Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-30, 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: \(y=x^{3}, y=0\), between \(x=-3\) and \(x=3\)

Step-by-Step Solution

Answer

The area of the region is zero, due to symmetry in the interval.

Step 1: Understand the Boundaries

The region is bounded by the curve defined by the equation \(y = x^3\) and the line \(y = 0\) (the x-axis), within the interval \([-3, 3]\) along the x-axis.

Step 2: Sketch the Region

Graph the curves \(y = x^3\) and the line \(y = 0\) on the same set of axes. Since the function \(y = x^3\) is symmetric about the origin, it results in a cubic curve crossing the x-axis at \(x = 0\). The region of interest is below the curve and above the x-axis, from \(x = -3\) to \(x = 3\).

Step 3: Show a Typical Slice

Visualize a thin vertical strip or rectangle ('slice') at an arbitrary point \(x\) within the interval \([-3, 3]\). The height of this slice is \(x^3 - 0 = x^3\), and the width is \(dx\), a small change in \(x\).

Step 4: Set Up the Integral

To find the total area of the region, integrate the height function \(x^3\) from \(x = -3\) to \(x = 3\). The integral is written as \[ \int_{-3}^{3} x^3 \, dx \].

Step 5: Compute the Integral

Evaluate \( \int_{-3}^{3} x^3 \, dx \). The antiderivative of \(x^3\) is \(\frac{x^4}{4}\). Apply the limits of integration: \[ \left[ \frac{x^4}{4} \right]_{-3}^{3} = \frac{3^4}{4} - \frac{(-3)^4}{4} = \frac{81}{4} - \frac{81}{4} = 0. \]

Step 6: Confirm the Result

Notice that the function \(y = x^3\) is symmetric about the origin. The areas above the x-axis and below the x-axis within \([-3, 0]\) and \([0, 3]\) respectively cancel each other, which is why the total net area is zero.

Key Concepts

Definite Integral

In calculus, the definite integral is a powerful tool for finding the total accumulation of a quantity over a specific interval. It often represents the area under a curve and above the x-axis within a given domain. For the function \(y = x^3\), we are looking to find this area between \(x = -3\) and \(x = 3\).
We set up this scenario as a definite integral \[ \int_{-3}^{3} x^3 \, dx \]. By evaluating it, we aim to calculate the precise area under the curve. This involves calculating the antiderivative followed by applying the limits of integration. The result, in this case, showcases the interesting property of symmetry in functions.

Area Under a Curve

The area under a curve refers to the space between the graph of a function and the x-axis, across a specified range. For the given problem, the curve \(y = x^3\) generates an area that must be computed from \(x = -3\) to \(x = 3\).
Drawing this curve reveals portions of the area both above and below the x-axis. Calculating this total area helps us understand the net accumulation of space, or differences above and below the axis. Since the function is cubic, these areas can offset each other, leading to a fascinating outcome when symmetrical boundaries are utilized in integration.

Antiderivative

An antiderivative, often called an indefinite integral, is the inverse operation to differentiation. It's used to determine the original function from its derivative. To calculate the area or integral, we need the antiderivative of the function.
For the equation \(y = x^3\), the antiderivative is \(\frac{x^4}{4}\). This function elevates \(x^3\) one degree higher and properly scales it to offset changes during differentiation. To compute the value of the definite integral \[ \left[ \frac{x^4}{4} \right]_{-3}^{3} = \frac{81}{4} - \frac{81}{4} = 0 \], we apply the given bounds and the antiderivative, demonstrating the perfect cancellation created by the symmetry of the function.

Symmetric Functions

A symmetric function is one that mirrors itself across a particular axis or point. In this case, the function \(y = x^3\) is symmetric about the origin. This symmetry plays a critical role in understanding why the net area between \(x = -3\) and \(x = 3\) results in zero.
Symmetry implies that for every positive area segment above the x-axis, there is an equally negative counterpart below the x-axis. Thus, when integrated over a symmetric interval around the origin, these segments cancel each other out.

• This concept helps in quickly assessing and calculating the area in problems involving symmetric functions.
• Recognizing symmetry can aide in checking the plausibility of results in integral calculations.

By understanding these symmetrical properties, students can save time and avoid unnecessary calculations.