Evaluate : `int_(0)^(1) (1-x)^(3//2) dx`

Answer

To evaluate the integral \( \int_{0}^{1} (1-x)^{\frac{3}{2}} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = 1 - x \). Then, differentiating both sides gives us:
\[ dx = -dt \]

### Step 2: Change the limits of integration
When \( x = 0 \), \( t = 1 - 0 = 1 \).
When \( x = 1 \), \( t = 1 - 1 = 0 \).
Thus, the limits change from \( x: 0 \to 1 \) to \( t: 1 \to 0 \).

### Step 3: Rewrite the integral
Substituting \( t \) into the integral, we have:
\[ \int_{0}^{1} (1-x)^{\frac{3}{2}} \, dx = \int_{1}^{0} t^{\frac{3}{2}} (-dt) = -\int_{1}^{0} t^{\frac{3}{2}} \, dt \]
This can be rewritten as:
\[ \int_{0}^{1} t^{\frac{3}{2}} \, dt \]

### Step 4: Evaluate the integral
Now we can evaluate the integral:
\[ \int t^{\frac{3}{2}} \, dt = \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} = \frac{t^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} t^{\frac{5}{2}} \]

### Step 5: Apply the limits
Now we apply the limits from 0 to 1:
\[ \left[ \frac{2}{5} t^{\frac{5}{2}} \right]_{0}^{1} = \frac{2}{5} (1^{\frac{5}{2}}) - \frac{2}{5} (0^{\frac{5}{2}}) = \frac{2}{5} - 0 = \frac{2}{5} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{1} (1-x)^{\frac{3}{2}} \, dx = \frac{2}{5} \]

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