Problem 14 - Calculus Early Transcendentals

Question

The following functions are negative on the given interval. a. Sketch the function on the given interval. b. Approximate the net area bounded by the graph off and the \(x\) -axis on the interval using a left, right, and midpoint Riemann sum with \(n=4\). $$f(x)=x^{3}-1 \text { on }[-2,0]$$

Step-by-Step Solution

Answer

1. Calculate the value of \(\Delta x\) for the given interval and number of sub-intervals. 2. Find the left, right, and midpoint Riemann sums by using the function values at the corresponding endpoints or midpoints of each sub-interval. 3. Multiply the sum of function values by \(\Delta x\) for each Riemann sum to approximate the net area. 4. Compare the results of the three Riemann sums to understand the different approximations.

Step 1: 1. Sketch the function on the given interval

To sketch the function \(f(x) = x^3 - 1\) on the interval \([-2, 0]\), we can start by finding some points on the graph and connecting them. For example, at \(x=-2\), \(x=-1\), \(x=0\), we can calculate \(f(x)\) as: - \(f(-2) = (-2)^3 -1 = -9\) - \(f(-1) = (-1)^3 -1 = -2\) - \(f(0) = 0^3 -1 = -1\) Now, plot these points on the Cartesian plane and connect them to form the graph. As you sketch the graph, you will find that the function is negative on the entire interval \([-2, 0]\) and crosses the x-axis lying entirely below it.

Step 2: 2. Approximate net area using Riemann sums

To approximate the net area, we will use left, right, and midpoint Riemann sums with \(n=4\). Let's first find the value of \(\Delta x\) for \(n=4\): $$\Delta x = \frac{b-a}{n} = \frac{0-(-2)}{4} = \frac{2}{4} = 0.5$$

Step 3: 2a. Approximate net area using left Riemann sum

For left Riemann sum, we use the function value at the left endpoint of each sub-interval. So, we have: $$ L_4 = \sum_{i=1}^{4} f(x_{i-1})\Delta x = (f(-2) + f(-1.5) + f(-1) + f(-0.5))\times 0.5 $$ Calculate \(f(-1.5)\) and \(f(-0.5)\), then sum the \(4\) function values and multiply by \(\Delta x = 0.5\) to get the left Riemann sum.

Step 4: 2b. Approximate net area using right Riemann sum

For the right Riemann sum, we use the function value at the right endpoint of each sub-interval. So, we have: $$ R_4 = \sum_{i=1}^{4} f(x_{i})\Delta x = (f(-1.5) + f(-1) + f(-0.5) + f(0)) \times 0.5 $$ Sum the \(4\) function values and multiply by \(\Delta x = 0.5\) to get the right Riemann sum.

Step 5: 2c. Approximate net area using midpoint Riemann sum

For the midpoint Riemann sum, we use the function value at the midpoint of each sub-interval. So, the midpoints are \(x^*_i = -1.75, -1.25, -0.75, -0.25\). Thus, we have: $$ M_4 = \sum_{i=1}^{4} f(x^*_i)\Delta x = (f(-1.75) + f(-1.25) + f(-0.75) + f(-0.25)) \times 0.5 $$ Calculate the function values at the midpoints, sum them up, and multiply by \(\Delta x = 0.5\) to get the midpoint Riemann sum. Finally, compare the results of these three Riemann sums to understand the different level of approximations they give to the net area bounded by the graph of the function and the \(x\)-axis within the given interval.

Key Concepts

Approximation of Area

Riemann sums provide a method to approximate the area under a curve. When dealing with a function that does not have a simple antiderivative, it becomes essential to use numerical methods like Riemann sums. In our example, we approximate the area under the curve - by using left, right, and midpoint Riemann sums. Each of these methods serves a distinct way of estimating area:

• Left Riemann Sum: Uses the value of the function at the left end of each subinterval. This method tends to underestimate the area if the function is increasing and overestimate if it's decreasing.
• Right Riemann Sum: Uses the value of the function at the right end of each subinterval. This often overestimates the area when the function is increasing and underestimates it when decreasing.
• Midpoint Riemann Sum: Uses the value of the function at the midpoint of each subinterval. This generally provides a better approximation as it balances the errors from over and underestimation.

The goal is to divide an interval into smaller pieces, sum up the areas of shape-like rectangles under the curve, and approximate the total area.

Net Area Under Curve

Net area considers both above and below the x-axis values, which can subtract from each other when evaluated. Given that we are examining a range where our function is negative, all approximations will return negative values. For the function \( f(x) = x^3 - 1\) being entirely negative on the interval \([-2, 0]\), the net area represents the algebraic sum of these negative values. To better understand, imagine the graph dipping entirely below the axis:

• Each portion above a horizontal slice counts as a negative contribution.
• The Riemann sums act as a way to calculate how much 'negative space' exists between the curve and the x-axis.

Thus, evaluating this net area provides insight into the nature of the slope and curvature differences across the graph interval.

Graph of Polynomial Function

Polynomial functions, such as \( f(x) = x^3 - 1\), create smooth, continuous curves with no breaks or jumps. Let's focus on essential characteristics of the graph:

• Cubic Function: The curve represents a degree of three, meaning it can potentially lift or fall three times over its domain.
• Smoothness: Cubics specifically result in a smooth 'S' curve when graphed, evident in our example from \([-2, 0]\).
• Negative Interval Influence:: Since values output negative results across our target [-2, 0] interval, the sketch will lie entirely below the x-axis.

When drawing or analyzing the curve, key anchor points (like the ones calculated: \(f(-2) = -9, f(-1) = -2, f(0) = -1\)) help establish the base shape of the plot, showing the flow of the function across specified intervals.