Evaluate: `int_(0)^(3) (2x^(2) +3x + 5)dx`, expressing as a limit of sum

Answer

To evaluate the integral \( \int_{0}^{3} (2x^{2} + 3x + 5) \, dx \) and express it as a limit of sums, we will follow these steps:

### Step 1: Define the integral as a limit of Riemann sums
The integral can be expressed as a limit of sums using the definition of the Riemann integral. We divide the interval \([0, 3]\) into \(n\) equal subintervals, each of width \(\Delta x = \frac{3 - 0}{n} = \frac{3}{n}\).

### Step 2: Determine the sample points
We can choose the right endpoints of the subintervals as our sample points. The right endpoint for the \(i\)-th subinterval is given by:
\[ x_i = 0 + i \Delta x = \frac{3i}{n} \]

### Step 3: Write the Riemann sum
The Riemann sum \(S_n\) for the function \(f(x) = 2x^2 + 3x + 5\) is given by:
\[ S_n = \sum_{i=1}^{n} f(x_i) \Delta x \]
Substituting \(f(x_i)\) and \(\Delta x\):
\[ S_n = \sum_{i=1}^{n} \left(2\left(\frac{3i}{n}\right)^{2} + 3\left(\frac{3i}{n}\right) + 5\right) \cdot \frac{3}{n} \]

### Step 4: Simplify the expression
Now we simplify \(S_n\):
\[ S_n = \sum_{i=1}^{n} \left(2 \cdot \frac{9i^2}{n^2} + 3 \cdot \frac{3i}{n} + 5\right) \cdot \frac{3}{n} \]
\[ = \sum_{i=1}^{n} \left(\frac{27i^2}{n^2} + \frac{9i}{n} + 15\right) \cdot \frac{3}{n} \]
\[ = \frac{3}{n} \sum_{i=1}^{n} \left(\frac{27i^2}{n^2} + \frac{9i}{n} + 15\right) \]

### Step 5: Distribute the sum
Distributing the sum gives:
\[ S_n = \frac{3}{n} \left(\frac{27}{n^2} \sum_{i=1}^{n} i^2 + \frac{9}{n} \sum_{i=1}^{n} i + 15n\right) \]

### Step 6: Use formulas for sums
Using the formulas:
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
we substitute these into our expression for \(S_n\):
\[ S_n = \frac{3}{n} \left(\frac{27}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{9}{n} \cdot \frac{n(n+1)}{2} + 15n\right) \]

### Step 7: Take the limit as \(n \to \infty\)
Now, we take the limit as \(n\) approaches infinity:
\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(\frac{3}{n} \left(\frac{27(n+1)(2n+1)}{6n} + \frac{9(n+1)}{2} + 15n\right)\right) \]

### Step 8: Evaluate the limit
After simplifying and evaluating the limit, we find:
\[ = \frac{3}{n} \left(\frac{27(2 + \frac{1}{n})(1 + \frac{1}{n})}{6} + \frac{9(1 + \frac{1}{n})}{2} + 15\right) \]
As \(n \to \infty\), the terms involving \(\frac{1}{n}\) vanish, leading to:
\[ = \frac{3}{n} \left(\frac{27 \cdot 2}{6} + \frac{9}{2} + 15\right) \]
Calculating this gives the final result.

### Final Result
The evaluated integral is:
\[ \int_{0}^{3} (2x^{2} + 3x + 5) \, dx = \frac{93}{2} \]