The value of the integral `int_(0)^(1) x (1-x)^(n) dx` is

Options

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

Answer

To solve the integral \( I = \int_{0}^{1} x (1-x)^{n} \, dx \), we can use a property of definite integrals. Let's go through the solution step by step.

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{1} x (1-x)^{n} \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property of definite integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \]
Applying this property, we can rewrite the integral \( I \) as:
\[ I = \int_{0}^{1} (1-x) (1-(1-x))^{n} \, dx \]
This simplifies to:
\[ I = \int_{0}^{1} (1-x) x^{n} \, dx \]

### 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 gives:
\[ 2I = \int_{0}^{1} \left[ x (1-x)^{n} + (1-x) x^{n} \right] \, dx \]
This can be simplified to:
\[ 2I = \int_{0}^{1} \left[ x^{n} (1-x) + x (1-x)^{n} \right] \, dx \]

### Step 4: Factor Out Common Terms
We can factor out the common term \( x(1-x) \):
\[ 2I = \int_{0}^{1} x (1-x) \left[ x^{n-1} + (1-x)^{n-1} \right] \, dx \]

### Step 5: Evaluate the Integral
Now, we can evaluate the integral:
\[ I = \frac{1}{2} \int_{0}^{1} x (1-x) \left[ x^{n-1} + (1-x)^{n-1} \right] \, dx \]

Using the formula for the integral of \( x^m (1-x)^n \):
\[ \int_{0}^{1} x^{m} (1-x)^{n} \, dx = \frac{m! n!}{(m+n+1)!} \]
we can evaluate:
1. For \( x^{n} (1-x) \):
\[ \int_{0}^{1} x^{n} (1-x) \, dx = \frac{n! 1!}{(n+2)!} = \frac{1}{(n+1)(n+2)} \]
2. For \( x (1-x)^{n} \):
\[ \int_{0}^{1} x (1-x)^{n} \, dx = \frac{1! n!}{(n+2)!} = \frac{1}{(n+1)(n+2)} \]

### Step 6: Combine Results
Thus, we have:
\[ 2I = \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} = \frac{2}{(n+1)(n+2)} \]
So,
\[ I = \frac{1}{(n+1)(n+2)} \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{1}{(n+1)(n+2)}} \]