Problem 19 - Calculus: Early Transcendentals

Question

Sketch the region enclosed by the given curves and find its area. \( y = \cos \pi x \) , \( y = 4x^2 - 1 \)

Step-by-Step Solution

Answer

The area is \(\frac{5}{6}\) and is enclosed between \(x = -0.5\) and \(x = 0.5\).

Step 1: Identify Intersection Points

To find the region enclosed by the curves, we first need to determine where they intersect. Set the equations equal to each other: \[ \cos(\pi x) = 4x^2 - 1 \]Solve for \(x\) by equating and finding the roots. This involves solving \[ \cos(\pi x) = 4x^2 - 1 \].Using numerical methods or graphically, find approximate intersection points at \( x \approx -0.5 \) and \( x \approx 0.5 \).

Step 2: Integrate to Find Area

The area between the curves from \( x = -0.5 \) to \( x = 0.5 \) is given by the integral: \[ A = \int_{-0.5}^{0.5} (\cos(\pi x) - (4x^2 - 1)) \, dx \].Simplify the integrand: \[ A = \int_{-0.5}^{0.5} \cos(\pi x) + 1 - 4x^2 \, dx \].

Step 3: Evaluate the Integral

Evaluate the integral: \[ A = \int_{-0.5}^{0.5} \cos(\pi x) \, dx + \int_{-0.5}^{0.5} 1 \, dx - \int_{-0.5}^{0.5} 4x^2 \, dx \].Each part simplifies as follows:1. \( \int_{-0.5}^{0.5} \cos(\pi x) \, dx = \frac{1}{\pi} [\sin(\pi x)]_{-0.5}^{0.5} = 0 \).2. \( \int_{-0.5}^{0.5} 1 \, dx = 0.5 - (-0.5) = 1 \).3. \( \int_{-0.5}^{0.5} 4x^2 \, dx = \frac{4}{3}[x^3]_{-0.5}^{0.5} = \frac{2}{3} [(0.5)^3 - (-0.5)^3] = \frac{2}{3} \times 0.25 = \frac{1}{6} \).Combine the results to find: \[ A = 0 + 1 - \frac{1}{6} = \frac{5}{6} \].

Step 4: Sketch the Region

Plot both curves on the same set of axes. The curve \( y = \cos(\pi x) \) oscillates between -1 and 1 with troughs and peaks at \( x = 0, -0.5, 0.5 \). The parabola \( y = 4x^2 - 1 \) is at its minimum at \( x = 0 \) where \( y = -1 \) and opens upwards. Shade the region between \( x = -0.5 \) and \( x = 0.5 \) where the two curves enclose the area we computed as \( \frac{5}{6} \).

Key Concepts

Area between curves

When understanding how to find the area between two curves, it's helpful to think of it as the space trapped between their paths. Imagine you draw two different curves on graph paper, and the goal is to find out how much paper lies between them.
To calculate this area, you start by identifying the curves on the graph and where they intersect, marking the start and end points of this area.

• Identify intersection points: These are the points where the two curves meet each other, and they form the boundaries of the enclosed area.
• Set up the integral: Once you know the intersection points, you set up an integral to calculate the area between the curves. In the context of the problem, the area is given by the integral of the upper curve minus the lower curve.

In our simplified problem, after determining intersection points at approximately \( x = -0.5 \) and \( x = 0.5 \), the integral helps us compute the area represented as the absolute difference of the curve values between these two points.By integrating from one intersection point to another, you add up a bunch of infinitesimally small rectangles between the curves to get the total enclosed area.

Integration techniques

Integration is a powerful calculus tool that allows us to find enclosed areas and other quantities. In this exercise, integration breaks down the complex task of finding area into simpler calculations.
A few key steps in integration techniques include:

• Simplifying the integrand: The problem simplifies the integrand \(\cos(\pi x) + 1 - 4x^2\), making it easier to integrate.
• Dividing the problem: Sometimes it's easier to break the integral into parts that are simpler to evaluate separately. Each part of the function is integrated from the first intersection point to the second.
• Finding antiderivatives: For each part, you calculate the antiderivative, essentially reversing differentiation.

In this example, we evaluated each part of the expression:- For \(\cos(\pi x)\), the antiderivative \([\sin(\pi x)/\pi]\) results in zero over this region.- For 1, it is straightforward and evaluated to 1.- For \(4x^2\), the antiderivative is \((4/3)x^3\), computed to yield \(1/6\).Combining these results correctly gives you the total area, which was found to be \(5/6\). Developing confidence in these techniques makes calculus more approachable and exciting!

Intersection of functions

Finding the intersection points of functions helps set boundaries for the enclosed area between curves.
For our exercise, the intersection occurs where the two curves equal each other, \( \cos(\pi x) = 4x^2 - 1 \). These points define the start and stop for our integration.There are usually a few ways to find these intersection points:

• Solving equations: Directly solving the equation by algebraic methods, though sometimes complex, provides exact intersection points.
• Graphical methods: Using a calculator or graphing software lets you visually inspect and estimate where these curves intersect. It's a good approach when algebraic solutions seem difficult.
• Numerical approximations: Techniques such as the Newton-Raphson method give you numerical solutions when algebra and graphical methods are unfeasible.

In this case, intersection points were found to be \( x \approx -0.5 \) and \( x \approx 0.5 \), indicating where the area between the curves lies. Recognizing intersections is crucial, because without them, setting up the integral and thus finding the area would be impossible.