Problem 37 - Essential Calculus

Question

Sketch the region bounded by the graphs of the functions, and find the area of the region. $$ f(x)=\cos x, \mathrm{~g}(x)=2-\cos x, 0 \leq x \leq 2 \pi $$

Step-by-Step Solution

Answer

The area of the region bounded by the two functions is \(4\pi.\)

Step 1: Plot the Functions

Plot the two functions \(f(x) = cos(x)\) and \(g(x) = 2 - cos(x)\) on the graph for the interval \(0 \leq x \leq 2\pi\). The first function has graph that follows the pattern of a cosine curve whilst the second function graph is just an upside-down cosine curve, shifted upwards by 2 units.

Step 2: Identify the Region

The region of interest is bounded by these two curves. Note that on the interval \([0, 2\pi]\), the graph of \(g(x) = 2 - cos(x)\) is always above the graph of \(f(x) = cos(x)\).

Step 3: Compute the Area of the Region

The area of the region can be computed as the integral of the absolute value of the difference between the two functions over the interval \([0, 2\pi]\). Here, the integral turns out to be \(\int_{0}^{2\pi} |g(x) - f(x)| dx = \int_{0}^{2\pi} (2 - cos(x) - cos(x)) dx = \int_{0}^{2\pi} (2 - 2cos(x)) dx.\)

Step 4: Evaluate the Integral

The integral equals \([2x - 2sin(x)]_{0}^{2\pi},\) which simplifies to \(4\pi.\)

Key Concepts

Graphing Functions

Graphing involves plotting the given functions on a coordinate plane to visualize their behavior over a particular interval. In this case, we have two functions:

• \( f(x) = \cos x \): This is the cosine function, which is a well-known periodic wave starting at 1, descending to -1, and returning to 1 within one complete period \([0, 2\pi]\).
• \( g(x) = 2 - \cos x \): This is a transformed version of the cosine function. Here, each point of the cosine wave is mirrored upside-down and then shifted up by 2 units.

By plotting these functions within the interval \([0, 2\pi]\), you will notice that the curve of \( g(x) \) is consistently positioned above that of \( f(x) \). This is due to both the negation and vertical shift of the \( \,2 - \cos x \, \) function.

Definite Integrals

Definite integrals are a powerful tool in calculus used to compute the accumulation of quantities, such as areas under curves. When finding the area between two curves, we specifically examine the integral of the difference between the two function values. For our problem, it involves:

• The integral from 0 to \(2\pi\) of \(|g(x) - f(x)|\).
• Since \(g(x)\) is always above \(f(x)\) between the stated limits, the absolute value can be omitted: \(\int_{0}^{2\pi} (g(x) - f(x)) \ dx \).
• Thus, the integral simplifies to \(\int_{0}^{2\pi} (2 - \cos x - \cos x) \ dx = \int_{0}^{2\pi} (2 - 2\cos x) \ dx \).

The outcome of this definite integral gives us the exact area between the curves over the specified interval.

Area Between Curves

The area between two curves is derived from evaluating the integral of their differences over a given range. To find the area between \( f(x) \) and \( g(x) \), follow these steps:

• Identify the bounds of the region, here \([0, 2\pi]\).
• With \( g(x) \) above \( f(x) \), compute the integral \( \int_{0}^{2\pi} [(2 - \cos x) - \cos x] \ dx \).
• This simplifies to \( \int_{0}^{2\pi} (2 - 2 \cos x) \ dx \), which represents the region's total area.
• After solving the definite integral, it results in \( [2x - 2\sin x]_{0}^{2\pi} = 4\pi \).

This final value, \(4\pi\), is the total area delimited by the functions \(f(x)\) and \(g(x)\) over the interval \([0, 2\pi]\).