, dx - \int_0^1 2^x \, dx \]

### Step 4: Calculate the first integral
Calculate \( \int_0^1 (x + 1) \, dx \):
\[ \int (x + 1) \, dx = \frac{x^2}{2} + x \]
Evaluating from 0 to 1:
\[ \left[ \frac{1^2}{2} + 1 \right] - \left[ \frac{0^2}{2} + 0 \right] = \left[ \frac{1}{2} + 1 \right] - 0 = \frac{3}{2} \]

### Step 5: Calculate the second integral
Calculate \( \int_0^1 2^x \, dx \):
\[ \int 2^x \, dx = \frac{2^x}{\ln 2} \]
Evaluating from 0 to 1:
\[ \left[ \frac{2^1}{\ln 2} \right] - \left[ \frac{2^0}{\ln 2} \right] = \frac{2}{\ln 2} - \frac{1}{\ln 2} = \frac{1}{\ln 2} \]

### Step 6: Combine the results to find the area
Now substitute back into the area formula:
\[ A = \frac{3}{2} - \frac{1}{\ln 2} \]

### Final Result
Thus, the area of the region bounded by the curves \( y = 2^x \) and \( y = |x + 1| \) in the first quadrant is:
\[ A = \frac{3}{2} - \frac{1}{\ln 2} \]