Evaluate the following integrals as limit of sums:
`int_(1)^(3)(3x^(2)-2x+4)dx`

Answer

To evaluate the integral \( \int_{1}^{3} (3x^2 - 2x + 4) \, dx \) as a limit of sums, we will follow these steps:

### Step 1: Define the integral and the function
We have the integral
\[ \int_{1}^{3} (3x^2 - 2x + 4) \, dx \]
Let \( f(x) = 3x^2 - 2x + 4 \).

### Step 2: Determine the interval and the width of subintervals
The limits of integration are \( a = 1 \) and \( b = 3 \). The width of each subinterval \( h \) is given by:
\[ h = \frac{b - a}{n} = \frac{3 - 1}{n} = \frac{2}{n} \]

### Step 3: Set up the Riemann sum
The Riemann sum for the integral can be expressed as:
\[ \lim_{n \to \infty} \sum_{i=0}^{n-1} f(a + ih) \cdot h \]
Substituting \( a = 1 \) and \( h = \frac{2}{n} \):
\[ \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(1 + i \cdot \frac{2}{n}\right) \cdot \frac{2}{n} \]

### Step 4: Evaluate \( f\left(1 + i \cdot \frac{2}{n}\right) \)
Now we calculate \( f\left(1 + i \cdot \frac{2}{n}\right) \):
\[ f\left(1 + i \cdot \frac{2}{n}\right) = 3\left(1 + \frac{2i}{n}\right)^2 - 2\left(1 + \frac{2i}{n}\right) + 4 \]
Calculating \( \left(1 + \frac{2i}{n}\right)^2 \):
\[ \left(1 + \frac{2i}{n}\right)^2 = 1 + \frac{4i}{n} + \frac{4i^2}{n^2} \]
Thus,
\[ f\left(1 + i \cdot \frac{2}{n}\right) = 3\left(1 + \frac{4i}{n} + \frac{4i^2}{n^2}\right) - 2\left(1 + \frac{2i}{n}\right) + 4 \]
Expanding this:
\[ = 3 + \frac{12i}{n} + \frac{12i^2}{n^2} - 2 - \frac{4i}{n} + 4 \]
Combining like terms:
\[ = 5 + \frac{8i}{n} + \frac{12i^2}{n^2} \]

### Step 5: Substitute back into the Riemann sum
Now substituting back into the Riemann sum:
\[ \lim_{n \to \infty} \sum_{i=0}^{n-1} \left(5 + \frac{8i}{n} + \frac{12i^2}{n^2}\right) \cdot \frac{2}{n} \]
This can be separated into three sums:
\[ = \lim_{n \to \infty} \left( \frac{2}{n} \sum_{i=0}^{n-1} 5 + \frac{2}{n} \sum_{i=0}^{n-1} \frac{8i}{n} + \frac{2}{n} \sum_{i=0}^{n-1} \frac{12i^2}{n^2} \right) \]

### Step 6: Evaluate each sum
1. The first sum:
\[ \frac{2}{n} \sum_{i=0}^{n-1} 5 = \frac{2 \cdot 5(n)}{n} = 10 \]

2. The second sum:
\[ \frac{2}{n} \sum_{i=0}^{n-1} \frac{8i}{n} = \frac{16}{n^2} \sum_{i=0}^{n-1} i = \frac{16}{n^2} \cdot \frac{(n-1)n}{2} \approx 8 \]

3. The third sum:
\[ \frac{2}{n} \sum_{i=0}^{n-1} \frac{12i^2}{n^2} = \frac{24}{n^3} \sum_{i=0}^{n-1} i^2 = \frac{24}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} \approx 4 \]

### Step 7: Combine the results
Combining these results:
\[ 10 + 8 + 4 = 22 \]

### Final Result
Thus, the value of the integral is:
\[ \int_{1}^{3} (3x^2 - 2x + 4) \, dx = 22 \]