Evaluate : `int_(0)^(1) (x+1)^(2)dx`

Answer

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

### Step 1: Expand the integrand
First, we expand \( (x + 1)^{2} \):
\[ (x + 1)^{2} = x^{2} + 2x + 1 \]

### Step 2: Set up the integral
Now, we can rewrite the integral:
\[ \int_{0}^{1} (x + 1)^{2} \, dx = \int_{0}^{1} (x^{2} + 2x + 1) \, dx \]

### Step 3: Split the integral
We can split the integral into three separate integrals:
\[ \int_{0}^{1} (x^{2} + 2x + 1) \, dx = \int_{0}^{1} x^{2} \, dx + \int_{0}^{1} 2x \, dx + \int_{0}^{1} 1 \, dx \]

### Step 4: Evaluate each integral
Now, we evaluate each integral separately.

1. **For \( \int_{0}^{1} x^{2} \, dx \)**:
\[ \int x^{2} \, dx = \frac{x^{3}}{3} \quad \text{(from 0 to 1)} \]
Evaluating from 0 to 1:
\[ \left[ \frac{x^{3}}{3} \right]_{0}^{1} = \frac{1^{3}}{3} - \frac{0^{3}}{3} = \frac{1}{3} \]

2. **For \( \int_{0}^{1} 2x \, dx \)**:
\[ \int 2x \, dx = x^{2} \quad \text{(from 0 to 1)} \]
Evaluating from 0 to 1:
\[ \left[ x^{2} \right]_{0}^{1} = 1^{2} - 0^{2} = 1 \]

3. **For \( \int_{0}^{1} 1 \, dx \)**:
\[ \int 1 \, dx = x \quad \text{(from 0 to 1)} \]
Evaluating from 0 to 1:
\[ \left[ x \right]_{0}^{1} = 1 - 0 = 1 \]

### Step 5: Combine the results
Now we can combine the results from the three integrals:
\[ \int_{0}^{1} (x^{2} + 2x + 1) \, dx = \frac{1}{3} + 1 + 1 = \frac{1}{3} + 2 = \frac{1}{3} + \frac{6}{3} = \frac{7}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{1} (x + 1)^{2} \, dx \) is:
\[ \frac{7}{3} \]

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