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

Options

A. `14/3`
B. `-14/3`
C. `16/3`
D. `-16/3`

Explanation

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

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

### Step 2: Identify the values of \( a \), \( b \), and \( f(x) \)
In our case:
- \( a = 0 \)
- \( b = 2 \)
- \( f(x) = x^2 + 1 \)

### Step 3: Calculate \( h \)
Using the formula for \( h \):
\[ h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n} \]

### Step 4: Write the Riemann sum
The Riemann sum becomes:
\[ \int_0^2 (x^2 + 1) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(0 + i \cdot \frac{2}{n}\right) \cdot \frac{2}{n} \]
This simplifies to:
\[ = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left( \left(\frac{2i}{n}\right)^2 + 1 \right) \cdot \frac{2}{n} \]

### Step 5: Substitute \( f(x) \) into the sum
Substituting \( f\left(\frac{2i}{n}\right) \):
\[ = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left( \frac{4i^2}{n^2} + 1 \right) \cdot \frac{2}{n} \]
\[ = \lim_{n \to \infty} \sum_{i=0}^{n-1} \left( \frac{8i^2}{n^3} + \frac{2}{n} \right) \]

### Step 6: Separate the sum into two parts
This can be separated into two sums:
\[ = \lim_{n \to \infty} \left( \frac{8}{n^3} \sum_{i=0}^{n-1} i^2 + \sum_{i=0}^{n-1} \frac{2}{n} \right) \]

### Step 7: Evaluate the sums
Using the formula for the sum of squares:
\[ \sum_{i=0}^{n-1} i^2 = \frac{(n-1)n(2n-1)}{6} \]
Thus,
\[ \lim_{n \to \infty} \left( \frac{8}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} + 2 \cdot \frac{n}{n} \right) \]
The second term simplifies to:
\[ \lim_{n \to \infty} 2 = 2 \]

### Step 8: Simplify the first sum
Now simplifying the first term:
\[ \frac{8}{n^3} \cdot \frac{(n-1)n(2n-1)}{6} \approx \frac{8 \cdot 2n^3}{6n^3} = \frac{16}{6} = \frac{8}{3} \]

### Step 9: Combine the results
Combining both results:
\[ \lim_{n \to \infty} \left( \frac{8}{3} + 2 \right) = \frac{8}{3} + \frac{6}{3} = \frac{14}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^2 (x^2 + 1) \, dx = \frac{14}{3} \]