Math Problem Statement

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(π‘₯) = 3π‘₯ βˆ’ 2 on [2,5]

Solution

To find the area under the curve f(x)=3xβˆ’2f(x) = 3x - 2 over the interval [2,5][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)f(x) over the interval [a,b][a, b] can be expressed as:

A=lim⁑nβ†’βˆžβˆ‘i=1nf(xiβˆ—)Ξ”xA = \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 within each subinterval.
  • nn is the number of subintervals.

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

Step 2: Compute Ξ”x\Delta x and the sample points xiβˆ—x_i^*

The width of each subinterval is:

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

We'll use the right endpoint for the sample points xiβˆ—x_i^*, so:

xiβˆ—=2+iΞ”x=2+iβ‹…3n=2+3inx_i^* = 2 + i \Delta x = 2 + i \cdot \frac{3}{n} = 2 + \frac{3i}{n}

Step 3: Set up the Riemann sum

Now, substitute xiβˆ—x_i^* into the function f(x)=3xβˆ’2f(x) = 3x - 2:

f(xiβˆ—)=3(2+3in)βˆ’2=6+9inβˆ’2=4+9inf(x_i^*) = 3\left( 2 + \frac{3i}{n} \right) - 2 = 6 + \frac{9i}{n} - 2 = 4 + \frac{9i}{n}

The Riemann sum becomes:

Sn=βˆ‘i=1n(4+9in)β‹…3nS_n = \sum_{i=1}^{n} \left( 4 + \frac{9i}{n} \right) \cdot \frac{3}{n}

Step 4: Simplify the sum

Now distribute 3n\frac{3}{n} inside the sum:

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

Split the sum into two parts:

Sn=12nβˆ‘i=1n1+27n2βˆ‘i=1niS_n = \frac{12}{n} \sum_{i=1}^{n} 1 + \frac{27}{n^2} \sum_{i=1}^{n} i

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 sum:

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

Simplify:

Sn=12+27(n+1)2nS_n = 12 + \frac{27(n+1)}{2n}

Step 5: Take the limit as nβ†’βˆžn \to \infty

Now take the limit as nβ†’βˆžn \to \infty:

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

The second term simplifies as follows:

27(n+1)2n=27n+272n=272+272n\frac{27(n+1)}{2n} = \frac{27n + 27}{2n} = \frac{27}{2} + \frac{27}{2n}

As nβ†’βˆžn \to \infty, 272nβ†’0\frac{27}{2n} \to 0, so the limit becomes:

A=12+272=242+272=512A = 12 + \frac{27}{2} = \frac{24}{2} + \frac{27}{2} = \frac{51}{2}

Final Answer:

The exact area under the curve f(x)=3xβˆ’2f(x) = 3x - 2 over the interval [2,5][2, 5] is:

A=512A = \frac{51}{2}

Would you like more details or have any questions?


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)=x2f(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!

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Limit Definition of Definite Integral
Riemann Sums

Formulas

A = lim_{n -> ∞} Ξ£ f(x_i^*) Ξ”x
Ξ”x = (b - a) / n
f(x) = 3x - 2
Ξ£_{i=1}^n i = n(n + 1)/2

Theorems

Limit Definition of the Definite Integral
Riemann Sum

Suitable Grade Level

Grades 11-12