Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the region \(R\) bounded by \(y=x+6, y=x^{3}\), and \(2 y+x=0 .\) Then find its area. Hint: Divide \(R\) into two pieces.

Step-by-Step Solution

Answer

The area of region R is approximately calculated by summing the areas from the definite integrals for each partitioned piece. This requires solving cube-root equations for precise boundary intersections.

Step 1: Identify the Equations

First, let's identify all equations that bound the region. We have three equations: 1. The line: \(y = x + 6\)2. The curve: \(y = x^3\)3. The line: \(2y + x = 0\) which can be rewritten as \(y = -\frac{x}{2}\).

Step 2: Find Intersection Points

Next, determine the points where each pair of curves intersect. 1. Intersection of \(y = x + 6\) and \(y = x^3\): Set \(x + 6 = x^3\) which simplifies to \(x^3 - x - 6 = 0\). Solve for real roots.2. Intersection of \(y = x + 6\) and \(y = -\frac{x}{2}\): Set \(x + 6 = -\frac{x}{2}\), which simplifies to \( x = -4\).3. Intersection of \(y = x^3\) and \(y = -\frac{x}{2}\): Set \(x^3 = -\frac{x}{2}\), which simplifies to the equation \(x^3 + \frac{x}{2} = 0\). Factor the equation to find intersections.

Step 3: Visualize and Divide the Region

Now plot the equations on the coordinate plane to visualize the bounded region. Identify where these curves enclose a region. Based on intersections, logically divide the region into two parts, as suggested by the hint. Typically, choose divisions where the curves switch from one upper curve to another, ensuring entire area coverage.

Step 4: Calculate the Area of Each Piece

Use definite integrals to find the area of each part. 1. Integrate the difference in \(y\)-values for each subregion.- For one section, determine limits from calculated intersections and integrate \(\int (x+6 - x^3)\text{d}x\).- For the other section, use \(\int (\frac{-x}{2} - x^3)\text{d}x\).

Step 5: Integrate and Solve

Evaluate all necessary integrals: 1. Calculate the definite integral for the first area between appropriate intersections. 2. Similarly, calculate for the second region. Sum both areas to obtain total area enclosed by the curves.

Step 6: Sum the Areas

Finally, sum up the areas calculated from both the steps to find the total area of the region R.

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that deals with the concept of integrals and accumulation of quantities. In this exercise, it's utilized to find the area of a region enclosed by curves. Integrals help us determine quantities like areas under a curve, total growth over time, and much more.
An integral can be understood as a mathematical tool that allows us to accumulate infinite small quantities to find a whole. When we integrate a function, we essentially compute the area under that function's graph over a given interval.

• Definite integrals are employed when computing areas, as they provide a specific numeric value.
• Indefinite integrals, on the other hand, represent a family of functions.

In the context of this problem, we're finding the area between curves. This requires calculating definite integrals since we need a precise result representing the area enclosed by the specified boundaries.

Intersection of Curves

The intersection of curves refers to the points where two or more graphs meet. Determining these points is crucial in problems like this, as they establish the boundaries of the regions under consideration.
To find intersections, we set the equations of two curves equal because at intersection points, the y-values (and x-values) must be the same for both functions. Solving these equations gives the x-coordinates of these intersections:

• For example, setting the line and the cubic function equal: \[x + 6 = x^3\] simplifies to \[x^3 - x - 6 = 0\].
• Additionally, solving \[x + 6 = -\frac{x}{2}\], which simplifies to \[x = -4\], gives another intersection point.

These intersection points serve as limits for our integrals and assist in visualizing the bounded region.

Definite Integrals

Definite integrals are used to calculate areas, lengths, and other quantities. They are definite because they result in a numerical value determined by the bounds of integration, usually the intersection points of curves.
To find the area between two functions, the definite integral is taken of the difference of the two functions over an interval. The upper curve’s function minus the lower curve’s function gives the height of the small slices of area we sum:

• For example, for one section of a region, you might calculate: \[\int_{a}^{b} (f(x) - g(x)) \, dx\]where \(a\) and \(b\) are the x-coordinates of our intersection points.

This integral gives the total area between the curves from \(x = a\) to \(x = b\). This approach is repeated for each identified region, and the results are summed to find the total area.

Area Between Curves

In finding the area between curves, the goal is to determine the space that graphs enclose on a plane. This space is found between two known curves, over a specific range of x-values.
Calculating this area requires setting up integrals that account for each curve's relative position.

• An approach involves identifying which function serves as the "upper" and which serves as the "lower" curve in the integration interval.
• Next, take the difference of these curves to find the height at any point along the x-axis.

Once these integrals are set, evaluating them provides the area of distinct regions. Adding these individual results yields the complete area enclosed by the given curves. The exercise here divides areas where different curves take turns as upper or lower bounds to ensure comprehensive coverage.