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

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi)/(16)`
D. `pi`

Answer

To solve the integral \( \int_{0}^{2} x^{2} \sqrt{4 - x^{2}} \, dx \), we will use the substitution method. Here are the steps:

### Step 1: Substitution
Let \( x = 2 \sin \theta \). Then, \( dx = 2 \cos \theta \, d\theta \).

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ 0 = 2 \sin \theta \implies \sin \theta = 0 \implies \theta = 0 \]
When \( x = 2 \):
\[ 2 = 2 \sin \theta \implies \sin \theta = 1 \implies \theta = \frac{\pi}{2} \]
So, the limits change from \( x = 0 \) to \( x = 2 \) to \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \).

### Step 3: Substitute in the integral
Now, substituting \( x \) and \( dx \) into the integral:
\[ \int_{0}^{2} x^{2} \sqrt{4 - x^{2}} \, dx = \int_{0}^{\frac{\pi}{2}} (2 \sin \theta)^{2} \sqrt{4 - (2 \sin \theta)^{2}} (2 \cos \theta) \, d\theta \]
This simplifies to:
\[ = \int_{0}^{\frac{\pi}{2}} 4 \sin^{2} \theta \sqrt{4 - 4 \sin^{2} \theta} (2 \cos \theta) \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} 4 \sin^{2} \theta \sqrt{4(1 - \sin^{2} \theta)} (2 \cos \theta) \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} 4 \sin^{2} \theta \cdot 2 \sqrt{4} \cos \theta \, d\theta \]
\[ = \int_{0}^{\frac{\pi}{2}} 16 \sin^{2} \theta \cos \theta \, d\theta \]

### Step 4: Simplify the integral
Now we can factor out constants:
\[ = 16 \int_{0}^{\frac{\pi}{2}} \sin^{2} \theta \cos \theta \, d\theta \]

### Step 5: Use the identity for sine
Using the identity \( \sin^{2} \theta = \frac{1 - \cos(2\theta)}{2} \):
\[ = 16 \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \cos \theta \, d\theta \]
\[ = 8 \int_{0}^{\frac{\pi}{2}} (1 - \cos(2\theta)) \cos \theta \, d\theta \]
\[ = 8 \left( \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta - \int_{0}^{\frac{\pi}{2}} \cos(2\theta) \cos \theta \, d\theta \right) \]

### Step 6: Evaluate the integrals
The first integral:
\[ \int_{0}^{\frac{\pi}{2}} \cos \theta \, d\theta = [\sin \theta]_{0}^{\frac{\pi}{2}} = 1 - 0 = 1 \]
The second integral can be computed using the product-to-sum formulas:
\[ \int_{0}^{\frac{\pi}{2}} \cos(2\theta) \cos \theta \, d\theta = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} \cos(\theta) \, d\theta + \int_{0}^{\frac{\pi}{2}} \cos(3\theta) \, d\theta \right) \]
The first part evaluates to \( 1 \) and the second part evaluates to \( 0 \) (as \( \sin(3\theta) \) evaluated at the limits gives \( 0 \)):
\[ = \frac{1}{2}(1 + 0) = \frac{1}{2} \]

### Step 7: Combine results
Thus, we have:
\[ = 8 \left( 1 - \frac{1}{2} \right) = 8 \cdot \frac{1}{2} = 4 \]

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