`int_(3)^(7)sqrt((x-3)(7-x))dx=`

Options

A. `pi`
B. `0`
C. `2pi`
D. `(pi)/2`

Answer

To solve the integral \( I = \int_{3}^{7} \sqrt{(x-3)(7-x)} \, dx \), we can use a substitution method. Here's a step-by-step solution:

### Step 1: Simplify the Integral
We start with the integral:
\[ I = \int_{3}^{7} \sqrt{(x-3)(7-x)} \, dx \]

### Step 2: Use Substitution
Let’s use the substitution:
\[ x = 5 + 2 \sin \theta \]
This substitution is chosen because it will transform the limits and simplify the square root.

### Step 3: Determine the New Limits
When \( x = 3 \):
\[ 3 = 5 + 2 \sin \theta \implies \sin \theta = -1 \implies \theta = -\frac{\pi}{2} \]
When \( x = 7 \):
\[ 7 = 5 + 2 \sin \theta \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \]

### Step 4: Change of Variables
Now, we need to find \( dx \):
\[ dx = 2 \cos \theta \, d\theta \]

### Step 5: Substitute into the Integral
Now substitute \( x \) and \( dx \) into the integral:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{(5 + 2 \sin \theta - 3)(7 - (5 + 2 \sin \theta))} \cdot 2 \cos \theta \, d\theta \]
This simplifies to:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{(2 + 2 \sin \theta)(2 - 2 \sin \theta)} \cdot 2 \cos \theta \, d\theta \]
\[ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{4 \cos^2 \theta} \cdot 2 \cos \theta \, d\theta \]
\[ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4 \cos^2 \theta \, d\theta \]

### Step 6: Use the Identity
Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \):
\[ I = 4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \]
\[ = 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta \]

### Step 7: Evaluate the Integral
Now evaluate the integral:
\[ = 2 \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \]
Calculating the limits:
\[ = 2 \left[ \left( \frac{\pi}{2} + 0 \right) - \left( -\frac{\pi}{2} + 0 \right) \right] \]
\[ = 2 \left[ \frac{\pi}{2} + \frac{\pi}{2} \right] = 2\pi \]

### Final Result
Thus, the value of the integral is:
\[ I = 2\pi \]