The value of `int_(0)^(1) x^(3) (1- x)^(11) dx` is equal to

Options

A. `(1)/( 11) + (3)/(12) + (3)/( 13) + (1)/( 14)`
B. `- (1)/( 11) + (3)/( 12) - (3)/( 13) + (1)/( 14)`
C. `(8)/( 13)`
D. `(1)/(11) - (3)/( 12) + (3)/(14) - (1)/( 14)`

Answer

To solve the integral \( I = \int_{0}^{1} x^{3} (1 - x)^{11} \, dx \), we can use the property of definite integrals that allows us to substitute \( x \) with \( 1 - x \).

### Step-by-Step Solution:

1. **Set Up the Integral:**
\[ I = \int_{0}^{1} x^{3} (1 - x)^{11} \, dx \]

2. **Use the Substitution Property:**
By substituting \( x \) with \( 1 - x \), we have:
\[ I = \int_{0}^{1} (1 - x)^{3} x^{11} \, dx \]

3. **Combine the Two Integrals:**
Now we can express \( I \) as:
\[ I = \int_{0}^{1} x^{3} (1 - x)^{11} \, dx + \int_{0}^{1} (1 - x)^{3} x^{11} \, dx \]
This gives us:
\[ 2I = \int_{0}^{1} \left( x^{3} (1 - x)^{11} + (1 - x)^{3} x^{11} \right) \, dx \]

4. **Simplify the Expression Inside the Integral:**
The expression can be simplified using the binomial expansion:
\[ (1 - x)^{11} = \sum_{k=0}^{11} \binom{11}{k} (-x)^{k} = \sum_{k=0}^{11} \binom{11}{k} (-1)^{k} x^{k} \]
Thus:
\[ x^{3} (1 - x)^{11} + (1 - x)^{3} x^{11} = x^{3} \sum_{k=0}^{11} \binom{11}{k} (-1)^{k} x^{k} + (1 - x)^{3} \sum_{j=0}^{11} \binom{11}{j} (-1)^{j} x^{j} \]

5. **Integrate Each Term:**
Now we can integrate term by term:
\[ 2I = \int_{0}^{1} \left( x^{3} (1 - x)^{11} + (1 - x)^{3} x^{11} \right) \, dx \]
This results in:
\[ 2I = \int_{0}^{1} (x^{3} (1 - x)^{11} + x^{11} (1 - x)^{3}) \, dx \]

6. **Evaluate the Integral:**
We can evaluate the integral using the Beta function:
\[ \int_{0}^{1} x^{m} (1 - x)^{n} \, dx = \frac{\Gamma(m + 1) \Gamma(n + 1)}{\Gamma(m + n + 2)} \]
For our case:
\[ I = \frac{\Gamma(4) \Gamma(12)}{\Gamma(16)} = \frac{3! \cdot 11!}{15!} \]

7. **Calculate the Values:**
\[ \Gamma(4) = 3! = 6, \quad \Gamma(12) = 11! = 39916800, \quad \Gamma(16) = 15! = 1307674368000 \]
Thus:
\[ I = \frac{6 \cdot 39916800}{1307674368000} = \frac{239500800}{1307674368000} = \frac{1}{5460} \]

8. **Final Result:**
Therefore, the value of the integral is:
\[ I = \frac{1}{5460} \]