Problem 8 - Calculus of a Single Variable

Question

Finding a Region In Exercises \(7-12\) , the integrand of the definite integral is a difference of two functions. Sketch the graph of each function and shade the region whose area is represented by the integral. $$ \int_{-1}^{1}\left[\left(2-x^{2}\right)-x^{2}\right] d x $$

Step-by-Step Solution

Answer

The region represented by the integral is the area between the graphs of \(y=2-x^{2}\) and \(y=x^{2}\) from \(-1\) to \(1\) on the x-axis. The area of this region can be calculated exactly with the integral \(\int_{-1}^{1} (2-2x^{2}) dx\).

Step 1: Plotting the Functions

Plot the two functions \(y=2-x^{2}\) and \(y=x^{2}\) on the same graph. The first is a downward-opening parabola with vertex at (0,2) and the second is an upward-opening parabola with vertex at the origin (0,0).

Step 2: Identify the Region

The area between these two curves from \(-1\) to \(1\) is the region whose area is represented by the given integral. It is bounded above by the graph of \(y = 2-x^2\) and below by \(y = x^2\). Shade this area on the graph.

Step 3: Calculate the Integral

The integral can be calculated as follows: \[ \int_{-1}^{1} (2-x^{2}-x^{2}) dx = \int_{-1}^{1} (2-2x^{2}) dx\]. By calculating this integral, the exact area of the shaded region can be obtained.

Key Concepts

Area Between Curves

Determining the area between curves is a fundamental topic in calculus that involves integrating the difference of two functions over a certain interval. This is often illustrated by considering two functions where one function, when plotted, lies on top of the other between two points of intersection. The integral of their difference computes the area enclosed by these two curves over the interval.

To find this area:

• Identify the two functions that form the boundary of the region. In the given exercise, these are the parabola \(y = 2 - x^2\), which opens downward, and the parabola \(y = x^2\), which opens upward.
• Determine the interval over which to compute the integral. The interval in this problem is from \(-1\) to \(1\).
• Set up the definite integral as the difference: \(\int_{-1}^{1} [(2-x^2) - x^2] dx\). This allows us to calculate the area between the curves.

The result of the definite integral can help solve numerous practical problems, including calculating material consumption and visualizing differences between theoretical and actual results.

Parabolas

Parabolas are one of the most common types of curves studied in mathematics, often described as a symmetrical, u-shaped curve. They can open upwards or downwards, and their shape is determined by the quadratic equation \(y = ax^2 + bx + c\).

In the exercise, there are two parabolas:

• \(y = 2-x^2\) is a downward-opening parabola. This shape occurs because the coefficient of \(x^2\) is negative, indicating the parabola's highest point is at the vertex (0, 2).
• \(y = x^2\) is an upward-opening parabola. With a positive \(x^2\) coefficient, it starts at its vertex (0, 0) and opens upwards.

The intersections and directions of these parabolas are vital when determining the area between them, which is further analyzed through graph sketching.

Graph Sketching

Sketching graphs is essential for visualizing mathematical relationships and solving calculus problems. When working with functions such as parabolas:

• Begin by understanding the basic shape and properties of each function.
• Identify key points such as the vertex and points of intersection.
• From our exercise, plot the graphs of \(y = 2-x^2\) and \(y = x^2\). Notice they intersect at points \(x = -1\) and \(x = 1\).
• Shade the area between the two curves on the plot over the interval of interest, which is essential for visualizing the area calculated by the definite integral.

Graph sketching helps to verify our understanding of mathematical concepts visually. In this particular problem, sketching aids in comprehending the position and the area between the curves more intuitively, leading to accurate calculations of the definite integral.