Use the limit definition to find the region's area bounded by the following function over the given interval on the x-axis. Give an exact answer. 𝑓(𝑥) = 3𝑥 − 2 on [2,5]

To find the area under the curve \( f(x) = 3x - 2 \) over the interval \([2, 5]\) using the limit definition of the definite integral, we will apply the following steps:

### Step 1: Set up the formula for the area using the limit definition of a definite integral
The area under the curve of a function \( f(x) \) over the interval \([a, b]\) can be expressed as:

\[
A = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x
\]

Where:
- \( \Delta x = \frac{b - a}{n} \) is the width of each subinterval.
- \( x_i^* \) is a sample point within each subinterval.
- \( n \) is the number of subintervals.

In this case, \( a = 2 \), \( b = 5 \), and \( f(x) = 3x - 2 \).

### Step 2: Compute \( \Delta x \) and the sample points \( x_i^* \)
The width of each subinterval is:

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

We'll use the right endpoint for the sample points \( x_i^* \), so:

\[
x_i^* = 2 + i \Delta x = 2 + i \cdot \frac{3}{n} = 2 + \frac{3i}{n}
\]

### Step 3: Set up the Riemann sum
Now, substitute \( x_i^* \) into the function \( f(x) = 3x - 2 \):

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

The Riemann sum becomes:

\[
S_n = \sum_{i=1}^{n} \left( 4 + \frac{9i}{n} \right) \cdot \frac{3}{n}
\]

### Step 4: Simplify the sum
Now distribute \( \frac{3}{n} \) inside the sum:

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

Split the sum into two parts:

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

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 sum:

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

Simplify:

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

### Step 5: Take the limit as \( n \to \infty \)
Now take the limit as \( n \to \infty \):

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

The second term simplifies as follows:

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

As \( n \to \infty \), \( \frac{27}{2n} \to 0 \), so the limit becomes:

\[
A = 12 + \frac{27}{2} = \frac{24}{2} + \frac{27}{2} = \frac{51}{2}
\]

### Final Answer:
The exact area under the curve \( f(x) = 3x - 2 \) over the interval \([2, 5]\) is:

\[
A = \frac{51}{2}
\]

Would you like more details or have any questions?

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Here are 5 related questions you might explore:

1. How does the limit definition of a definite integral differ from the Fundamental Theorem of Calculus?
2. What is the geometric interpretation of the Riemann sum used in this problem?
3. How can we apply different sampling methods (midpoint, left endpoint) to calculate areas?
4. What happens if the function is non-linear? How does that change the Riemann sum?
5. Can you find the area under the curve of \( f(x) = x^2 \) over the interval [1, 4]?

**Tip**: Using the limit definition of a definite integral is useful for understanding the rigorous foundation of integration, but using the Fundamental Theorem of Calculus is usually faster!

Learn how to find the area under the curve for f(x) = 3x - 2 over the interval [2,5] using the limit definition of a definite integral. This problem uses Riemann sums and the limit process for calculating the exact area, suitable for Grades 11-12 calculus students.

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