The positive value of the parmeter 'a' for which the area of the figure bounded by `y=sinas, y=0, x=pi//a and x=pi//3a` is 3, is equal to

Options

A. 2
B. `1//2`
C. `(2+sqrt3)/(3)`
D. `sqrt3`

Answer

To find the positive value of the parameter 'a' for which the area of the figure bounded by the curves \( y = \sin(ax) \), \( y = 0 \), \( x = \frac{\pi}{a} \), and \( x = \frac{\pi}{3a} \) is equal to 3, we can follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) bounded by the curves can be expressed as:
\[ A = \int_{\frac{\pi}{3a}}^{\frac{\pi}{a}} \sin(ax) \, dx \]
We know that the area is given to be 3, so we set up the equation:
\[ \int_{\frac{\pi}{3a}}^{\frac{\pi}{a}} \sin(ax) \, dx = 3 \]

### Step 2: Evaluate the integral
The integral of \( \sin(ax) \) is:
\[ \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C \]
Now, we can evaluate the definite integral:
\[ \int_{\frac{\pi}{3a}}^{\frac{\pi}{a}} \sin(ax) \, dx = \left[-\frac{1}{a} \cos(ax)\right]_{\frac{\pi}{3a}}^{\frac{\pi}{a}} \]
Substituting the limits:
\[ = -\frac{1}{a} \left( \cos\left(a \cdot \frac{\pi}{a}\right) - \cos\left(a \cdot \frac{\pi}{3a}\right) \right) \]
This simplifies to:
\[ = -\frac{1}{a} \left( \cos(\pi) - \cos\left(\frac{\pi}{3}\right) \right) \]
Using the values \( \cos(\pi) = -1 \) and \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ = -\frac{1}{a} \left( -1 - \frac{1}{2} \right) = -\frac{1}{a} \left( -\frac{3}{2} \right) = \frac{3}{2a} \]

### Step 3: Set the area equal to 3
Now we set the area equal to 3:
\[ \frac{3}{2a} = 3 \]

### Step 4: Solve for 'a'
To solve for \( a \), we multiply both sides by \( 2a \):
\[ 3 = 6a \]
Dividing both sides by 6 gives:
\[ a = \frac{1}{2} \]

### Final Answer
The positive value of the parameter \( a \) is:
\[ \boxed{\frac{1}{2}} \]