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 and some basic integration techniques. Here’s a step-by-step solution:

### Step 1: Use the property of definite integrals
We know that:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \]
For our case, \( a = 1 \), so we can write:
\[ I = \int_{0}^{1} x(1-x)^{n} \, dx = \int_{0}^{1} (1-x)(x)^{n} \, dx \]

### Step 2: Rewrite the integral
Now we can express the integral \( I \) in terms of the new variable:
\[ I = \int_{0}^{1} (1-x)x^{n} \, dx \]

### Step 3: Combine the two expressions for \( I \)
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 simplifies to:
\[ 2I = \int_{0}^{1} \left( x(1-x)^{n} + x^{n}(1-x) \right) \, dx = \int_{0}^{1} x(1-x)^{n} + x^{n}(1-x) \, dx \]

### Step 4: Factor out common terms
Factoring out \( x(1-x) \):
\[ 2I = \int_{0}^{1} x(1-x) \left( (1-x)^{n-1} + x^{n-1} \right) \, dx \]

### Step 5: Integrate \( x(1-x) \)
Now we can integrate:
\[ I = \int_{0}^{1} x(1-x)^{n} \, dx \]
Using the formula for the integral of \( x(1-x)^{n} \):
\[ \int x(1-x)^{n} \, dx = \frac{1}{n+1} - \frac{1}{n+2} \]

### Step 6: Evaluate the limits
Evaluating at the limits \( 0 \) and \( 1 \):
\[ I = \left[ \frac{x^{2}}{2} \cdot \frac{(1-x)^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{(n+1)(n+2)} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{1}{(n+1)(n+2)} \]