Two lines drawn through the point P (4, 0) divide the area bounded by the curve `y=sqrt(2)sin.(pi x)/(4)` and x-axis, between the lines x = 2 and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to:

Options

A. `(-2sqrt(2))/(pi)`
B. `(-sqrt(2))/(pi)`
C. `(-2)/(pi)`
D. `(-4sqrt(2))/(pi)`

Answer

To solve the problem, we need to find the sum of the slopes of two lines drawn through the point P(4, 0) that divide the area bounded by the curve \( y = \sqrt{2} \sin\left(\frac{\pi x}{4}\right) \) and the x-axis between the lines \( x = 2 \) and \( x = 4 \) into three equal parts.

### Step 1: Find the area under the curve from \( x = 2 \) to \( x = 4 \)

We need to calculate the area \( A \) under the curve from \( x = 2 \) to \( x = 4 \):

\[ A = \int_{2}^{4} \sqrt{2} \sin\left(\frac{\pi x}{4}\right) \, dx \]

### Step 2: Compute the integral

To compute the integral, we can use the substitution method. Let \( u = \frac{\pi x}{4} \), then \( du = \frac{\pi}{4} dx \) or \( dx = \frac{4}{\pi} du \).

Changing the limits:
- When \( x = 2 \), \( u = \frac{\pi \cdot 2}{4} = \frac{\pi}{2} \)
- When \( x = 4 \), \( u = \frac{\pi \cdot 4}{4} = \pi \)

Now, substituting into the integral:

\[ A = \int_{\frac{\pi}{2}}^{\pi} \sqrt{2} \sin(u) \cdot \frac{4}{\pi} \, du \]

This simplifies to:

\[ A = \frac{4\sqrt{2}}{\pi} \int_{\frac{\pi}{2}}^{\pi} \sin(u) \, du \]

### Step 3: Evaluate the integral of \( \sin(u) \)

The integral of \( \sin(u) \) is \( -\cos(u) \):

\[ \int \sin(u) \, du = -\cos(u) \]

Now, evaluate it from \( \frac{\pi}{2} \) to \( \pi \):

\[ A = \frac{4\sqrt{2}}{\pi} \left[-\cos(u)\right]_{\frac{\pi}{2}}^{\pi} = \frac{4\sqrt{2}}{\pi} \left[-\cos(\pi) + \cos\left(\frac{\pi}{2}\right)\right] \]

Calculating the values:

\[ = \frac{4\sqrt{2}}{\pi} \left[1 - 0\right] = \frac{4\sqrt{2}}{\pi} \]

### Step 4: Divide the area into three equal parts

The area \( A \) is \( \frac{4\sqrt{2}}{\pi} \). Each part will be:

\[ \frac{A}{3} = \frac{4\sqrt{2}}{3\pi} \]

### Step 5: Determine the heights of the triangles formed by the lines

Let the heights of the triangles formed by the lines be \( h_1 \) and \( h_2 \). The area of the triangle formed by the first line is:

\[ \text{Area}_1 = \frac{1}{2} \times \text{base} \times h_1 = \frac{1}{2} \times 2 \times h_1 = h_1 \]

Setting this equal to \( \frac{4\sqrt{2}}{3\pi} \):

\[ h_1 = \frac{4\sqrt{2}}{3\pi} \]

For the second triangle:

\[ \text{Area}_2 = \frac{1}{2} \times 2 \times h_2 = h_2 \]

Setting this equal to \( \frac{8\sqrt{2}}{3\pi} \):

\[ h_2 = \frac{8\sqrt{2}}{3\pi} \]

### Step 6: Calculate the slopes of the lines

The slopes of the lines through point \( P(4, 0) \) are given by:

\[ m_1 = \frac{h_1 - 0}{2 - 4} = \frac{\frac{4\sqrt{2}}{3\pi}}{-2} = -\frac{2\sqrt{2}}{3\pi} \]

\[ m_2 = \frac{h_2 - 0}{4 - 2} = \frac{\frac{8\sqrt{2}}{3\pi}}{2} = \frac{4\sqrt{2}}{3\pi} \]

### Step 7: Sum of the slopes

Now, we can find the sum of the slopes:

\[ m_1 + m_2 = -\frac{2\sqrt{2}}{3\pi} + \frac{4\sqrt{2}}{3\pi} = \frac{2\sqrt{2}}{3\pi} \]

### Final Answer

Thus, the sum of the slopes of the lines is:

\[ \boxed{\frac{2\sqrt{2}}{3\pi}} \]