The maximum area (in sq. units) bounded by `y=sinx, y=ax(AA a in [1, 4])` and then line `pi-2x=0` is

Options

A. `pi^(2)`
B. `(pi^(2))/(2)-1`
C. `pi+2`
D. `pi^(2)-4`

Answer

To find the maximum area bounded by the curves \( y = \sin x \), \( y = ax \) (where \( a \) is in the range [1, 4]), and the line \( \pi - 2x = 0 \) (which simplifies to \( x = \frac{\pi}{2} \)), we can follow these steps:

### Step 1: Identify the curves and their intersections
The curves we are dealing with are:
1. \( y = \sin x \)
2. \( y = ax \) (with \( a \) varying from 1 to 4)
3. The vertical line \( x = \frac{\pi}{2} \)

### Step 2: Determine the area bounded by the curves
The area \( A \) bounded by these curves from \( x = 0 \) to \( x = \frac{\pi}{2} \) can be expressed as:
\[ A = \int_{0}^{\frac{\pi}{2}} (ax - \sin x) \, dx \]

### Step 3: Calculate the integral
We need to compute the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} ax \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

Calculating the first integral:
\[ \int_{0}^{\frac{\pi}{2}} ax \, dx = a \left[ \frac{x^2}{2} \right]_{0}^{\frac{\pi}{2}} = a \cdot \frac{(\frac{\pi}{2})^2}{2} = a \cdot \frac{\pi^2}{8} \]

Calculating the second integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[ -\cos x \right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

Putting it all together:
\[ A = a \cdot \frac{\pi^2}{8} - 1 \]

### Step 4: Maximize the area with respect to \( a \)
To find the maximum area, we need to maximize the expression \( A = a \cdot \frac{\pi^2}{8} - 1 \) for \( a \) in the interval [1, 4]. Since the area is a linear function of \( a \), it will be maximized at the upper endpoint of the interval.

Thus, substituting \( a = 4 \):
\[ A_{\text{max}} = 4 \cdot \frac{\pi^2}{8} - 1 = \frac{4\pi^2}{8} - 1 = \frac{\pi^2}{2} - 1 \]

### Final Answer
The maximum area bounded by the curves is:
\[ \boxed{\frac{\pi^2}{2} - 1} \]