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: Define the integral as a limit of a sum
The integral can be expressed as a limit of a Riemann sum. The formula for the Riemann sum is given by:
\[ \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{r=1}^{n} f\left(a + \frac{b-a}{n} r\right) \cdot \frac{b-a}{n} \]
In our case, \( a = 0 \) and \( b = 2 \). Thus, we have:
\[ \int_0^2 (x^2 + 1) \, dx = \lim_{n \to \infty} \sum_{r=1}^{n} f\left(0 + \frac{2-0}{n} r\right) \cdot \frac{2-0}{n} \]

### Step 2: Determine the function and the width of each subinterval
The width of each subinterval is:
\[ \Delta x = \frac{b-a}{n} = \frac{2}{n} \]
The function we are integrating is \( f(x) = x^2 + 1 \). Therefore, we need to evaluate \( f\left(\frac{2r}{n}\right) \):
\[ f\left(\frac{2r}{n}\right) = \left(\frac{2r}{n}\right)^2 + 1 = \frac{4r^2}{n^2} + 1 \]

### Step 3: Substitute into the Riemann sum
Now we can substitute this into the sum:
\[ \int_0^2 (x^2 + 1) \, dx = \lim_{n \to \infty} \sum_{r=1}^{n} \left(\frac{4r^2}{n^2} + 1\right) \cdot \frac{2}{n} \]
This can be split into two separate sums:
\[ = \lim_{n \to \infty} \left( \sum_{r=1}^{n} \left(\frac{4r^2}{n^2}\right) \cdot \frac{2}{n} + \sum_{r=1}^{n} 1 \cdot \frac{2}{n} \right) \]

### Step 4: Simplify the sums
The first sum becomes:
\[ \sum_{r=1}^{n} \frac{8r^2}{n^3} = \frac{8}{n^3} \sum_{r=1}^{n} r^2 \]
Using the formula for the sum of squares, \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \):
\[ \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{8(n(n+1)(2n+1))}{6n^3} \]
As \( n \to \infty \), this simplifies to:
\[ \frac{8 \cdot 2}{6} = \frac{16}{6} = \frac{8}{3} \]

The second sum becomes:
\[ \sum_{r=1}^{n} \frac{2}{n} = 2 \]

### Step 5: Combine the results
Now, combining both parts:
\[ \int_0^2 (x^2 + 1) \, dx = \lim_{n \to \infty} \left( \frac{8}{3} + 2 \right) = \frac{8}{3} + \frac{6}{3} = \frac{14}{3} \]

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