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: Set up the integral
Let \( I = \int_{1}^{2} (3x^{2} + 2x + 1) \, dx \).

### Step 2: Integrate the function
We will integrate each term in the integrand separately:
- The integral of \( 3x^{2} \) is \( \frac{3}{3} x^{3} = x^{3} \).
- The integral of \( 2x \) is \( \frac{2}{2} x^{2} = x^{2} \).
- The integral of \( 1 \) is \( x \).

Thus, we have:
\[ I = \left[ x^{3} + x^{2} + x \right]_{1}^{2} \]

### Step 3: Evaluate the definite integral at the limits
Now we will evaluate the expression at the upper limit (2) and the lower limit (1):
1. Calculate at \( x = 2 \):
\[ 2^{3} + 2^{2} + 2 = 8 + 4 + 2 = 14 \]
2. Calculate at \( x = 1 \):
\[ 1^{3} + 1^{2} + 1 = 1 + 1 + 1 = 3 \]

### Step 4: Subtract the lower limit from the upper limit
Now we subtract the value at the lower limit from the value at the upper limit:
\[ I = 14 - 3 = 11 \]

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

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