Evaluate the following integrals as limit of sums:
`int_(0)^(1)(3x^(2)+2x+1)dx`

Answer

To evaluate the integral \( \int_{0}^{1} (3x^2 + 2x + 1) \, dx \) as a limit of sums, we can follow these steps:

### Step 1: Set up the limit of sums
The integral can be expressed as a limit of sums using the formula:
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(a + i \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n} \]
In our case, \( a = 0 \), \( b = 1 \), and \( f(x) = 3x^2 + 2x + 1 \).

### Step 2: Determine \( \Delta x \)
Here, \( \Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n} \).

### Step 3: Substitute into the sum
We substitute \( x_i = a + i \Delta x = 0 + i \cdot \frac{1}{n} = \frac{i}{n} \) into the function:
\[ f\left(\frac{i}{n}\right) = 3\left(\frac{i}{n}\right)^2 + 2\left(\frac{i}{n}\right) + 1 \]
Thus,
\[ f\left(\frac{i}{n}\right) = 3\frac{i^2}{n^2} + 2\frac{i}{n} + 1 \]

### Step 4: Write the Riemann sum
The Riemann sum becomes:
\[ \sum_{i=0}^{n-1} f\left(\frac{i}{n}\right) \cdot \Delta x = \sum_{i=0}^{n-1} \left(3\frac{i^2}{n^2} + 2\frac{i}{n} + 1\right) \cdot \frac{1}{n} \]
This simplifies to:
\[ \sum_{i=0}^{n-1} \left(3\frac{i^2}{n^3} + 2\frac{i}{n^2} + \frac{1}{n}\right) \]

### Step 5: Separate the sums
We can separate the sums:
\[ \frac{1}{n^3} \sum_{i=0}^{n-1} 3i^2 + \frac{1}{n^2} \sum_{i=0}^{n-1} 2i + \sum_{i=0}^{n-1} \frac{1}{n} \]

### Step 6: Use formulas for sums
Using the formulas:
- \( \sum_{i=0}^{n-1} i = \frac{(n-1)n}{2} \)
- \( \sum_{i=0}^{n-1} i^2 = \frac{(n-1)n(2n-1)}{6} \)

We can substitute these into our expression:
\[ \frac{1}{n^3} \cdot 3 \cdot \frac{(n-1)n(2n-1)}{6} + \frac{1}{n^2} \cdot 2 \cdot \frac{(n-1)n}{2} + \frac{n}{n} \]

### Step 7: Simplify each term
1. For the first term:
\[ \frac{3(n-1)(2n-1)}{6n} = \frac{(n-1)(2n-1)}{2n} \]
2. For the second term:
\[ \frac{(n-1)}{n} = 1 - \frac{1}{n} \]
3. The third term is simply:
\[ 1 \]

### Step 8: Combine and take the limit
Combining these terms:
\[ \frac{(n-1)(2n-1)}{2n} + 1 - \frac{1}{n} = \frac{2n^2 - 3n + 1}{2n} + 1 - \frac{1}{n} \]
Taking the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \left( \frac{2n^2 - 3n + 1}{2n} + 1 - \frac{1}{n} \right) = \lim_{n \to \infty} \left( n - \frac{3}{2} + \frac{1}{2n} + 1 - \frac{1}{n} \right) = 1 + 1 - \frac{3}{2} = 3 \]

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