Use the limit process to find the area under the following curves on the given interval. 
𝑦 = 3𝑥 − 2, [2,5]



To find the area under the curve \( y = 3x - 2 \) over the interval \([2, 5]\) using the limit process, we can approximate the area as the sum of the areas of rectangles under the curve, then take the limit as the number of rectangles approaches infinity.

Here's the step-by-step breakdown:

### Step 1: Divide the interval into subintervals
Let’s divide the interval \([2, 5]\) into \(n\) equal subintervals. The width of each subinterval, \( \Delta x \), is given by:

\[
\Delta x = \frac{b - a}{n} = \frac{5 - 2}{n} = \frac{3}{n}
\]

### Step 2: Identify the sample points
Let \( x_i^* \) be the \(i\)-th sample point in the interval. We will use the right endpoints of the subintervals for this example. So, the right endpoint \(x_i^*\) is given by:

\[
x_i^* = a + i \Delta x = 2 + i \left(\frac{3}{n}\right)
\]

### Step 3: Express the function at each sample point
For each subinterval, we evaluate the function \( y = 3x - 2 \) at the right endpoint \(x_i^*\):

\[
f(x_i^*) = 3\left(2 + i\frac{3}{n}\right) - 2 = 6 + 9\frac{i}{n} - 2 = 4 + 9\frac{i}{n}
\]

### Step 4: Approximate the area using Riemann sums
The area under the curve can be approximated by summing the areas of the rectangles. The area of each rectangle is given by \(f(x_i^*) \cdot \Delta x\), so the total area is approximated by:

\[
A_n = \sum_{i=1}^{n} f(x_i^*) \Delta x = \sum_{i=1}^{n} \left( 4 + 9\frac{i}{n} \right) \cdot \frac{3}{n}
\]

Expanding the sum:

\[
A_n = \sum_{i=1}^{n} \left( \frac{12}{n} + \frac{27i}{n^2} \right)
\]

\[
A_n = \frac{12}{n} \sum_{i=1}^{n} 1 + \frac{27}{n^2} \sum_{i=1}^{n} i
\]

### Step 5: Simplify the sums
We know that:

\[
\sum_{i=1}^{n} 1 = n \quad \text{and} \quad \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
\]

Substituting these into the expression for \(A_n\):

\[
A_n = \frac{12}{n} \cdot n + \frac{27}{n^2} \cdot \frac{n(n+1)}{2}
\]

Simplifying:

\[
A_n = 12 + \frac{27(n+1)}{2n}
\]

### Step 6: Take the limit as \(n \to \infty\)
Now, we take the limit as \(n\) approaches infinity to find the exact area:

\[
A = \lim_{n \to \infty} \left( 12 + \frac{27(n+1)}{2n} \right)
\]

As \(n \to \infty\), \( \frac{n+1}{n} \to 1 \), so:

\[
A = 12 + \frac{27}{2} = 12 + 13.5 = 25.5
\]

### Final Answer
The area under the curve \( y = 3x - 2 \) over the interval \([2, 5]\) is **25.5 square units**.

---

Would you like more details on any step or have any questions?

Here are 5 related questions to further your understanding:

1. How would the result change if the interval was \([1, 4]\)?
2. Can you find the area under a different curve using the same limit process, e.g., \(y = x^2\)?
3. How does using left endpoints change the approximation process?
4. What would happen if the curve was below the x-axis on a certain interval?
5. Can you derive the area under the curve using definite integrals?

**Tip**: The limit process for finding the area is essentially the foundation of integral calculus, where Riemann sums become definite integrals.

Use the limit process to find the area under the following curves on the given interval. y = 3x − 2, [2,5]

Learn how to use the limit process to find the area under the curve y = 3x − 2 over the interval [2, 5]. This problem involves key calculus concepts, including Riemann sums and the limit process, and provides a detailed step-by-step solution. Suitable for students in Grades 11-12 or college-level calculus.

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