Area of the region bounded by the curves `y=2^(x),y=2x-x^(2),x=0" and "x=2` is given by :

Options

A. `3/(log2)-4/3`
B. `3/(log2)+4/3`
C. `3 log 2-4/3`
D. None of these

Answer

To find the area of the region bounded by the curves \( y = 2^x \), \( y = 2x - x^2 \), \( x = 0 \), and \( x = 2 \), we will follow these steps:

### Step 1: Identify the curves and points of intersection
We have two curves:
1. \( y = 2^x \)
2. \( y = 2x - x^2 \)

We need to find the points of intersection of these curves between \( x = 0 \) and \( x = 2 \).

### Step 2: Set the equations equal to find points of intersection
Set \( 2^x = 2x - x^2 \).
This may not have a simple algebraic solution, so we can evaluate the functions at specific points:
- At \( x = 0 \): \( 2^0 = 1 \) and \( 2(0) - 0^2 = 0 \) (Point: (0, 1))
- At \( x = 1 \): \( 2^1 = 2 \) and \( 2(1) - 1^2 = 1 \) (Point: (1, 2))
- At \( x = 2 \): \( 2^2 = 4 \) and \( 2(2) - 2^2 = 0 \) (Point: (2, 0))

### Step 3: Determine which curve is on top
From the evaluations:
- At \( x = 0 \), \( 2^x \) is above \( 2x - x^2 \).
- At \( x = 1 \), \( 2^x \) is above \( 2x - x^2 \).
- At \( x = 2 \), \( 2^x \) is above \( 2x - x^2 \).

Thus, \( y = 2^x \) is the upper curve and \( y = 2x - x^2 \) is the lower curve in the interval \( [0, 2] \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 2 \) is given by:
\[ A = \int_{0}^{2} (2^x - (2x - x^2)) \, dx \]

### Step 5: Simplify the integral
This can be rewritten as:
\[ A = \int_{0}^{2} (2^x - 2x + x^2) \, dx \]

### Step 6: Compute the integral
Now we compute each part of the integral:
1. The integral of \( 2^x \) is \( \frac{2^x}{\ln(2)} \).
2. The integral of \( -2x \) is \( -x^2 \).
3. The integral of \( x^2 \) is \( \frac{x^3}{3} \).

Thus,
\[ A = \left[ \frac{2^x}{\ln(2)} - x^2 + \frac{x^3}{3} \right]_{0}^{2} \]

### Step 7: Evaluate the definite integral
Now we evaluate from \( 0 \) to \( 2 \):
\[ A = \left( \frac{2^2}{\ln(2)} - 2^2 + \frac{2^3}{3} \right) - \left( \frac{2^0}{\ln(2)} - 0 + 0 \right) \]
\[ = \left( \frac{4}{\ln(2)} - 4 + \frac{8}{3} \right) - \left( \frac{1}{\ln(2)} \right) \]
\[ = \frac{4}{\ln(2)} - 4 + \frac{8}{3} - \frac{1}{\ln(2)} \]
\[ = \frac{3}{\ln(2)} - 4 + \frac{8}{3} \]

### Step 8: Combine and simplify
Now, we can combine the terms:
\[ A = \left( \frac{3}{\ln(2)} - 4 + \frac{8}{3} \right) \]
Calculating the numerical value will give us the area.

### Final Result
The area of the region bounded by the curves is:
\[ A = \frac{3}{\ln(2)} - 4 + \frac{8}{3} \text{ square units.} \]