Question

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

Short answer

The integral \( \int_{0}^{4\pi} |\cos x| \, dx = 8. \)

Step-by-step solution

Identify the Periodicity of the Function

The function \( |\cos x| \) is periodic with period \( \pi \) because after every interval of \( \pi \), the pattern of the absolute value function repeats due to the symmetry of the cosine wave above and below the x-axis.

Set the Bounds According to the Periodicity

Since we know \( |\cos x| \) repeats every \( \pi \), we can consider the interval from \( 0 \) to \( 4\pi \) as four intervals of \( [0, \pi] \), \( [\pi, 2\pi] \), \( [2\pi, 3\pi] \), and \( [3\pi, 4\pi] \). The integral over each interval is the same.

Calculate the Integral Over One Period

Calculate the integral of \( |\cos x| \) from \( 0 \) to \( \pi \): \[ \int_0^{\pi} |\cos x| \, dx = \int_0^{\pi} \cos x \, dx. \]This equates to:\[ = \left[\sin x\right]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0. \]

Use Symmetry for Calculation Over Full Range

Although the result of one period \( [0, \pi] \) is zero, the symmetry we calculated is incorrect because the expression for \( |\cos x| \) is positive over these intervals. Let's correct it:\[ \int_0^{\pi} |\cos x| \, dx = \int_0^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} -\cos x \, dx. \]This resolves to:- \( \int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = 1 - 0 = 1 \)- \( \int_{\pi/2}^{\pi} -\cos x \, dx = [-\sin x]_{\pi/2}^{\pi} = 0 + 1 = 1 \)Thus, each integral on \([0, \pi]\) yields \( 2 \).

Calculate Integral Over All Intervals

Since the integral over one period \([0, \pi]\) is \(2\), the integral over \([0, 4\pi]\) is:\[ 4 \times 2 = 8. \]

Final Answer

The total integral of \( |\cos x| \) from \( 0 \) to \( 4\pi \) is \( 8 \).

Key concepts

Periodicity

Periodicity is a fascinating property of certain functions where the pattern repeats over regular intervals. In the context of the function \(|\cos x|\), we observe that it exhibits a period of \(\pi\). This means that every \(\pi\) units along the x-axis, the function's shape recurs. Recognizing periodicity can simplify integral calculus problems, as it reduces the scope to manage repetitive sections. For example:

• \(|\cos x| \) duplicates its cycle over every interval like \([0, \pi]\), \([\pi, 2\pi]\), and so on.
• This type of function generally has consistent maximums and minimums across these intervals.

Understanding how \(|\cos x|\) behaves lets you use one cycle to predict and solve for multiple cycles, which is a very efficient approach in calculus. When you detect periodicity, you can reduce the calculation workload dramatically.

Definite Integral

A definite integral represents the accumulated area under a curve within a specified interval. When solved, it delivers a numeric value that quantifies this area. The definite integral of a function like \(|\cos x|\) from 0 to \(4\pi\) can be approached by:

• Recognizing the repeated interval pattern due to periodicity.
• Breaking the larger interval into manageable sub-intervals, like several cycles between \([0, \pi]\).

Each period in our exercise is \(|\cos x|\) over \([0, \pi]\), which initially miscomputes to zero due to symmetry error. But realigning by integrating different segments of the interval yields accurate partial results. Add these equally accumulated areas across all periods to obtain the total area beneath \(|\cos x|\) from 0 to \(4\pi\). This gives a precise measurement using integral calculus.

Absolute Value Functions

The absolute value function is crucial in various fields in mathematics, as it manipulates the direction of the original function to always be non-negative. For \(|\cos x|\), it affects the symmetry of the function:

• A cosine wave without an absolute value has both positive and negative portions.
• Applying \(|\cdot|\) results in the negative parts being flipped above the x-axis, hence becoming non-negative.

This transformation ensures that when calculating the integral of \(|\cos x|\), we reflect on full value, factoring positive areas correctly. This is essential in integrating across periodic intervals, where every bump or cycle counts fully because any we previously "missed" due to a negative value is now counted correctly as a positive contribution. Understanding this feature of absolute value functions is vital for precise integral solutions in periodic scenarios.