Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=\sqrt[3]{x}, y=0\), between \(x=-2\) and \(x=2\)

Short answer

The area of the region is approximately 3.78 square units.

Step-by-step solution

Sketch the Graphs

First, plot the graph of the equation \( y = \sqrt[3]{x} \). This is the cube root function, which is symmetric with respect to the origin. The line \( y = 0 \) is the x-axis. Plot these functions from \( x = -2 \) to \( x = 2 \). These curves will bound the region whose area we want to find.

Identify the Bounded Region

Identify the region bounded above by \( y = \sqrt[3]{x} \) and below by \( y = 0 \) from \( x = -2 \) to \( x = 2 \). This region lies below the curve and above the x-axis.

Choose a Typical Slice

Consider a vertical slice at position \( x \) within the interval \([-2, 2]\). The height of this slice is given by \( y = \sqrt[3]{x} \), and the width of the slice, \( \Delta x \), is an infinitesimally small change in \( x \). The area of this slice is \( y \cdot \Delta x = \sqrt[3]{x} \cdot \Delta x \).

Set Up the Integral

To find the total area, sum the areas of all these slices by integrating from \( x = -2 \) to \( x = 2 \). The integral is:\[\int_{-2}^{2} \sqrt[3]{x} \, dx\].

Calculate the Integral

Use the integral formula for \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n = \frac{1}{3} \). The antiderivative is:\[ \frac{3}{4}x^{\frac{4}{3}} \]Evaluate this from \( x = -2 \) to \( x = 2 \):\[ \left[ \frac{3}{4}x^{\frac{4}{3}} \right]_{-2}^{2} = \frac{3}{4}(2^{\frac{4}{3}}) - \frac{3}{4}((-2)^{\frac{4}{3}}) \].

Evaluate and Simplify

Calculate \( 2^{\frac{4}{3}} \) and \( (-2)^{\frac{4}{3}} \) both equal \( 2^{\frac{4}{3}} \) since raising to an even power negates negative signs within the base. So:\[ 2^{\frac{4}{3}} = 2.5198421 \]Therefore the area becomes:\[ 2 \times \frac{3}{4} \times 2.5198421 = \frac{3}{2} \times 2.5198421 \approx 3.7798 \].

Estimate Area to Confirm

To estimate, consider that the region is roughly a trapezoidal shape with the top arc being approximated by line segments. Using an average height of \( y \approx 1 \) from \( x = -2 \) to \( x = 2 \), the estimated area is about \( 2 \times 2 = 4 \). This estimation confirms our calculated area is reasonable.

Key concepts

Curve Sketching

Curve sketching is an essential aspect of calculus that helps in visualizing the behavior of a function over a specific interval. The exercise involves sketching the graph of the cube root function, given by the equation \( y = \sqrt[3]{x} \), from \( x = -2 \) to \( x = 2 \). This function has a characteristic shape, symmetric about the origin. As you draw this graph, remember that the curve passes through negative and positive values of \( x \), depicting the behavior of odd roots over real numbers.
Sketching helps to identify important features of the function, such as intercepts and symmetry. Here, since \( y = 0 \) represents the x-axis, our task is to examine how the curve interacts with this boundary. This step sets the stage for identifying the area enclosed by the curve and the x-axis, which is critical for finding the definite integral.

Definite Integrals

Defining a definite integral involves computing the signed area between a curve and the x-axis on a given interval. In our problem, we focus on the integral of \( y = \sqrt[3]{x} \) from \( x = -2 \) to \( x = 2 \). The integral is denoted as \( \int_{-2}^{2} \sqrt[3]{x} \, dx \), conveying the sum of infinitely small products of the function’s value and an infinitesimally small width, \( dx \).
This operation is crucial because it allows us to determine the exact area under the curve, accounting for the function's behavior over the interval. Calculating this definite integral gives a precise measure of the region's size and helps confirm the initial sketch's assumptions. This approach moves from a visual estimate to a mathematical calculation, ensuring accuracy.

Area Under Curve

The area under a curve is a concept closely linked with definite integrals. In this specific case, it refers to the region bounded by \( y = \sqrt[3]{x} \) and \( y = 0 \) across \( x = -2 \) to \( x = 2 \). This area can be visualized as the accumulation of the function’s output over the specified range of \( x \).
Integral calculus gives us the tools to calculate this area precisely, transcending basic geometric shapes. By setting up the definite integral, we parameterize this problem into manageable mathematical operations. This approach not only explains the area in a concrete manner but also verifies our manual estimations by performing calculations with recognized mathematical operations. An estimate supports that the calculations are within a reasonable range, enhancing understanding and reassurance.

Antiderivative Calculation

Calculating the antiderivative is a fundamental step in solving integrals. It allows us to go backward from the derivative of a function to the original function itself. For our function, \( y = \sqrt[3]{x} \), we find its antiderivative using the basic rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \). Here, \( n = \frac{1}{3} \), so the antiderivative becomes \( \frac{3}{4}x^{\frac{4}{3}} \).
The next step involves evaluating the antiderivative at the limits \( x = -2 \) and \( x = 2 \). The results give us the net area by subtracting the evaluated results at these points. This operation simplifies the earlier setup into a calculated value, precisely \( \approx 3.7798 \), consolidating the understanding of how integrals work and are derived. Understanding this concept deeply enhances your skill in solving complex calculus problems.