`int_(0)^(4)sqrt(16-x^(2))dx=`

Options

A. `-4pi`
B. `2pi`
C. `4pi`
D. `-2pi`

Answer

To solve the integral \( \int_{0}^{4} \sqrt{16 - x^2} \, dx \), we can use a known formula for the integral of the form \( \int \sqrt{a^2 - x^2} \, dx \).

### Step 1: Identify the integral form
The integral can be recognized as a specific case of the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C \]
where \( a = 4 \) in our case, since \( 16 = 4^2 \).

### Step 2: Apply the formula
Using the formula, we have:
\[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2} \sqrt{16 - x^2} + \frac{16}{2} \sin^{-1} \left( \frac{x}{4} \right) + C \]
This simplifies to:
\[ \int \sqrt{16 - x^2} \, dx = \frac{x}{2} \sqrt{16 - x^2} + 8 \sin^{-1} \left( \frac{x}{4} \right) + C \]

### Step 3: Evaluate the definite integral from 0 to 4
Now we need to evaluate this from 0 to 4:
\[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{16 - x^2} + 8 \sin^{-1} \left( \frac{x}{4} \right) \right]_{0}^{4} \]

### Step 4: Calculate the upper limit (x = 4)
Substituting \( x = 4 \):
\[ = \frac{4}{2} \sqrt{16 - 4^2} + 8 \sin^{-1} \left( \frac{4}{4} \right) \]
\[ = 2 \cdot \sqrt{16 - 16} + 8 \cdot \sin^{-1}(1) \]
\[ = 2 \cdot 0 + 8 \cdot \frac{\pi}{2} = 4\pi \]

### Step 5: Calculate the lower limit (x = 0)
Now substituting \( x = 0 \):
\[ = \frac{0}{2} \sqrt{16 - 0^2} + 8 \sin^{-1} \left( \frac{0}{4} \right) \]
\[ = 0 + 8 \cdot 0 = 0 \]

### Step 6: Combine results
Now, combining the results from the upper and lower limits:
\[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = 4\pi - 0 = 4\pi \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = 4\pi \]