The value of `int_(3)^(6)(sqrt(x+sqrt(12x-36))+sqrt(x-sqrt(12x- 36)))dx` is equal to

Options

A. `6sqrt3`
B. `4sqrt3`
C. `12sqrt3`
D. None of these

Answer

To solve the definite integral

\[ \int_{3}^{6} \left( \sqrt{x + \sqrt{12x - 36}} + \sqrt{x - \sqrt{12x - 36}} \right) dx, \]

we will follow these steps:

### Step 1: Simplify the expression under the square roots
First, we notice that \( \sqrt{12x - 36} \) can be simplified. We can factor out the 12:

\[ \sqrt{12x - 36} = \sqrt{12(x - 3)} = \sqrt{12} \sqrt{x - 3}. \]

### Step 2: Substitute
Let \( t = \sqrt{12(x - 3)} \). Then, squaring both sides gives:

\[ t^2 = 12(x - 3) \implies x - 3 = \frac{t^2}{12} \implies x = \frac{t^2}{12} + 3. \]

### Step 3: Find dx in terms of dt
Differentiating both sides with respect to \( t \):

\[ dx = \frac{1}{12} \cdot 2t \, dt = \frac{t}{6} \, dt. \]

### Step 4: Change the limits of integration
When \( x = 3 \):

\[ t = \sqrt{12(3 - 3)} = 0. \]

When \( x = 6 \):

\[ t = \sqrt{12(6 - 3)} = \sqrt{36} = 6. \]

So the limits change from \( x = 3 \) to \( x = 6 \) into \( t = 0 \) to \( t = 6 \).

### Step 5: Substitute into the integral
Now, substituting \( x \) and \( dx \) into the integral:

\[ \int_{0}^{6} \left( \sqrt{\frac{t^2}{12} + 3 + t} + \sqrt{\frac{t^2}{12} + 3 - t} \right) \cdot \frac{t}{6} \, dt. \]

### Step 6: Simplify the integrand
We simplify the terms inside the square roots:

\[ \sqrt{\frac{t^2}{12} + 3 + t} = \sqrt{\frac{t^2 + 36 + 12t}{12}} = \frac{\sqrt{t^2 + 12t + 36}}{\sqrt{12}} = \frac{\sqrt{(t + 6)^2}}{\sqrt{12}} = \frac{t + 6}{\sqrt{12}}. \]

Similarly,

\[ \sqrt{\frac{t^2}{12} + 3 - t} = \frac{\sqrt{(t - 6)^2}}{\sqrt{12}} = \frac{6 - t}{\sqrt{12}}. \]

### Step 7: Combine the terms
Now, substituting back into the integral:

\[ \int_{0}^{6} \left( \frac{t + 6}{\sqrt{12}} + \frac{6 - t}{\sqrt{12}} \right) \cdot \frac{t}{6} \, dt. \]

This simplifies to:

\[ \int_{0}^{6} \frac{12}{\sqrt{12}} \cdot \frac{t}{6} \, dt = \int_{0}^{6} \sqrt{12} \cdot \frac{t}{6} \, dt = \frac{\sqrt{12}}{6} \int_{0}^{6} t \, dt. \]

### Step 8: Evaluate the integral
The integral \( \int_{0}^{6} t \, dt \) is:

\[ \left[ \frac{t^2}{2} \right]_{0}^{6} = \frac{6^2}{2} - 0 = 18. \]

### Step 9: Final calculation
Now, substituting back:

\[ \frac{\sqrt{12}}{6} \cdot 18 = 3\sqrt{12} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}. \]

Thus, the value of the definite integral is:

\[ \boxed{6\sqrt{3}}. \]