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

Answer

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

### Step 1: Define the integral as a limit of sums
The integral can be expressed as a limit of Riemann sums. We divide the interval \([0, 2]\) into \(n\) equal parts. The width of each subinterval is given by:
\[ h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n} \]

### Step 2: Identify the sample points
The sample points can be taken as the right endpoints of each subinterval. The \(i\)-th sample point is:
\[ x_i = a + ih = 0 + i \cdot \frac{2}{n} = \frac{2i}{n} \]

### Step 3: Write the Riemann sum
The Riemann sum for the function \(f(x) = x^2 + 3\) over the interval \([0, 2]\) is:
\[ S_n = \sum_{i=1}^{n} f(x_i) \cdot h = \sum_{i=1}^{n} \left( \left(\frac{2i}{n}\right)^2 + 3 \right) \cdot \frac{2}{n} \]

### Step 4: Simplify the Riemann sum
Substituting \(f(x_i)\) into the sum:
\[ S_n = \sum_{i=1}^{n} \left( \frac{4i^2}{n^2} + 3 \right) \cdot \frac{2}{n} \]
\[ = \sum_{i=1}^{n} \left( \frac{8i^2}{n^3} + \frac{6}{n} \right) \]
\[ = \frac{8}{n^3} \sum_{i=1}^{n} i^2 + \frac{6}{n} \sum_{i=1}^{n} 1 \]

### Step 5: Use the formula for the sum of squares
We know that:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
Substituting this into our sum:
\[ S_n = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{6n}{n} \]
\[ = \frac{8(n+1)(2n+1)}{6n^2} + 6 \]

### Step 6: Take the limit as \(n \to \infty\)
Now we take the limit as \(n\) approaches infinity:
\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{8(n+1)(2n+1)}{6n^2} + 6 \right) \]
\[ = \lim_{n \to \infty} \left( \frac{8(2n^2 + 3n + 1)}{6n^2} + 6 \right) \]
\[ = \lim_{n \to \infty} \left( \frac{16n^2 + 24n + 8}{6n^2} + 6 \right) \]
\[ = \frac{16}{6} + 6 = \frac{8}{3} + 6 = \frac{8}{3} + \frac{18}{3} = \frac{26}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{2} (x^{2} + 3) \, dx = \frac{26}{3} \]