Evaluate the following integrals as limit of sums:
`int_(2)^(4)(2x+1)dx`

Answer

To evaluate the integral \( \int_{2}^{4} (2x + 1) \, dx \) using the limit of sums, we will follow these steps:

### Step 1: Define the Integral
We start with the integral:
\[ I = \int_{2}^{4} (2x + 1) \, dx \]

### Step 2: Set Up the Limit of Sums
We will express the integral as a limit of Riemann sums. The formula for the definite integral can be expressed as:
\[ I = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(x_i^*) \Delta x \]
where \( \Delta x = \frac{b - a}{n} \) and \( x_i^* = a + i \Delta x \).

### Step 3: Determine \( a \), \( b \), and \( \Delta x \)
Here, \( a = 2 \), \( b = 4 \), and thus:
\[ \Delta x = \frac{4 - 2}{n} = \frac{2}{n} \]

### Step 4: Determine \( x_i^* \)
The points \( x_i^* \) can be defined as:
\[ x_i^* = 2 + i \Delta x = 2 + i \left(\frac{2}{n}\right) = 2 + \frac{2i}{n} \]

### Step 5: Evaluate \( f(x_i^*) \)
Now, we evaluate the function \( f(x) = 2x + 1 \) at \( x_i^* \):
\[ f(x_i^*) = 2\left(2 + \frac{2i}{n}\right) + 1 = 4 + \frac{4i}{n} + 1 = 5 + \frac{4i}{n} \]

### Step 6: Set Up the Riemann Sum
The Riemann sum can now be expressed as:
\[ \sum_{i=0}^{n-1} f(x_i^*) \Delta x = \sum_{i=0}^{n-1} \left(5 + \frac{4i}{n}\right) \cdot \frac{2}{n} \]
This expands to:
\[ \sum_{i=0}^{n-1} \left(5 \cdot \frac{2}{n} + \frac{8i}{n^2}\right) = \frac{10}{n} \sum_{i=0}^{n-1} 1 + \frac{8}{n^2} \sum_{i=0}^{n-1} i \]

### Step 7: Calculate the Sums
1. The first sum:
\[ \sum_{i=0}^{n-1} 1 = n \]
2. The second sum:
\[ \sum_{i=0}^{n-1} i = \frac{(n-1)n}{2} \]

### Step 8: Substitute Back into the Riemann Sum
Substituting these sums back gives:
\[ \frac{10}{n} \cdot n + \frac{8}{n^2} \cdot \frac{(n-1)n}{2} = 10 + \frac{8(n-1)}{2n} = 10 + \frac{4(n-1)}{n} \]

### Step 9: Take the Limit as \( n \to \infty \)
Now we take the limit:
\[ I = \lim_{n \to \infty} \left(10 + \frac{4(n-1)}{n}\right) = 10 + \lim_{n \to \infty} \left(4 - \frac{4}{n}\right) = 10 + 4 = 14 \]

### Final Result
Thus, the value of the integral is:
\[ \int_{2}^{4} (2x + 1) \, dx = 14 \]