What is `int_(0)^(1)x (1-x)^(9) dx` equal to ?

Options

A. `1/110`
B. `1/132`
C. `1/148`
D. `1/240`

Explanation

To solve the integral \( \int_{0}^{1} x (1-x)^{9} \, dx \), we can use a property of definite integrals. Let's go through the steps:

### Step 1: Use the property of definite integrals
We can use the property that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, \( a = 1 \), so we can write:
\[ \int_{0}^{1} x (1-x)^{9} \, dx = \int_{0}^{1} (1-x) (1-(1-x))^{9} \, dx \]
This simplifies to:
\[ \int_{0}^{1} (1-x) x^{9} \, dx \]

### Step 2: Combine the integrals
Now we have two integrals:
\[ \int_{0}^{1} x (1-x)^{9} \, dx + \int_{0}^{1} (1-x) x^{9} \, dx = \int_{0}^{1} (x (1-x)^{9} + (1-x) x^{9}) \, dx \]
This can be simplified to:
\[ \int_{0}^{1} (x^{9} (1-x) + x (1-x)^{9}) \, dx \]

### Step 3: Factor out common terms
Notice that:
\[ x (1-x)^{9} + x^{9} (1-x) = x (1-x) (x^{8} + (1-x)^{8}) \]
Thus, we can rewrite the integral as:
\[ \int_{0}^{1} x (1-x) (x^{8} + (1-x)^{8}) \, dx \]

### Step 4: Evaluate the integral
Now we can evaluate the integral:
\[ \int_{0}^{1} x (1-x) \, dx \]
This can be calculated as:
\[ \int_{0}^{1} x \, dx - \int_{0}^{1} x^{2} \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} - \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Step 5: Combine results
Now we can combine our results:
\[ \int_{0}^{1} x (1-x)^{9} \, dx = \frac{1}{2} \cdot \frac{1}{10} = \frac{1}{20} \]

### Final Result
Thus, the value of the integral \( \int_{0}^{1} x (1-x)^{9} \, dx \) is:
\[ \frac{1}{110} \]