Area bounded by the curves `y = 1 - 2^(1 + sin x), x in [0, (pi)/(2)]` where `underset(0)overset(pi//2)(int) 2^(sin x) dx = k`

Options

A. `(pi - 4k)/(2)`
B. `(pi - 2 k)/(2)`
C. `(2 pi - k)/(2)`
D. none of these

Answer

To find the area bounded by the curve \( y = 1 - 2^{1 + \sin x} \) for \( x \) in the interval \( [0, \frac{\pi}{2}] \), we will follow these steps:

### Step 1: Set Up the Area Integral
The area \( A \) bounded by the curve from \( x = 0 \) to \( x = \frac{\pi}{2} \) can be expressed as:
\[ A = \int_0^{\frac{\pi}{2}} (1 - 2^{1 + \sin x}) \, dx \]

### Step 2: Split the Integral
We can split the integral into two parts:
\[ A = \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} 2^{1 + \sin x} \, dx \]

### Step 3: Evaluate the First Integral
The first integral is straightforward:
\[ \int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Step 4: Evaluate the Second Integral
The second integral can be simplified:
\[ \int_0^{\frac{\pi}{2}} 2^{1 + \sin x} \, dx = 2 \int_0^{\frac{\pi}{2}} 2^{\sin x} \, dx \]
Given that \( \int_0^{\frac{\pi}{2}} 2^{\sin x} \, dx = k \), we have:
\[ \int_0^{\frac{\pi}{2}} 2^{1 + \sin x} \, dx = 2k \]

### Step 5: Combine the Results
Now, substituting back into our area formula:
\[ A = \frac{\pi}{2} - 2k \]

### Final Result
Thus, the area bounded by the curve \( y = 1 - 2^{1 + \sin x} \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) is:
\[ A = \frac{\pi}{2} - 2k \]