Problem 47 - Calculus

Question

Use periodicity to calculate \(\int_{0}^{4 \pi}|\cos x| d x.\)

Step-by-Step Solution

Answer

The integral evaluates to 8.

Step 1: Recognize the Periodicity

Recognize that the function \(|\cos x|\) is periodic with period \(\pi\) because \(\cos(x + 2\pi) = \cos x\) implies \(|\cos(x + \pi)| = |\cos x|\). This means that the integral over any interval of length \(\pi\) will have the same value.

Step 2: Break Down the Interval

The integral is from \(0\) to \(4\pi\). Since the period is \(\pi\), we can break the interval into four sections of length \(\pi\): \([0, \pi], \ [\pi, 2\pi], \ [2\pi, 3\pi], \ [3\pi, 4\pi]\).

Step 3: Evaluate the Integral Over One Period

Calculate the integral of \(|\cos x|\) from \(0\) to \(\pi\). Since \(\cos x\) is non-negative in \([0, \pi/2]\) and negative in \([\pi/2, \pi]\), we have:\[\int_{0}^{\pi}|\cos x|\, dx = \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} -\cos x \, dx.\]Both of these integrals result in \( \int_{0}^{\pi/2} \cos x \, dx = \int_{\pi/2}^{\pi} -\cos x \, dx = 1\). Thus, \( \int_{0}^{\pi}|\cos x|\, dx = 1 + 1 = 2\).

Step 4: Sum the Integrals

Since each sub-interval contributes equally to the total integral due to periodicity, multiply the integral over one period by four. Thus, the final result is:\[\int_{0}^{4\pi}|\cos x|\, dx = 4 \times 2 = 8.\]

Key Concepts

Periodicity of Functions

Periodicity in functions refers to the repeating nature of certain mathematical patterns over regular intervals. A function is said to be periodic if there is a positive number \( p \) such that \( f(x + p) = f(x) \) for all \( x \) in the domain of \( f \). In this exercise, we worked with the cosine function, which is inherently periodic.

The standard cosine function \( \cos(x) \) repeats every \( 2\pi \) because of the circular nature of the trigonometric circle. However, when examining \(|\cos(x)|\), which is the absolute value, the periodicity becomes \( \pi \). This is because the absolute value makes the function's negative portions positive, effectively halving the period. This insight allows us to simplify integrals over extended intervals by calculating within a single period and then multiplying the result accordingly.

Understanding periodicity is crucial as it helps in breaking down complex integrals into manageable pieces, especially when dealing with trigonometric functions.

Definite Integrals

Definite integrals represent the area under a curve within a specific interval, providing a way to accumulate quantities continuously. In mathematical terms, the definite integral of a function \( f(x) \) from \( a \) to \( b \) is given by \( \int_{a}^{b} f(x) \, dx \). It is used often when wanting to find total quantities, such as distance, area, or in this case, the total of \(|\cos(x)|\) over a set interval.

By recognizing the periodicity of \(|\cos(x)|\), we split the interval between 0 and \(4\pi\) into four equal parts, each representing one complete period. Calculating the definite integral over just one of these periods reduces the complexity. As demonstrated in the exercise, you find the integral over one period and then multiply by the number of periods within the larger interval.

This process not only simplifies calculations but also illustrates the important principle of using symmetry and repetition in mathematical problems to achieve solutions efficiently.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that contain trigonometric functions like sine, cosine, and tangent. These integrals can require special techniques due to the unique properties of trigonometric functions. In our scenario with \(|\cos x|\), applying the periodicity of the cosine function simplifies the integration process.

The absolute value \(|\cos x|\) modifies the behavior of the function. Between 0 and \(\pi\), we had to divide the interval further into \([0, \pi/2]\) and \([\pi/2, \pi]\). During these intervals, recognizing when \(\cos x\) is positive or negative allowed us to correctly evaluate \(|\cos x|\) by adjusting signs.

This example highlights two key techniques in working with trigonometric integrals:

• **Breaking Down Intervals**: Splitting the interval into manageable sections where the function behaves predictably.
• **Utilizing Symmetry**: Because trigonometric functions like cosine have symmetrical properties, use them to assess integration limits efficiently.

These approaches are not only applicable here but also useful in more complex trigonometric integration scenarios, making them a fundamental part of calculus studies.