`int_(0)^(1)x^(2)(1-x)^(3)dx` is equal to

Answer

To solve the integral \( \int_{0}^{1} x^{2} (1-x)^{3} \, dx \), we can follow these steps:

### Step 1: Expand the integrand
First, we need to expand \( (1-x)^{3} \) using the binomial theorem:
\[ (1-x)^{3} = 1 - 3x + 3x^{2} - x^{3} \]
So, we can rewrite the integral as:
\[ \int_{0}^{1} x^{2} (1 - 3x + 3x^{2} - x^{3}) \, dx \]

### Step 2: Distribute \( x^{2} \)
Now, distribute \( x^{2} \) across the expanded polynomial:
\[ \int_{0}^{1} (x^{2} - 3x^{3} + 3x^{4} - x^{5}) \, dx \]

### Step 3: Separate the integral
We can separate the integral into individual terms:
\[ \int_{0}^{1} x^{2} \, dx - 3 \int_{0}^{1} x^{3} \, dx + 3 \int_{0}^{1} x^{4} \, dx - \int_{0}^{1} x^{5} \, dx \]

### Step 4: Evaluate each integral
Now we can evaluate each integral:
1. \( \int_{0}^{1} x^{2} \, dx = \left[ \frac{x^{3}}{3} \right]_{0}^{1} = \frac{1}{3} \)
2. \( \int_{0}^{1} x^{3} \, dx = \left[ \frac{x^{4}}{4} \right]_{0}^{1} = \frac{1}{4} \)
3. \( \int_{0}^{1} x^{4} \, dx = \left[ \frac{x^{5}}{5} \right]_{0}^{1} = \frac{1}{5} \)
4. \( \int_{0}^{1} x^{5} \, dx = \left[ \frac{x^{6}}{6} \right]_{0}^{1} = \frac{1}{6} \)

### Step 5: Substitute back into the expression
Substituting these values back into our separated integrals:
\[ \frac{1}{3} - 3 \cdot \frac{1}{4} + 3 \cdot \frac{1}{5} - \frac{1}{6} \]

### Step 6: Simplify the expression
Now we need to simplify this expression. First, we find a common denominator, which is 60:
\[ \frac{1}{3} = \frac{20}{60}, \quad -3 \cdot \frac{1}{4} = -\frac{45}{60}, \quad 3 \cdot \frac{1}{5} = \frac{36}{60}, \quad -\frac{1}{6} = -\frac{10}{60} \]

Putting it all together:
\[ \frac{20}{60} - \frac{45}{60} + \frac{36}{60} - \frac{10}{60} = \frac{20 - 45 + 36 - 10}{60} = \frac{1}{60} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{1} x^{2} (1-x)^{3} \, dx = \frac{1}{60} \]