Math Problem Statement

Use the limit definition to find the area of the region bounded by the graph of the following function. Give an exact answer. 𝑓(π‘₯) = 3π‘₯ βˆ’ 2 on [2,5]

Solution

To find the area of the region bounded by the graph of the function f(x)=3xβˆ’2f(x) = 3x - 2 on the interval [2,5][2,5], we will use the limit definition of a definite integral.

The limit definition of the area under a curve is given by:

Area=lim⁑nβ†’βˆžβˆ‘i=1nf(xiβˆ—)Ξ”x\text{Area} = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x

Where:

  • Ξ”x=bβˆ’an\Delta x = \frac{b - a}{n} is the width of each subinterval,
  • xiβˆ—x_i^* is a sample point in each subinterval [xiβˆ’1,xi][x_{i-1}, x_i],
  • f(xiβˆ—)f(x_i^*) is the function value at the sample point,
  • aa and bb are the endpoints of the interval.

For the given problem:

  • f(x)=3xβˆ’2f(x) = 3x - 2,
  • a=2a = 2, b=5b = 5.

Step 1: Set up the partition

Ξ”x=5βˆ’2n=3n\Delta x = \frac{5 - 2}{n} = \frac{3}{n}

We will choose xiβˆ—x_i^* as the right endpoint of each subinterval: xiβˆ—=a+iΞ”x=2+i3nx_i^* = a + i\Delta x = 2 + i\frac{3}{n}

Step 2: Calculate the sum

The area under the curve is approximated by the sum: βˆ‘i=1nf(xiβˆ—)Ξ”x=βˆ‘i=1n(3(2+i3n)βˆ’2)β‹…3n\sum_{i=1}^{n} f(x_i^*) \Delta x = \sum_{i=1}^{n} \left( 3(2 + i\frac{3}{n}) - 2 \right) \cdot \frac{3}{n} Simplifying f(xiβˆ—)f(x_i^*): f(xiβˆ—)=3(2+i3n)βˆ’2=6+9inβˆ’2=4+9inf(x_i^*) = 3(2 + i\frac{3}{n}) - 2 = 6 + \frac{9i}{n} - 2 = 4 + \frac{9i}{n}

Thus, the sum becomes: βˆ‘i=1n(4+9in)β‹…3n\sum_{i=1}^{n} \left( 4 + \frac{9i}{n} \right) \cdot \frac{3}{n} Distribute 3n\frac{3}{n}: βˆ‘i=1n(12n+27in2)\sum_{i=1}^{n} \left( \frac{12}{n} + \frac{27i}{n^2} \right) Now split the sum into two parts: βˆ‘i=1n12n+βˆ‘i=1n27in2\sum_{i=1}^{n} \frac{12}{n} + \sum_{i=1}^{n} \frac{27i}{n^2}

Step 3: Use sum formulas

For the first part: βˆ‘i=1n12n=12nβ‹…n=12\sum_{i=1}^{n} \frac{12}{n} = \frac{12}{n} \cdot n = 12

For the second part, use the formula for the sum of the first nn integers: βˆ‘i=1ni=n(n+1)2\sum_{i=1}^{n} i = \frac{n(n+1)}{2} Thus: βˆ‘i=1n27in2=27n2β‹…n(n+1)2=27(n+1)2n\sum_{i=1}^{n} \frac{27i}{n^2} = \frac{27}{n^2} \cdot \frac{n(n+1)}{2} = \frac{27(n+1)}{2n}

Step 4: Take the limit as nβ†’βˆžn \to \infty

The total sum is: 12+27(n+1)2n12 + \frac{27(n+1)}{2n} As nβ†’βˆžn \to \infty, n+1nβ†’1\frac{n+1}{n} \to 1, so: lim⁑nβ†’βˆž(12+27(n+1)2n)=12+272=12+13.5=25.5\lim_{n \to \infty} \left( 12 + \frac{27(n+1)}{2n} \right) = 12 + \frac{27}{2} = 12 + 13.5 = 25.5

Thus, the exact area under the curve is:

25.5\boxed{25.5}

This is the area of the region bounded by the graph of f(x)=3xβˆ’2f(x) = 3x - 2 on the interval [2,5][2, 5].

Would you like more details on any step or further clarification?


Here are 5 related questions:

  1. How does the limit definition of an integral differ from a Riemann sum?
  2. How would the method change if we had chosen left endpoints instead of right endpoints?
  3. What if the function were non-linear, such as f(x)=x2f(x) = x^2? How would the steps differ?
  4. How can we use the Fundamental Theorem of Calculus to compute this integral more efficiently?
  5. How is this method applied in finding areas between two curves?

Tip: When evaluating sums involving ii, remember common sum formulas like βˆ‘i=1ni=n(n+1)2\sum_{i=1}^n i = \frac{n(n+1)}{2}. This can simplify calculations significantly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Definite Integrals
Limits
Riemann Sums

Formulas

Limit definition of a definite integral: lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x
Sum of first n integers: \sum_{i=1}^{n} i = \frac{n(n+1)}{2}

Theorems

Limit Definition of Definite Integral
Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, Early College