`int_(3)^(6) (sqrt""x)/(sqrt""(9-x) + sqrt""x) dx`=

Options

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

Answer

To solve the integral

\[ I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx, \]

we will use a property of definite integrals. The property states that:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx. \]

### Step 1: Apply the property

In our case, \( a = 3 \) and \( b = 6 \), so \( a + b = 9 \). We can rewrite the integral as:

\[ I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx. \]

### Step 2: Write the second integral

Let’s denote this new integral as \( I_2 \):

\[ I_2 = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx. \]

### Step 3: Add the two integrals

Now, we add \( I \) and \( I_2 \):

\[ I + I_2 = \int_{3}^{6} \left( \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} + \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \right) dx. \]

### Step 4: Simplify the expression

Notice that the denominators are the same, so we can combine the fractions:

\[ I + I_2 = \int_{3}^{6} \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{9-x} + \sqrt{x}} \, dx = \int_{3}^{6} 1 \, dx. \]

### Step 5: Calculate the integral of 1

The integral of 1 from 3 to 6 is simply the length of the interval:

\[ \int_{3}^{6} 1 \, dx = 6 - 3 = 3. \]

### Step 6: Relate \( I \) and \( I_2 \)

Since \( I = I_2 \), we can write:

\[ 2I = 3. \]

### Step 7: Solve for \( I \)

Dividing both sides by 2 gives:

\[ I = \frac{3}{2}. \]

Thus, the value of the integral is:

\[ \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx = \frac{3}{2}. \]

### Final Answer:

\[ \boxed{\frac{3}{2}}. \]