Problem 39 - Precalculus with Limits

Question

In Exercises 37-48, use the limit process to find the area of the region between the graph of the function and the x-axis over the specified interval. $$ f(x) = -2x + 3 $$ Interval \( [0, 1] \)

Step-by-Step Solution

Answer

The area between the function \(f(x) = -2x + 3\) and the x-axis over the interval [0, 1] is 1.

Step 1: Write out the function and the interval

We are given the function \(f(x)\) and its interval: \(f(x) = -2x + 3\) and [0, 1].

Step 2: Expression for Rectangle height and width

Set up the formula for a definite integral as a limit of Riemann sums. The height of each rectangle is equivalent to the function's value at that point or \(f(x = x_i) = -2x_i + 3\). The width of each rectangle is the length of the interval divided by the number of rectangles or \(\Delta x = \frac{1 - 0}{n} = \frac{1}{n}\).

Step 3: Sum of rectange areas

Set up the sum of the areas as \(\lim_{n \rightarrow \infty} \sum_{i = 0}^{n-1} f(x_i) \Delta x\), where \(i\) indexes the rectangles from 0 to \((n-1)\). Substitute \(f(x_i) = -2x_i + 3\) and \(\Delta x = \frac{1}{n}\) to get \(\lim_{n \rightarrow \infty} \sum_{i = 0}^{n-1} (-2x_i + 3) \frac{1}{n}\).

Step 4: Express xi in terms of i

Express \(x_i\) in terms of \(i\) and \(\Delta x\) as \(x_i = i \Delta x\). Substituting \(\Delta x = \frac{1}{n}\) gives \(x_i = \frac{i}{n}\). Then substitute \(-2x_i + 3\) as \(-2(i/n) + 3\). So, the sum is now \(\lim_{n \rightarrow \infty} \sum_{i = 0}^{n-1} (-2 \cdot \frac{i}{n} + 3) \cdot \frac{1}{n}\).

Step 5: Distribute the sum

Distribute the sum inside the parentheses, resulting in two separate sums: \(\lim_{n \rightarrow \infty} \sum_{i = 0}^{n-1} (-2 \cdot \frac{i}{n^2} + \frac{3}{n})\) = \(\lim_{n \rightarrow \infty} ( -2 \cdot \sum_{i=0}^{n-1}\frac{i}{n^2} + \sum_{i=0}^{n-1}\frac{3}{n} )\).

Step 6: Find the limit of each sum

Evaluate the limit as \(n\) approaches infinity to obtain the area under the curve. Use the formulas that \(\sum_{i=0}^{n-1} i = \frac{{n\cdot(n - 1)}}{2}\) and \(\sum_{i=0}^{n-1} 1 = n\). Thus, the area becomes \(-2 \cdot \lim_{n \rightarrow \infty} \frac{\frac{n \cdot (n-1)}{2}}{n^2} + \lim_{n \rightarrow \infty} \frac{3n}{n}\). Simplifying this gives us the answer \(= -2(\frac{1}{2}) + 3 = 1\).

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus, representing the accumulation of quantities or the net area between a function and the x-axis over a specific interval. This is achieved using limits, transforming infinite Riemann sums into a precise value. In this exercise, the function

• Given: \( f(x) = -2x + 3 \)
• Interval: \([0, 1]\)

becomes closely examined using the Riemann sum. As the number of rectangles (**n**) approaches infinity, these sums precisely calculate the area under \( f(x) \). Essentially, the definite integral encapsulates not just area but can also help interpret sum values in various applications, such as physics or engineering, where calculating things like total distance or space is crucial. Here, you're leveraging the definite integral to study the bounded region, honing the abstract sum into a usable answer.

Limit Process

The concept of a limit process is key when we discuss the transition from Riemann sums to a definitive integral. This means progressively adding finer partitions (rectangles) under the curve, bringing more accuracy to our area approximation. For the function \( f(x) = -2x + 3 \), observed over [0, 1], we perform these steps:

• Partition the interval into \( n \) rectangles, each with a width of \( \Delta x = \frac{1}{n} \).
• Calculated at each slice, the height or rectangle value is \( f(x_i) = -2x_i + 3 \), where \( x_i = \frac{i}{n} \).
• The actual sum becomes \( \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(-2 \cdot \frac{i}{n} + 3\right) \cdot \frac{1}{n} \), as we refine further partitions.

As \( n \) increases indefinitely, the sum reaches a limit, symbolizing the precise measure of accumulated areas – thus providing the area accurately. This process becomes an integral concept in calculus that allows complex curves to be resolved with detailed precision.

Area Under a Curve

One of the primary applications of the definite integral and limit process is to calculate the area under a curve. This captures the space between a curve and the x-axis over a set interval. For the given function \( f(x) = -2x + 3 \) over [0, 1], this means summing the areas of infinitely many rectangles:

• Each rectangle's area: height \( \times \) width, where the width \( \Delta x = \frac{1}{n} \) and height is the curve's value at that point, or \( f(x_i) \).
• Combining all these mini-areas results in a sum \( \lim_{n \rightarrow \infty} \sum_{i = 0}^{n-1} (-2 \cdot \frac{i}{n} + 3) \cdot \frac{1}{n} \), chasing towards the exact area.
• Performing this calculation in the example yielded a total area of \( 1 \).

This method systematically deals with any continuous function by such an approach, showcasing the power of calculus. It's a mathematical method to encapsulate the idea of 'accumulating' infinitely small parts, providing an exact total of instrumental use.