Evaluate : `int_(0)^(1)x(1-x)^(n)dx`

Answer

To evaluate the integral \( I = \int_0^1 x(1-x)^n \, dx \), we can use a property of definite integrals. Let's go through the steps systematically.

### Step 1: Use the property of definite integrals
We can use the property that states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, \( a = 1 \). Thus, we can rewrite the integral as:
\[ I = \int_0^1 x(1-x)^n \, dx = \int_0^1 (1-x)(1-(1-x))^n \, dx \]

### Step 2: Substitute and simplify
Now let's substitute \( 1 - x \) in place of \( x \):
\[ I = \int_0^1 (1-x)x^n \, dx \]
This gives us:
\[ I = \int_0^1 (1-x)x^n \, dx \]

### Step 3: Combine the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_0^1 x(1-x)^n \, dx \)
2. \( I = \int_0^1 (1-x)x^n \, dx \)

Adding these two equations:
\[ 2I = \int_0^1 [x(1-x)^n + (1-x)x^n] \, dx \]

### Step 4: Factor the integrand
The integrand can be factored:
\[ x(1-x)^n + (1-x)x^n = x(1-x) \left( (1-x)^{n-1} + x^{n-1} \right) \]
Thus:
\[ 2I = \int_0^1 x(1-x) \left( (1-x)^{n-1} + x^{n-1} \right) \, dx \]

### Step 5: Evaluate the integral
Now we can evaluate the integral:
\[ 2I = \int_0^1 x(1-x) \, dx \cdot \int_0^1 \left( (1-x)^{n-1} + x^{n-1} \right) \, dx \]

The integral \( \int_0^1 x(1-x) \, dx \) can be calculated as follows:
\[ \int_0^1 x(1-x) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]

### Step 6: Calculate the remaining integrals
The integrals \( \int_0^1 (1-x)^{n-1} \, dx \) and \( \int_0^1 x^{n-1} \, dx \) can be computed using the beta function or directly:
\[ \int_0^1 (1-x)^{n-1} \, dx = \frac{1}{n} \]
\[ \int_0^1 x^{n-1} \, dx = \frac{1}{n} \]

Thus:
\[ 2I = \frac{1}{6} \left( \frac{1}{n} + \frac{1}{n} \right) = \frac{1}{6} \cdot \frac{2}{n} = \frac{1}{3n} \]

### Step 7: Solve for \( I \)
Finally, we can solve for \( I \):
\[ I = \frac{1}{6n} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_0^1 x(1-x)^n \, dx = \frac{1}{6n} \]