Problem 8 - Thomas Calculus

Question

Using rectangles each of whose height is given by the value of the function at the midpoint of the rectangle's base (the midpoint rule), estimate the area under the graphs of the following functions, using first two and then four rectangles. $$ f(x)=4-x^{2} \text { between } x=-2 \text { and } x=2 $$

Step-by-Step Solution

Answer

With 2 rectangles, area is 12; with 4 rectangles, area is 11.

Step 1: Identify the Interval and Function

We want to estimate the area under the function \( f(x) = 4 - x^2 \) from \( x = -2 \) to \( x = 2 \). This is the interval over which we will calculate the area using the midpoint rule.

Step 1: Divide the Interval into Subintervals

First, we use two rectangles. The total interval length is 4 (from -2 to 2), so each rectangle will have a width of 2. The subintervals are \([-2, 0]\) and \([0, 2]\).

Step 2: Calculate Midpoints for Two Rectangles

The midpoints of the subintervals \([-2, 0]\) and \([0, 2]\) are calculated as: Midpoint for \([-2, 0]\): \(x = -2 + \frac{2}{2} = -1\)Midpoint for \([0, 2]\): \(x = 0 + \frac{2}{2} = 1\)

Step 3: Evaluate the Function at Midpoints for Two Rectangles

Compute the function values at the midpoints:- \(f(-1) = 4 - (-1)^2 = 3\)- \(f(1) = 4 - (1)^2 = 3\)

Step 4: Calculate Areas of Two Rectangles

The area of each rectangle is given by the product of its width and height:- Area from \([-2, 0]\) is \(2 \times 3 = 6\)- Area from \([0, 2]\) is \(2 \times 3 = 6\)Total area for two rectangles: \(6 + 6 = 12\).

Step 5: Divide the Interval into More Subintervals

Now, use four rectangles. Each rectangle now has a width of \(1\) since the total length of \(4\) is divided by \(4\) rectangles. Subintervals are \([-2, -1]\), \([-1, 0]\), \([0, 1]\), and \([1, 2]\).

Step 6: Calculate Midpoints for Four Rectangles

Find the midpoints for the new subintervals:- \([-2, -1]\): midpoint at \(-1.5\)- \([-1, 0]\): midpoint at \(-0.5\)- \([0, 1]\): midpoint at \(0.5\)- \([1, 2]\): midpoint at \(1.5\)

Step 7: Evaluate the Function at Midpoints for Four Rectangles

Compute function values at these midpoints:- \(f(-1.5) = 4 - (-1.5)^2 = 1.75\)- \(f(-0.5) = 4 - (-0.5)^2 = 3.75\)- \(f(0.5) = 4 - (0.5)^2 = 3.75\)- \(f(1.5) = 4 - (1.5)^2 = 1.75\)

Step 8: Calculate Areas of Four Rectangles

Calculate the area for each rectangle (width \(1\)) and sum them:- Area from \([-2, -1]\) is \(1 \times 1.75 = 1.75\)- Area from \([-1, 0]\) is \(1 \times 3.75 = 3.75\)- Area from \([0, 1]\) is \(1 \times 3.75 = 3.75\)- Area from \([1, 2]\) is \(1 \times 1.75 = 1.75\)Total area for four rectangles: \(1.75 + 3.75 + 3.75 + 1.75 = 11\).

Key Concepts

Riemann Sums

Riemann sums are a way to approximate the area under a curve, often used in calculus to find integrals. Imagine you have a curve, and you'd like to find out the area beneath it. Instead of complex calculations, you can break it down into simpler parts using shapes like rectangles.

A Riemann sum involves dividing the area under the curve into small rectangles, then summing their areas. You have flexibility in choosing how to set up these rectangles: you can make them wide or narrow and decide how to place them under the curve. This setup can significantly affect your approximation's accuracy.

For example:

• Left and Right Endpoints: You place the rectangle's corner on either the left or right end of a subinterval. This means the height of the rectangle is determined by the function's value at that point.
• Midpoint Rule: This technique uses the function value at the midpoint of the rectangle's base as the height. It often provides a balance between left and right endpoint approximations, offering greater accuracy.

Ultimately, Riemann sums are foundational tools in integral calculus for estimating the total area under a curve, or actually, the integral of the function.

Integral Approximation

Integral approximation is a practical technique used when calculating an exact integral is difficult or unnecessary. In real-world scenarios, many functions might be far too complex to integrate precisely, and that's where approximations save the day.

The Midpoint Rule helps in approximating the integrals using average values from certain intervals. Each step focuses on selecting a point within each subinterval (typically at its midpoint), evaluates the function there, and assumes the rectangle's height to be the function value. By summing these individual rectangle areas, we get an approximation of the total integral over the interval.

Here's why it’s useful:

• Simplicity: It converts complex integrals into manageable computations.
• Precision Control: You can increase accuracy by using more subdivisions (narrower rectangles).
• Versatility: It applies to numerous functions, even those not easily solvable through standard integration methods.

Though not perfect, integral approximations like the Midpoint Rule quickly deliver results, making them essential tools for engineers, scientists, and mathematicians.

Area Under a Curve

Finding the area under a curve is a fundamental concept in calculus that helps to determine the integral of a function over a specified interval. This area represents accumulated quantities such as distance traveled, total growth, or even economic activity over time.

Here’s why calculating the area under a curve is essential:

• Practical Applications: It has real-world uses in physics (for calculating displacement and work), economics (determining consumer surplus), and biology (modeling population growth).
• Analysis: It aids in understanding the behavior of a function over an interval, showing total increments or decrements as the function progresses.

Using the Midpoint Rule as shown in the provided solution, the area is estimated by rectangles, where each rectangle’s height is the function value at its midpoint. As you split the interval into smaller segments and more rectangles, the approximation becomes increasingly precise. Applying this rule to the specific function \( f(x) = 4 - x^2 \), allows for an intuitive and direct method to estimate the area, even when exact analytical computation is challenging.