**Evaluate the integral**

**3**

**x2  3x 10 dx as limit of sums**

To evaluate the integral \(\int_0^3 (x^2 - 3x + 10) \, dx\) as a limit of sums, we use the concept of Riemann sums. Here are the steps:

### Step 1: Partition the interval
Let's partition the interval \([0, 3]\) into \(n\) subintervals of equal length. The length of each subinterval, \(\Delta x\), is:
\[
\Delta x = \frac{3 - 0}{n} = \frac{3}{n}
\]
The \(i\)-th subinterval will be \([x_{i-1}, x_i]\) where:
\[
x_i = 0 + i\Delta x = \frac{3i}{n}
\]

### Step 2: Choose sample points
We'll choose the right endpoint of each subinterval as the sample point. So, the sample point \(x_i^*\) in the \(i\)-th subinterval is:
\[
x_i^* = \frac{3i}{n}
\]

### Step 3: Write the Riemann sum
The Riemann sum for the integral is given by:
\[
\sum_{i=1}^n f(x_i^*) \Delta x
\]
Substituting \(f(x) = x^2 - 3x + 10\), \(x_i^* = \frac{3i}{n}\), and \(\Delta x = \frac{3}{n}\), the Riemann sum becomes:
\[
\sum_{i=1}^n \left(\left(\frac{3i}{n}\right)^2 - 3\left(\frac{3i}{n}\right) + 10\right) \cdot \frac{3}{n}
\]
Simplifying the expression inside the sum:
\[
\sum_{i=1}^n \left(\frac{9i^2}{n^2} - \frac{9i}{n} + 10\right) \cdot \frac{3}{n}
\]
Distribute \(\frac{3}{n}\):
\[
\sum_{i=1}^n \left(\frac{27i^2}{n^3} - \frac{27i}{n^2} + \frac{30}{n}\right)
\]

### Step 4: Take the limit as \(n \to \infty\)
The integral is the limit of the Riemann sum as \(n\) approaches infinity:
\[
\int_0^3 (x^2 - 3x + 10) \, dx = \lim_{n \to \infty} \sum_{i=1}^n \left(\frac{27i^2}{n^3} - \frac{27i}{n^2} + \frac{30}{n}\right)
\]
We can evaluate the sum term by term:

1. **First term:**
   \[
   \lim_{n \to \infty} \frac{27}{n^3} \sum_{i=1}^n i^2 = \frac{27}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}
   \]
   Simplifying:
   \[
   \lim_{n \to \infty} \frac{27n^3 + 27n^2 + 9n}{6n^3} = \lim_{n \to \infty} \left(\frac{27}{6} + \frac{27}{6n} + \frac{9}{6n^2}\right) = \frac{27}{6} = 4.5
   \]

2. **Second term:**
   \[
   \lim_{n \to \infty} \frac{-27}{n^2} \sum_{i=1}^n i = \frac{-27}{n^2} \cdot \frac{n(n+1)}{2}
   \]
   Simplifying:
   \[
   \lim_{n \to \infty} \frac{-27n^2 - 27n}{2n^2} = \lim_{n \to \infty} \left(\frac{-27}{2} - \frac{27}{2n}\right) = \frac{-27}{2} = -13.5
   \]

3. **Third term:**
   \[
   \lim_{n \to \infty} \sum_{i=1}^n \frac{30}{n} = 30 \cdot 1 = 30
   \]

### Final Result:
Summing up the results of these limits:
\[
\int_0^3 (x^2 - 3x + 10) \, dx = 4.5 - 13.5 + 30 = 21
\]

So, the value of the integral is \(\boxed{21}\).

Would you like to discuss any part of this calculation in more detail?

Here are 5 related questions to explore:
1. How does the choice of sample points (left, right, or midpoint) affect the Riemann sum?
2. Can you apply the same Riemann sum approach to an integral with a different function?
3. How can we generalize this method for non-uniform partitions?
4. What is the geometric interpretation of the integral as the area under the curve?
5. How does the limit process work in Riemann sums, and why does it lead to the exact value of the integral?

**Tip:** Practicing the evaluation of integrals using Riemann sums helps reinforce the concept of integration as the sum of areas of infinitesimally small rectangles.

Learn how to evaluate the integral ∫(x^2 - 3x + 10) dx from 0 to 3 using Riemann sums. Understand the step-by-step process, including partitioning the interval, choosing sample points, and taking the limit as n approaches infinity. Suitable for college-level calculus students.

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