Evaluate : `int_(0)^(4) x(4-x)^(3//2)dx`

Answer

To evaluate the integral \( I = \int_{0}^{4} x (4 - x)^{\frac{3}{2}} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{4} x (4 - x)^{\frac{3}{2}} \, dx \]

### Step 2: Use substitution
Let \( t = 4 - x \). Then, differentiating both sides gives:
\[ dx = -dt \]
We also need to change the limits of integration. When \( x = 0 \), \( t = 4 \) and when \( x = 4 \), \( t = 0 \). Thus, the integral becomes:
\[ I = \int_{4}^{0} (4 - t) t^{\frac{3}{2}} (-dt) = \int_{0}^{4} (4 - t) t^{\frac{3}{2}} \, dt \]

### Step 3: Expand the integrand
Now, we expand the integrand:
\[ I = \int_{0}^{4} (4t^{\frac{3}{2}} - t^{\frac{5}{2}}) \, dt \]

### Step 4: Split the integral
We can split the integral into two parts:
\[ I = \int_{0}^{4} 4t^{\frac{3}{2}} \, dt - \int_{0}^{4} t^{\frac{5}{2}} \, dt \]

### Step 5: Integrate each term
Now we will integrate each term separately.

1. For the first term:
\[ \int t^{\frac{3}{2}} \, dt = \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} = \frac{t^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} t^{\frac{5}{2}} \]
Thus,
\[ \int_{0}^{4} 4t^{\frac{3}{2}} \, dt = 4 \cdot \left[ \frac{2}{5} t^{\frac{5}{2}} \right]_{0}^{4} = \frac{8}{5} [4^{\frac{5}{2}} - 0] = \frac{8}{5} \cdot 32 = \frac{256}{5} \]

2. For the second term:
\[ \int t^{\frac{5}{2}} \, dt = \frac{t^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} = \frac{t^{\frac{7}{2}}}{\frac{7}{2}} = \frac{2}{7} t^{\frac{7}{2}} \]
Thus,
\[ \int_{0}^{4} t^{\frac{5}{2}} \, dt = \left[ \frac{2}{7} t^{\frac{7}{2}} \right]_{0}^{4} = \frac{2}{7} [4^{\frac{7}{2}} - 0] = \frac{2}{7} \cdot 128 = \frac{256}{7} \]

### Step 6: Combine the results
Now substituting back into our expression for \( I \):
\[ I = \frac{256}{5} - \frac{256}{7} \]

### Step 7: Find a common denominator
The common denominator of 5 and 7 is 35. Thus, we rewrite the fractions:
\[ I = \frac{256 \cdot 7}{35} - \frac{256 \cdot 5}{35} = \frac{1792 - 1280}{35} = \frac{512}{35} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{512}{35}} \]