Evaluate `int_0^2(x^2+3) dx` as limit of sums.

Explanation

To evaluate the integral \( \int_0^2 (x^2 + 3) \, dx \) as a limit of sums, we will follow these steps:

### Step 1: Set up the integral as a limit of sums
The definite integral can be expressed as:
\[ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{r=0}^{n-1} f(a + r \cdot h) \cdot h \]
where \( h = \frac{b - a}{n} \).

For our integral:
- \( a = 0 \)
- \( b = 2 \)
- \( f(x) = x^2 + 3 \)

Thus, we have:
\[ h = \frac{2 - 0}{n} = \frac{2}{n} \]

### Step 2: Substitute into the limit of sums
Now, substituting into the limit of sums:
\[ \int_0^2 (x^2 + 3) \, dx = \lim_{n \to \infty} \sum_{r=0}^{n-1} \left( (0 + r \cdot h)^2 + 3 \right) \cdot h \]
Substituting \( h = \frac{2}{n} \):
\[ = \lim_{n \to \infty} \sum_{r=0}^{n-1} \left( \left( r \cdot \frac{2}{n} \right)^2 + 3 \right) \cdot \frac{2}{n} \]

### Step 3: Simplify the expression
Now, simplifying the expression inside the summation:
\[ = \lim_{n \to \infty} \sum_{r=0}^{n-1} \left( \frac{4r^2}{n^2} + 3 \right) \cdot \frac{2}{n} \]
\[ = \lim_{n \to \infty} \sum_{r=0}^{n-1} \left( \frac{8r^2}{n^3} + \frac{6}{n} \right) \]
This can be separated into two sums:
\[ = \lim_{n \to \infty} \left( \frac{8}{n^3} \sum_{r=0}^{n-1} r^2 + \frac{6}{n} \sum_{r=0}^{n-1} 1 \right) \]

### Step 4: Calculate the sums
Using the formula for the sum of squares:
\[ \sum_{r=0}^{n-1} r^2 = \frac{(n-1)n(2n-1)}{6} \]
And the sum of ones:
\[ \sum_{r=0}^{n-1} 1 = n \]

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

### Step 5: Simplify further
As \( n \to \infty \):
\[ \frac{8(n-1)(2n-1)}{6n} \approx \frac{8 \cdot 2n^2}{6n} = \frac{16n}{6} = \frac{8n}{3} \]
Thus, we have:
\[ = \lim_{n \to \infty} \left( \frac{8n}{3} + 6 \right) = \frac{8}{3} \cdot 2 + 6 = \frac{16}{3} + 6 = \frac{16 + 18}{3} = \frac{34}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^2 (x^2 + 3) \, dx = \frac{34}{3} \]