Problem 30 - Algebra 2

Question

a. Graph the curve \(y=\frac{1}{3} x^{3}\) . b. Use inscribed rectangles to approximate the area under the curve for the interval \(0 \leq x \leq 3\) and rectangle width of 1 unit. c. Repeat part (b) using circumscribed rectangles. d. Find the mean of the areas you found in parts \((b)\) and (c). Of the three estimates, which best approximates the area for the interval? Explain.

Step-by-Step Solution

Answer

After calculating, you might find that the area using inscribed rectangles is a lower-end estimate, the area using circumscribed rectangles is an upper-end estimate, and the mean of these two provides the best approximation of the actual area under the curve for the interval \(0 \leq x \leq 3\).

Step 1: Graph the Curve

Sketch the curve \(y=\frac{1}{3} x^{3}\). Using different values for \(x\), chart out points on a graph. You should see that the graph rises from left to right, with a point at the origin (0, 0).

Step 2: Approximate the Area using Inscribed Rectangles

For the interval \(0 \leq x \leq 3\), draw rectangles inscribed under the curve where the rectangle's top right corner touches the curve and the width of each rectangle is 1 unit. Estimate the area of each rectangle by multiplying the width (1 unit) with the height (the y-value of the function at that x-value), counting from 0 to 3. Sum these areas to get the total area under the curve using inscribed rectangles.

Step 3: Approximate the Area using Circumscribed Rectangles

For the same interval, draw rectangles circumscribed over the curve where the rectangle's top left corner touches the curve. Determine the area of each rectangle same as Step 2. Sum these areas to get the total area under the curve using circumscribed rectangles.

Step 4: Calculate the Mean of the Areas

Calculate the mean of the areas found using inscribed rectangles and circumscribed rectangles. Simply add the two areas together and divide by 2.

Step 5: Determine the Best Estimate

Compare the three estimates: the area using inscribed rectangles, the area using circumscribed rectangles, and the mean of the two. Discuss which one would give the best approximation of the actual area under the curve.

Key Concepts

Inscribed Rectangles

When approximating the area under a curve like the cubic function \(y = \frac{1}{3}x^3\), inscribed rectangles are a useful tool. Inscribed rectangles lie entirely below the curve. This means that their top right corner is just touching the curve. As a result, inscribed rectangles tend to underestimate the true area under the curve.
Let's break it down: we divide the interval from 0 to 3 into smaller segments, with each segment being 1 unit wide. For each segment, you determine the height of the rectangle by evaluating the function at the left endpoint of each segment. This ensures that the entire rectangle lies underneath the curve.

• For example, when \(x = 0\), the height of the rectangle is \(y = \frac{1}{3}(0)^3 = 0\).
• For \(x = 1\), the height is calculated as \(y = \frac{1}{3}(1)^3 = \frac{1}{3}\).
• Continue this process for \(x = 2\) and \(x = 3\).

To find the total approximate area, sum up the areas of all the rectangles, each calculated as the width (1 unit) multiplied by its height.
This method provides a simple way to approximate the area, knowing it will likely be less than the actual area.

Circumscribed Rectangles

Circumscribed rectangles provide another way to approximate the area under a curve, and they tend to overestimate the true area. These rectangles are different from inscribed ones because their top left corner touches the curve, which means they extend slightly above the curve.
Once again, consider the interval from 0 to 3, divided into segments 1 unit wide. Unlike inscribed rectangles, the height for circumscribed rectangles is determined by evaluating the function at the right endpoint of each segment.

• For \(x = 1\), the height is \(y = \frac{1}{3}(1)^3 = \frac{1}{3}\).
• For \(x = 2\), the height is \(y = \frac{1}{3}(2)^3 = \frac{8}{3}\).
• The last height for \(x = 3\) will be \(y = \frac{1}{3}(3)^3 = 9\).

By calculating the areas of these rectangles (width times height) and summing them up, you'll get an estimate that is likely to be higher than the true area because the rectangles overshoot the curve.
Finding both inscribed and circumscribed rectangle areas gives a lower and upper bound estimate for the integral, helping to refine your approximation.

Graphing Cubic Functions

Graphing cubic functions can be rewarding as they often form interesting shapes. The function \(y=\frac{1}{3}x^3\) is a simple cubic equation where the coefficient before \(x^3\) affects the graph’s steepness. Here's how you can approach graphing such a function.
Start by identifying key points. Since \(y = \frac{1}{3}x^3\) is a cubic function, it's symmetric about the origin, (0, 0). Cubic functions can have critical points where their slopes change, but for this specific function, they create a smooth S-curve without bends.

• Calculate a few values: When \(x = 0\), \(y = 0\), marking the origin.
• Try positive values: For \(x = 1\), \(y = \frac{1}{3}\) and for \(x = 3\), \(y= 9\).
• Negative values similarly result in negative y-values: \(x = -1\), \(y = -\frac{1}{3}\).

Use these points to sketch the curve’s general trajectory on a graph. It rises to the right and descends to the left, reflecting typical cubic function behavior.
Practicing this graphing not only helps visualize integrals but deepens understanding of how cubic shapes behave.