Evaluate: `int_(1)^(2)(3x^(2)+2x+1)dx`

Explanation

To evaluate the integral \( \int_{1}^{2} (3x^{2} + 2x + 1) \, dx \), we will follow these steps:

### Step 1: Identify the integral
We need to evaluate the definite integral of the polynomial \( 3x^{2} + 2x + 1 \) from 1 to 2.

### Step 2: Find the antiderivative
To find the antiderivative of \( 3x^{2} + 2x + 1 \), we integrate each term separately:
- The integral of \( 3x^{2} \) is \( x^{3} \) (since \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \)).
- The integral of \( 2x \) is \( x^{2} \).
- The integral of \( 1 \) is \( x \).

Thus, the antiderivative of \( 3x^{2} + 2x + 1 \) is:
\[ x^{3} + x^{2} + x + C \]

### Step 3: Evaluate the definite integral
Now we evaluate the definite integral from 1 to 2:
\[ \int_{1}^{2} (3x^{2} + 2x + 1) \, dx = \left[ x^{3} + x^{2} + x \right]_{1}^{2} \]

### Step 4: Substitute the limits
Now we substitute the upper limit (2) and the lower limit (1) into the antiderivative:
1. For the upper limit (2):
\[ (2^{3} + 2^{2} + 2) = 8 + 4 + 2 = 14 \]
2. For the lower limit (1):
\[ (1^{3} + 1^{2} + 1) = 1 + 1 + 1 = 3 \]

### Step 5: Calculate the result
Now we subtract the value at the lower limit from the value at the upper limit:
\[ 14 - 3 = 11 \]

### Final Answer
The value of the integral \( \int_{1}^{2} (3x^{2} + 2x + 1) \, dx \) is \( 11 \).

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