Problem 18 - Calculus Early Transcendentals

Question

The following functions are positive and negative on the given interval. a. Sketch the function on the given interval. b. Approximate the net area bounded by the graph of \(f\) and the \(x\) -axis on the interval using a left, right, and midpoint Riemann sum with \(n=4\) c. Use the sketch in part (a) to show which intervals of \([a, b]\) make positive and negative contributions to the net area. $$f(x)=x^{3} \text { on }[-1,2]$$

Step-by-Step Solution

Answer

Answer: Left Riemann sum approximation is \(1.3125\), the right Riemann sum approximation is \(6.609375\), and the midpoint Riemann sum approximation is \(3.392334\). The graph contributes negatively on the interval \([-1, 0]\) and positively on the interval \([0, 2]\).

Step 1: Sketch the Function

To sketch the function \(f(x)=x^3\) on the interval \([-1,2]\), plot some representative points within the interval and apply the cubic function. After plotting, connect the points and draw the curve of the function. The representative points can be: - \((-1,-1)\) - \((0,0)\) - \((1,1)\) - \((2,8)\) The graph of the function will pass through these points, with the endpoints being \((-1,-1)\) and \((2,8)\). Use these points to draw the curve of the cubic function.

Step 2: Calculate Left, Right, and Midpoint Riemann Sums

To approximate the net area bounded by the graph of \(f(x)\) and the x-axis on the interval \([-1,2]\) with \(n=4\), we need to partition the interval into 4 equal subintervals. a. Left Riemann Sum: Left Riemann sum uses the left endpoint of each subinterval to approximate the area. Subintervals: \([-1,-0.25]\), \([-0.25,0.5]\), \([0.5,1.25]\), \([1.25,2]\). Left Riemann Sum: \(\Delta x[(f(-1))+(f(-0.25))+(f(0.5))+(f(1.25))]=(0.75)[-1+(-0.015625)+0.125+1.953125]=1.3125\). b. Right Riemann Sum: Right Riemann sum uses the right endpoint of each subinterval. Right Riemann Sum: \(\Delta x[(f(-0.25))+(f(0.5))+(f(1.25))+(f(2))]=(0.75)[(-0.015625)+0.125+1.953125+8]=6.609375\). c. Midpoint Riemann Sum: Midpoint Riemann sum uses the midpoint of each subinterval. Midpoints: \([-0.625, 0.125, 0.875, 1.625]\). Midpoint Riemann Sum: \(\Delta x[(f(-0.625))+(f(0.125))+(f(0.875))+(f(1.625))]=(0.75)[(-0.244140625)+0.001953125+0.669921875+4.300048828]=3.392334\).

Step 3: Identify Intervals for Positive and Negative Contributions

By analyzing the sketch in step 1, we can identify where the graph is above or below the x-axis. On the interval \([-1,0]\), the graph of the function is below the x-axis, contributing negatively to the net area. On the interval \([0,2]\), the graph of the function is above the x-axis, contributing positively to the net area.

Key Concepts

Cubic function

A cubic function is a polynomial of degree three, generally represented as \( f(x) = ax^3 + bx^2 + cx + d \). In simpler terms, it's an equation where the highest exponent of \( x \) is three. The function we've worked with, \( f(x) = x^3 \), is a basic cubic function with no other terms besides \( x^3 \), making it a smoother learning model.

Characteristics of cubic functions include their unique shape, which can have up to two bends or turns. For \( f(x) = x^3 \), this unique shape will pass through the origin at \( (0,0) \) and feature its turning points as smooth curves, not sharp turns.

Understanding how to sketch these functions is vital, as it helps visualize where the function is above or below the x-axis, crucial for solving problems involving Riemann sums and analyzing areas.

Net area

The concept of net area involves calculating the "net" sum between a curve and the x-axis. Rather than simply adding up all the areas under the curve, it considers whether parts of the curve lie above or below the x-axis.In our case, the function \( f(x) = x^3 \) is evaluated over the interval \([-1,2]\), producing both positive and negative sections. These sections balance each other out to form a "net" value of the total area.

Calculating the net area using Riemann sums conceptually involves:

• Breaking down the entire domain into subintervals,
• Evaluating the function at strategic points within each subinterval,
• Multiplying these evaluated values by the width of each subinterval.

This process gives a numerical estimate of the area, helping us grasp how portions of the graph contribute to the entire net area.

Positive and negative contributions

Positive and negative contributions to the net area are essential in understanding the true nature of the area enclosed by a function over a particular interval.Consider our function \( f(x) = x^3 \), which spans the interval \([-1,2]\). The key here is knowing when the graph is above or below the x-axis:

• On the interval \([-1,0]\), \( f(x) = x^3 \) provides negative area contributions as the entire graph is below the x-axis.
• On the interval \([0,2]\), \( f(x) = x^3 \) contributes positively because the curve is above the x-axis.

This division of positive and negative segments reflects the integral's powerful role in determining the genuine sum of these areas, offering a more insightful analysis than merely calculating total area.

Subintervals

To accurately estimate the net area using Riemann sums, the interval over which a function is being integrated must be divided into smaller sections called subintervals. These divisions are crucial for precision in calculation.For \( f(x) = x^3 \), we're focused on the interval \([-1,2]\), broken into four equal subintervals using the process of:

• Defining subintervals such as \([-1,-0.25]\), \([-0.25,0.5]\), \([0.5,1.25]\), and \([1.25,2]\),
• Evaluating the function at either the left, right, or midpoint of each subinterval,
• Calculating an approximate area under the curve for each subinterval and summing them up.

This approach, using different types of Riemann sums, not only estimates the net area but enhances understanding of how widely the function’s graph varies and the sum's dependency on chosen subintervals.