What is `int_(0)^(1) x(1-x)^(n)dx` equal to ?

Options

A. `(1)/(n(n+1))`
B. `1/((n+1)(n+2))`
C. `1`
D. `0`

Explanation

To solve the integral \( I = \int_{0}^{1} x(1-x)^{n} \, dx \), we will use a substitution method.

### Step-by-step Solution:

1. **Substitution**: Let \( t = 1 - x \). Then, \( x = 1 - t \) and \( dx = -dt \).

2. **Change of Limits**: When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \). Thus, the limits of integration change from \( 0 \) to \( 1 \) into \( 1 \) to \( 0 \).

3. **Rewrite the Integral**:
\[ I = \int_{1}^{0} (1-t) t^{n} (-dt) = \int_{0}^{1} (1-t) t^{n} dt \]

4. **Expand the Integral**:
\[ I = \int_{0}^{1} (t^{n} - t^{n+1}) dt \]

5. **Integrate**:
\[ I = \int_{0}^{1} t^{n} dt - \int_{0}^{1} t^{n+1} dt \]
The integrals can be computed as follows:
\[ \int_{0}^{1} t^{n} dt = \frac{1}{n+1} \]
\[ \int_{0}^{1} t^{n+1} dt = \frac{1}{n+2} \]

6. **Combine the Results**:
\[ I = \frac{1}{n+1} - \frac{1}{n+2} \]

7. **Find a Common Denominator**:
\[ I = \frac{(n+2) - (n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} \]

### Final Result:
Thus, the value of the integral \( \int_{0}^{1} x(1-x)^{n} \, dx \) is:
\[ \frac{1}{(n+1)(n+2)} \]