Problem 8 - A Graphical Approach to Precalculus with Limits

Question

Approximate the area under the graph of \(f(x)\) and above the \(x\) -axis, using each of the following methods with \(n=4\). (a) Use left endpoints. (b) Use right endpoints. (c) Average the answers in parts ( \(a\) ) and ( \(b\) ). (d) Use midpoints. $$f(x)=-x^{2}+4 \text { from } x=-2 \text { to } x=2$$

Step-by-Step Solution

Answer

The approximate area under the curve is 10 using endpoint methods, and 11 using midpoints.

Step 1: Determine the Interval Width Δx

The function is given from \(x = -2\) to \(x = 2\), and we need to use \(n = 4\) intervals. The width of each sub-interval \(\Delta x\) is determined by dividing the total interval length by the number of intervals, \(\Delta x = \frac{2 - (-2)}{4} = 1\).

Step 2(a): Calculate Area Using Left Endpoints

For left endpoints, we use the leftmost point of each sub-interval to calculate the function value. The endpoints are \(-2, -1, 0, 1\). Calculate \(f(x)\) for each point:- \(f(-2) = -(-2)^2 + 4 = 0\)- \(f(-1) = -(-1)^2 + 4 = 3\)- \(f(0) = -(0)^2 + 4 = 4\)- \(f(1) = -(1)^2 + 4 = 3\)Approximate the area: \(A = \Delta x \cdot [f(-2) + f(-1) + f(0) + f(1)] = 1 \cdot (0 + 3 + 4 + 3) = 10\).

Step 2(b): Calculate Area Using Right Endpoints

For right endpoints, use the rightmost point of each sub-interval. The right endpoints are \(-1, 0, 1, 2\). Calculate \(f(x)\) for each point:- \(f(-1) = 3\)- \(f(0) = 4\)- \(f(1) = 3\)- \(f(2) = -(2)^2 + 4 = 0\)Approximate the area: \(A = \Delta x \cdot [f(-1) + f(0) + f(1) + f(2)] = 1 \cdot (3 + 4 + 3 + 0) = 10\).

Step 2(c): Average the Left and Right Endpoint Approximations

Since both left and right endpoint approximations yielded an area of 10, their average is: \(\frac{10 + 10}{2} = 10\).

Step 2(d): Calculate Area Using Midpoints

For midpoints, calculate the midpoint for each sub-interval: \(-1.5, -0.5, 0.5, 1.5\). Evaluate \(f(x)\) at each midpoint:- \(f(-1.5) = -(-1.5)^2 + 4 = 1.75\)- \(f(-0.5) = -(-0.5)^2 + 4 = 3.75\)- \(f(0.5) = -(0.5)^2 + 4 = 3.75\)- \(f(1.5) = -(1.5)^2 + 4 = 1.75\)Approximate the area: \(A = \Delta x \cdot [f(-1.5) + f(-0.5) + f(0.5) + f(1.5)] = 1 \cdot (1.75 + 3.75 + 3.75 + 1.75) = 11\).

Key Concepts

Left Endpoint Approximation

The Left Endpoint Approximation is a simple method to estimate the area under a curve by using the leftmost point of each sub-interval. When you have a function like \(f(x) = -x^2 + 4\) from \(x = -2\) to \(x = 2\), with \(4\) intervals, this method is applied using sub-intervals of width \(\Delta x = 1\).

To compute the approximation, you calculate the function value at the left endpoint of each interval. Here, the points are \(-2, -1, 0,\) and \(1\). Evaluate \(f(x)\) at these points:

• \(f(-2) = 0\)
• \(f(-1) = 3\)
• \(f(0) = 4\)
• \(f(1) = 3\)

To find the area, multiply each function value by \(\Delta x\) and sum them up:\[A = \sum_{i=1}^{4} f(x_i) \cdot \Delta x = 0 + 3 + 4 + 3 = 10\].

This gives a good initial estimate of the area using plainly the left edge of each segment of the curve.

Right Endpoint Approximation

The Right Endpoint Approximation follows a similar principle to the left endpoint method, but instead uses the rightmost point of each sub-interval. In this context, we're approximating the area under \(f(x) = -x^2 + 4\) from \(x = -2\) to \(x = 2\), again using \(4\) sub-intervals, each \(\Delta x = 1\) wide.

The right endpoints in this example are \(-1, 0, 1,\) and \(2\). Calculate \(f(x)\) for these points:

• \(f(-1) = 3\)
• \(f(0) = 4\)
• \(f(1) = 3\)
• \(f(2) = 0\)

Multiply these values by the interval width \(\Delta x\) and sum them to approximate the area:\[A = \sum_{i=1}^{4} f(x_i) \cdot \Delta x = 3 + 4 + 3 + 0 = 10\].

Again, just like with the left endpoint, this provides an equally reasonable approximation of the area.

Midpoint Rule

The Midpoint Rule is a bit more precise for estimating areas, as it uses the midpoint of each sub-interval, rather than the edges. Continuing with the function \(f(x) = -x^2 + 4\) from \(x = -2\) to \(x = 2\) with \(4\) equal intervals, each \(\Delta x = 1\), we calculate the midpoint for each section.

The midpoints we focus on here are \(-1.5, -0.5, 0.5,\) and \(1.5\). Evaluate the function at each of these points:

• \(f(-1.5) = 1.75\)
• \(f(-0.5) = 3.75\)
• \(f(0.5) = 3.75\)
• \(f(1.5) = 1.75\)

Multiply these values by \(\Delta x\), and sum them up to get the area:\[A = \sum_{i=1}^{4} f(x_i) \cdot \Delta x = 1.75 + 3.75 + 3.75 + 1.75 = 11\].

This midpoint method often gives a significantly better estimation of the true area, balancing the inaccuracies of the left and right endpoint approximations.

Riemann Sums

Riemann Sums are a fundamental concept in mathematics used to estimate the area under curves. They are the basis for all the methods discussed above: Left Endpoint, Right Endpoint, and Midpoint Approximations. All these are different forms of Riemann Sums.

When integrating numerically, the function is divided into a number of small "strips" or "rectangles" over which the function is estimated.

• The Left Endpoint method uses the left corner of each rectangle.
• The Right Endpoint method uses the right corner.
• The Midpoint Rule utilizes the center point for a more balanced approach.

The width of these rectangles (\(\Delta x\)) and their height (\(f(x)\) evaluated at chosen points) form the basis of estimating the area under the curve.

Riemann Sums approximate the integral:\[\int_{a}^{b} f(x) \, dx \approx \sum_{i=1}^{n} f(x_i^{*}) \cdot \Delta x\]where \(x_i^{*}\) can be any point within the \(i^{th}\) sub-interval. This flexibility is what allows variations like the left, right, and midpoint approaches, each offering unique perspectives on approximating area.