What is the value of k if the area bounded by the curve `y=sinkx,y=0,x=pi//(3k)` is 3 sq. unit?

Options

A. `1//2`
B. 1
C. `3//2`
D. 2

Answer

To find the value of \( k \) such that the area bounded by the curve \( y = \sin(kx) \), \( y = 0 \), and the vertical lines \( x = \frac{\pi}{3k} \) and \( x = \frac{\pi}{k} \) is equal to 3 square units, we can follow these steps:

### Step 1: Set up the integral for the area
The area \( A \) bounded by the curve can be expressed as:
\[ A = \int_{\frac{\pi}{3k}}^{\frac{\pi}{k}} \sin(kx) \, dx \]
We know that this area is given to be 3 square units:
\[ \int_{\frac{\pi}{3k}}^{\frac{\pi}{k}} \sin(kx) \, dx = 3 \]

### Step 2: Evaluate the integral
To evaluate the integral, we can use the antiderivative of \( \sin(kx) \):
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) \]
Thus, we can write:
\[ A = \left[-\frac{1}{k} \cos(kx)\right]_{\frac{\pi}{3k}}^{\frac{\pi}{k}} \]

### Step 3: Substitute the limits into the integral
Now, substituting the limits:
\[ A = -\frac{1}{k} \left( \cos\left(k \cdot \frac{\pi}{k}\right) - \cos\left(k \cdot \frac{\pi}{3k}\right) \right) \]
This simplifies to:
\[ A = -\frac{1}{k} \left( \cos(\pi) - \cos\left(\frac{\pi}{3}\right) \right) \]
Since \( \cos(\pi) = -1 \) and \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ A = -\frac{1}{k} \left( -1 - \frac{1}{2} \right) = -\frac{1}{k} \left( -\frac{3}{2} \right) = \frac{3}{2k} \]

### Step 4: Set the area equal to 3 square units
Now we set the area equal to 3:
\[ \frac{3}{2k} = 3 \]

### Step 5: Solve for \( k \)
To solve for \( k \), we multiply both sides by \( 2k \):
\[ 3 = 6k \]
Dividing both sides by 6 gives:
\[ k = \frac{1}{2} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\frac{1}{2}} \]