Problem 22 - Applied Calculus

Question

Use a definite integral to find the area under each curve between the given \(x\) -values. For Exercises \(19-24\) also make a sketch of the curve showing the region. $$ f(x)=1-x^{2} \text { from } x=-1 \text { to } x=1 $$

Step-by-Step Solution

Answer

The area is \( \frac{4}{3} \).

Step 1: Understand the Function and Region

We are given the function \( f(x) = 1 - x^2 \) and need to find the area under this curve from \( x = -1 \) to \( x = 1 \). This function is a downward-opening parabola with vertex at (0,1). We'll be calculating the area of the region enclosed by this parabola and the x-axis over the interval [-1, 1].

Step 2: Set Up the Definite Integral

The area under the curve \( f(x) = 1 - x^2 \) from \( x = -1 \) to \( x = 1 \) is given by the definite integral \( \int_{-1}^{1} (1 - x^2) \, dx \). This will give us the precise area of the desired region.

Step 3: Integrate the Function

To evaluate \( \int_{-1}^{1} (1 - x^2) \, dx \), first, integrate the function: 1. The antiderivative of \( 1 \) is \( x \). 2. The antiderivative of \( -x^2 \) is \( -\frac{x^3}{3} \). Thus, the integrated function is \( x - \frac{x^3}{3} \).

Step 4: Evaluate the Definite Integral

Substitute the limits \( x = -1 \) and \( x = 1 \) into the integrated result \( x - \frac{x^3}{3} \): - For \( x = 1 \): \[ \left( 1 - \frac{1^3}{3} \right) = \left( 1 - \frac{1}{3} \right) = \frac{2}{3} \]- For \( x = -1 \): \[ \left( -1 - \frac{(-1)^3}{3} \right) = \left( -1 + \frac{1}{3} \right) = -\frac{2}{3} \]Calculate the difference: \[ \frac{2}{3} - (-\frac{2}{3}) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

Step 5: Sketch the Curve and Region

To visualize, draw the parabola \( y = 1 - x^2 \) which opens downwards with its vertex at \( (0, 1) \) and intersects the x-axis at \( x = -1 \) and \( x = 1 \). The area below this curve and above the x-axis from \( x = -1 \) to \( x = 1 \) forms a symmetric region about the y-axis.

Key Concepts

Area Under the Curve

When we talk about finding the area under the curve, we're essentially referring to the space that lies between a function graph and the x-axis within a specific interval of the x-values. This concept is significant because it allows us to calculate quantities such as distance, profit, or even accumulated change over time. In our particular example, we examine the area under the curve for the function \(f(x) = 1 - x^2\) between \(x = -1\) and \(x = 1\). This function represents a downward-opening parabola. By determining the area under this curve, we create a method to understand its impact over the specified range. The area can be measured precisely using a definite integral, which is a mathematical expression smoothing out the curve's fluctuations and providing an exact value, unlike approximations that might result from drawing squares or rectangles under the curve. With integral calculus, this calculation becomes systematic and accurate.

Integral Calculus

Integral calculus plays a critical role in mathematics, and it's primarily concerned with the concept of integrals. Unlike derivative calculus, which deals with rates of change, integral calculus focuses on the accumulation of quantities. Whether calculating areas, volumes, or other accumulative measures, integrals provide the solution.
The integral itself can be visualized as the reverse process of differentiation. If we differentiate a function and get another function,

• integration works to determine the original function given its derivative.

In our case, the definite integral \( \int_{-1}^{1} (1 - x^2) \, dx \) calculates the precise area confined by \(f(x) = 1 - x^2\) from \(x = -1\) to \(x = 1\). By integrating, we find the function that describes all accumulated values under the curve. Evaluating it between the given limits quantifies the exact area, yielding \( \frac{4}{3} \). This precise calculation method allows us to leverage the robust framework of integral calculus for various scientific and engineering applications.

Parabola

Within the broader study of conic sections, the parabola is a fascinating shape you'll often encounter in mathematics. In the Cartesian plane, a parabola takes on various forms, depending on its equation. Specifically, the function \(f(x) = 1 - x^2\) describes a downward-opening parabola. Here's a breakdown of its essential properties:

• Vertex at point (0, 1), where it reaches its maximum value because it opens downward.
• Symmetrical about the y-axis, meaning it mirrors itself on either side.
• Intersects the x-axis at \(x = -1\) and \(x = 1\), which are also its roots.

At these intersection points with the x-axis, the parabola bounds the region whose area we calculated earlier. Understanding these key characteristics makes interpreting the graph of a parabola more intuitive. They're not just theoretical; parabolas appear in the real world, from the paths of projectiles to satellite dish shapes, thanks to their unique reflective properties.