Problem 36 - Precalculus with Limits

Question

In Exercises 29-36, complete the table using the function \( f(x) \), over the specified interval [a, b], to approximate the area of the region bounded by the graph of the \( y = f(x) \), the x-axis, and the vertical lines \(x=a\) and \(x=b\) and using the indicated number of rectangles. Then find the exact area as \( n \to \infty \). $$ f(x) = \frac{1}{2}x + 1 $$ Interval \( [-2, 2] \)

Step-by-Step Solution

Answer

The approximate area under the curve \( f(x) = \frac{1}{2}x + 1 \) from [-2, 2] is the sum of the areas of the rectangles, and the exact area as \( n \to \infty \) is 2.

Step 1: Define the function and interval

The function given is \( f(x) = \frac{1}{2}x + 1 \). The interval on which we are finding the area under the curve is [-2, 2].

Step 2: Approximate the area using rectangles

Divide the interval [-2, 2] into equally sized subintervals. The width of each subinterval is \( \Delta x = \frac{b - a}{n} = \frac{2 - (-2)}{n} = \frac{4}{n} \). The height of the rectangle on each subinterval is the function value at the right endpoint of the subinterval. The total area, \( A \), of all the rectangles is the sum of the areas of all rectangles, \( A \approx \sum_{i = 1}^{n} f(x_{i})\Delta x \), where \( x_{i} \) are the right endpoints. As \( n \) tends towards infinity, this sum becomes an infinite sum, or a definite integral.

Step 3: Find the exact area

As \( n \to \infty \), the sum of the areas of the rectangles approaches the exact area under the curve. This is given by the definite integral \( \int_{a}^{b} f(x) \, dx \). Here, this is \( \int_{-2}^{2} f(x) \, dx = \int_{-2}^{2} (\frac{1}{2}x + 1) \, dx = [\frac{1}{4}x^2 + x]_{-2}^{2} = (\frac{1}{4}(2)^2 + 2) - (\frac{1}{4}(-2)^2 + (-2)) = 2 \).

Key Concepts

Riemann Sums

Before diving into definite integrals, it's essential to understand Riemann Sums. These are methods to approximate the area under a curve using multiple rectangles. Imagine a curved line on a graph, and we want to find the area it encloses with the x-axis.

• We start by dividing the area under the curve from point \(a\) to \(b\) into smaller sections or rectangles.
• The width of these rectangles, denoted as \(\Delta x\), is calculated by \(\frac{b-a}{n}\), where \(n\) is the number of rectangles.
• The height of each rectangle is the value of the function at a specific point within the interval, often at the right endpoint.

The Riemann sum is the sum of the areas of all these rectangles. Mathematically, this is expressed as:\[A \approx \sum_{i=1}^{n} f(x_{i})\Delta x\]As you increase \(n\), making the sections smaller, the approximation becomes more accurate, approaching the actual area when \(n\) goes to infinity.

Area Under a Curve

To find the exact area under a curve, Riemann sums transition into definite integrals. Consider a function like our example, \( f(x) = \frac{1}{2}x + 1 \), on an interval \([-2, 2]\). Initially, we approximate by adding the areas of rectangles (Riemann sums). But to get the exact answer, we'll use integrals.

• This involves calculating the definite integral of \(f(x)\) from \(a\) to \(b\), denoted as \( \int_{a}^{b} f(x) \, dx \).
• For our function, this translates to:\[\int_{-2}^{2} \left(\frac{1}{2}x + 1\right) \, dx\]

The integration gives us the net area, capturing all intricacies of the curve. Solving this specific integral, you perform the calculations to get:\[[\frac{1}{4}x^2 + x]_{-2}^{2} = 2\]This means the area enclosed by the curve, the x-axis, from \(-2\) to \(2\), is exactly \(2\). Integrals help convert approximations into precise measurements.

Function Approximation

Approximating functions means finding ways to represent complex shapes with simpler components, which is crucial for definite integrals. Understanding this concept helps to simplify the process of finding the area under the curve.

• By using rectangles (in Riemann sums) or other geometric shapes, we make the function simpler or more manageable.
• For instance, although the function \( f(x) = \frac{1}{2}x + 1 \) is linear, approximating it as piecewise constant over subintervals shows how its behavior fits into rectangles.
• This non-exact approach allows us to use approximate methods before applying exact calculation methods like integrals.

Function approximation leads to simpler calculations, enabling students to handle more complex functions by breaking them into understandable parts. This foundational understanding is vital before applying definite integrals, as it builds on this decomposition of complex areas into simpler, solvable segments.