Evaluate: `int_0^2 (x^2+x+1)dx` as the limit of a sum.

Answer

To evaluate the integral \(\int_0^2 (x^2 + x + 1) \, dx\) as the limit of a sum, we can follow these steps:

### Step 1: Set up the integral as a limit of a sum
We know that the definite integral can be expressed as a limit of a Riemann sum. The formula for the integral from \(a\) to \(b\) is given by:
\[ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(a + i \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n} \]
In our case, \(a = 0\), \(b = 2\), and \(f(x) = x^2 + x + 1\).

### Step 2: Determine \(h\) and \(n\)
Here, we have:
\[ b - a = 2 - 0 = 2 \]
Let \(h = \frac{b - a}{n} = \frac{2}{n}\). Thus, \(nh = 2\).

### Step 3: Write the Riemann sum
The Riemann sum can be expressed as:
\[ \int_0^2 (x^2 + x + 1) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(0 + i \cdot \frac{2}{n}\right) \cdot \frac{2}{n} \]
Substituting \(f(x)\):
\[ = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left(\left(\frac{2i}{n}\right)^2 + \frac{2i}{n} + 1\right) \cdot \frac{2}{n} \]

### Step 4: Simplify the expression
Now we simplify the expression inside the sum:
\[ = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left(\frac{4i^2}{n^2} + \frac{2i}{n} + 1\right) \cdot \frac{2}{n} \]
Distributing \(\frac{2}{n}\):
\[ = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left(\frac{8i^2}{n^3} + \frac{4i}{n^2} + \frac{2}{n}\right) \]

### Step 5: Split the sum
Now we can split the sum:
\[ = \lim_{n \to \infty} \left(\frac{8}{n^3} \sum_{i=0}^{n-1} i^2 + \frac{4}{n^2} \sum_{i=0}^{n-1} i + \sum_{i=0}^{n-1} 2\right) \]

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

Substituting these into the limit:
\[ = \lim_{n \to \infty} \left(\frac{8}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} + \frac{4}{n^2} \cdot \frac{(n-1)n}{2} + 2n\right) \]

### Step 7: Simplify each term
1. The first term:
\[ \frac{8(n-1)n(2n-1)}{6n^3} \approx \frac{16}{3} \text{ as } n \to \infty \]
2. The second term:
\[ \frac{4(n-1)n}{2n^2} \approx 2 \text{ as } n \to \infty \]
3. The third term:
\[ 2n \text{ remains as } 2 \text{ as } n \to \infty \]

### Step 8: Combine the results
Combining these results:
\[ = \frac{16}{3} + 2 + 2 = \frac{16}{3} + \frac{6}{3} + \frac{6}{3} = \frac{28}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^2 (x^2 + x + 1) \, dx = \frac{28}{3} \]