Evaluate : `int_(0)^(2)x^(3//2)sqrt(2-x)dx`.

Answer

To evaluate the integral \( \int_{0}^{2} x^{\frac{3}{2}} \sqrt{2-x} \, dx \), we will use the substitution method. Here are the steps:

### Step 1: Substitution
Let \( x = 2 \cos^2 \theta \). Then, we differentiate to find \( dx \):
\[ dx = 2 \cdot 2 \cos \theta (-\sin \theta) \, d\theta = -4 \cos \theta \sin \theta \, d\theta \]

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

### Step 3: Substitute into the integral
Now substitute \( x \) and \( dx \) into the integral:
\[ \int_{0}^{2} x^{\frac{3}{2}} \sqrt{2-x} \, dx = \int_{\frac{\pi}{2}}^{0} (2 \cos^2 \theta)^{\frac{3}{2}} \sqrt{2 - 2 \cos^2 \theta} (-4 \cos \theta \sin \theta) \, d\theta \]

### Step 4: Simplify the integrand
First, simplify \( (2 \cos^2 \theta)^{\frac{3}{2}} \):
\[ (2 \cos^2 \theta)^{\frac{3}{2}} = 2^{\frac{3}{2}} \cos^3 \theta \]
Next, simplify \( \sqrt{2 - 2 \cos^2 \theta} \):
\[ \sqrt{2 - 2 \cos^2 \theta} = \sqrt{2(1 - \cos^2 \theta)} = \sqrt{2 \sin^2 \theta} = \sqrt{2} \sin \theta \]

### Step 5: Combine everything
Now substitute these back into the integral:
\[ \int_{\frac{\pi}{2}}^{0} 2^{\frac{3}{2}} \cos^3 \theta \cdot \sqrt{2} \sin \theta \cdot (-4 \cos \theta \sin \theta) \, d\theta \]
This simplifies to:
\[ -4 \cdot 2^{2} \int_{\frac{\pi}{2}}^{0} \cos^4 \theta \sin^2 \theta \, d\theta \]
or
\[ -16 \int_{\frac{\pi}{2}}^{0} \cos^4 \theta \sin^2 \theta \, d\theta \]

### Step 6: Change the limits
Using the property of integrals, we can change the limits:
\[ 16 \int_{0}^{\frac{\pi}{2}} \cos^4 \theta \sin^2 \theta \, d\theta \]

### Step 7: Use the known integral formula
We use the formula:
\[ \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{(n-1)(n-3)(n-5)...}{(m+n)(m+n-1)(m+n-2)...} \cdot \frac{\pi}{2} \]
Here, \( m = 2 \) and \( n = 4 \):
\[ = \frac{(4-1)}{(2+4)} \cdot \frac{(4-3)}{(2+4-2)} \cdot \frac{\pi}{2} = \frac{3}{6} \cdot \frac{1}{4} \cdot \frac{\pi}{2} \]
This simplifies to:
\[ \frac{3}{24} \cdot \frac{\pi}{2} = \frac{3\pi}{48} \]

### Step 8: Final calculation
Putting it all together:
\[ 16 \cdot \frac{3\pi}{48} = \frac{48\pi}{48} = \pi \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{2} x^{\frac{3}{2}} \sqrt{2-x} \, dx = \pi \]