`int_(0)^(1) x (1 -x)^(n) dx=?`

Answer

To solve the integral \( \int_{0}^{1} x (1 - x)^{n} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = 1 - x \). Then, we differentiate to find \( dx \):
\[ dx = -dt \]
Now, we need to change the limits of integration. When \( x = 0 \), \( t = 1 \), and when \( x = 1 \), \( t = 0 \). Thus, the integral becomes:
\[ \int_{0}^{1} x (1 - x)^{n} \, dx = \int_{1}^{0} (1 - t) t^{n} (-dt) = \int_{0}^{1} (1 - t) t^{n} \, dt \]

### Step 2: Expand the integrand
Now, we expand the integrand:
\[ (1 - t) t^{n} = t^{n} - t^{n+1} \]
So the integral can be rewritten as:
\[ \int_{0}^{1} (t^{n} - t^{n+1}) \, dt \]

### Step 3: Integrate term by term
Now we can integrate each term separately:
\[ \int_{0}^{1} t^{n} \, dt - \int_{0}^{1} t^{n+1} \, dt \]
Using the formula for the integral of \( t^{k} \):
\[ \int t^{k} \, dt = \frac{t^{k+1}}{k+1} + C \]
we find:
\[ \int_{0}^{1} t^{n} \, dt = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1} \]
and
\[ \int_{0}^{1} t^{n+1} \, dt = \frac{1^{n+2}}{n+2} - \frac{0^{n+2}}{n+2} = \frac{1}{n+2} \]

### Step 4: Combine the results
Now substituting back into our expression:
\[ \int_{0}^{1} (t^{n} - t^{n+1}) \, dt = \frac{1}{n+1} - \frac{1}{n+2} \]

### Step 5: Simplify the expression
To simplify \( \frac{1}{n+1} - \frac{1}{n+2} \), we find a common denominator:
\[ \frac{(n+2) - (n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} \]

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

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