Math Problem Statement

Use the limit process to find the area under the following curves on the given interval. 𝑦 = 3π‘₯ βˆ’ 2, [2,5]

Solution

To find the area under the curve y=3xβˆ’2y = 3x - 2 over the interval [2,5][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][2, 5] into nn equal subintervals. The width of each subinterval, Ξ”x\Delta x, is given by:

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

Step 2: Identify the sample points

Let xiβˆ—x_i^* be the ii-th sample point in the interval. We will use the right endpoints of the subintervals for this example. So, the right endpoint xiβˆ—x_i^* is given by:

xiβˆ—=a+iΞ”x=2+i(3n)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=3xβˆ’2y = 3x - 2 at the right endpoint xiβˆ—x_i^*:

f(xiβˆ—)=3(2+i3n)βˆ’2=6+9inβˆ’2=4+9inf(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(xiβˆ—)β‹…Ξ”xf(x_i^*) \cdot \Delta x, so the total area is approximated by:

An=βˆ‘i=1nf(xiβˆ—)Ξ”x=βˆ‘i=1n(4+9in)β‹…3nA_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:

An=βˆ‘i=1n(12n+27in2)A_n = \sum_{i=1}^{n} \left( \frac{12}{n} + \frac{27i}{n^2} \right)

An=12nβˆ‘i=1n1+27n2βˆ‘i=1niA_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:

βˆ‘i=1n1=nandβˆ‘i=1ni=n(n+1)2\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 AnA_n:

An=12nβ‹…n+27n2β‹…n(n+1)2A_n = \frac{12}{n} \cdot n + \frac{27}{n^2} \cdot \frac{n(n+1)}{2}

Simplifying:

An=12+27(n+1)2nA_n = 12 + \frac{27(n+1)}{2n}

Step 6: Take the limit as nβ†’βˆžn \to \infty

Now, we take the limit as nn approaches infinity to find the exact area:

A=lim⁑nβ†’βˆž(12+27(n+1)2n)A = \lim_{n \to \infty} \left( 12 + \frac{27(n+1)}{2n} \right)

As nβ†’βˆžn \to \infty, n+1nβ†’1\frac{n+1}{n} \to 1, so:

A=12+272=12+13.5=25.5A = 12 + \frac{27}{2} = 12 + 13.5 = 25.5

Final Answer

The area under the curve y=3xβˆ’2y = 3x - 2 over the interval [2,5][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][1, 4]?
  2. Can you find the area under a different curve using the same limit process, e.g., y=x2y = 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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Calculus
Riemann Sums
Limits

Formulas

Ξ”x = (b - a) / n
x_i^* = a + i * Ξ”x
f(x_i^*) = 3x - 2
A_n = Ξ£ f(x_i^*) * Ξ”x
lim(n β†’ ∞) A_n

Theorems

Limit of Riemann Sums
Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12, College Calculus