Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

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. \(y=x^{3}, y=0\), between \(x=-3\) and \(x=3\)

Step-by-Step Solution

Answer

The area of the region bounded by y = x^3 and y = 0 from x = -3 to x = 3 is zero.

Step 1: Sketch the Region

To sketch the region, plot the equation \( y = x^3 \) from \( x = -3 \) to \( x = 3 \). The graph is an odd function, symmetric about the origin, forming an S-shape curve starting from point (-3, -27) to (3, 27). The line \( y = 0 \) is the x-axis, which serves as the lower boundary of the region.

Step 2: Show a Typical Slice

A typical slice of the area is a vertical line segment extending from the curve \( y = x^3 \) to the x-axis. Each slice has a small width \( \Delta x \). The height of the slice at a particular \( x \) is given by the value of \( y = x^3 \).

Step 3: Approximate the Area

To approximate the area, consider sub-regions under the curve by summing the areas of rectangles with the height \( f(x) = x^3 \) and width \( \Delta x \). As \( \Delta x \) becomes smaller, the approximation improves.

Step 4: Set Up the Integral

The area under the curve from \( x = -3 \) to \( x = 3 \) can be found by setting up an integral: \[\int_{-3}^{3} x^3 \, dx\] This integral represents the sum of all the infinitesimal slices' areas.

Step 5: Calculate the Integral

Calculate the definite integral: \[\int_{-3}^{3} x^3 \, dx = \left[\frac{x^4}{4}\right]_{-3}^{3}\] Evaluate it from \( -3 \) to \( 3 \), obtaining: \[ \left( \frac{3^4}{4} \right) - \left( \frac{(-3)^4}{4} \right) = 20.25 - 20.25 = 0\] The area is zero because the contributions from \(-3\) to \(0\) and from \(0\) to \(3\) cancel out, given the symmetry of the function around the y-axis.

Step 6: Estimate Area for Validation

Estimate the area by observing the symmetry: The function \( y = x^3 \) is odd and symmetric about the origin, providing equal positive and negative area contributions up to the x-axis over equal positive and negative intervals \(-3\) to \(3\). This implies that any estimated area would also suggest a net area of zero.

Key Concepts

area under a curve

When we look at the curve for the function \(y = x^3\), we're interested in finding the total area between this curve and the x-axis over a specified interval. This space is referred to as the "area under the curve." Understanding this helps us know how much space a given function covers above or below the x-axis over a certain range. However, it's worth noting that if a curve dips below the x-axis, the area is considered negative. This can lead to total cancellation, especially if the curve is symmetric, which is the case here.

definite integral

The concept of a definite integral is central to calculating areas under curves. For a function \(f(x)\), the definite integral over an interval \([a, b]\) is written as \(\int_{a}^{b} f(x) \, dx\). This represents the sum of an infinite number of infinitesimally small rectangles under the curve from \(x = a\) to \(x = b\).
In the example problem, we calculated \(\int_{-3}^{3} x^3 \, dx\) to find the area under the curve \(y = x^3\) between \(x = -3\) and \(x = 3\). It's important to realize that this computes the net area, considering both positive and negative contributions, even if these areas might cancel each other out.

graph sketching

Graphing a function gives us a tangible view of the relationship described by an equation. In sketching the graph of \(y = x^3\) from \(x = -3\) to \(x = 3\), the shape is crucial. The graph shows an S-like curve starting from negative coordinates and moving upwards, symmetric around the origin. This sketch can guide our understanding of how the area under the curve is distributed and help visualize properties like symmetry or sections above or below the x-axis.

odd function symmetry

In mathematics, functions are categorized as even or odd based on their symmetry properties. An odd function has the property \(f(-x) = -f(x)\). For our function \(y = x^3\), if you replace \(x\) with \(-x\), you get \((-x)^3 = -x^3\), confirming its odd nature.
The significance of odd symmetry is that the graph is symmetric about the origin, meaning the area contributions on the negative-x side cancel those on the positive-x side. This symmetry property helped us understand why the integral of \(y = x^3\) from \(-3\) to \(3\) resulted in zero: the positive area from 0 to 3 cancels out the negative area from -3 to 0.