Problem 48
Question
Calculate \(\int_{0}^{4 \pi}|\sin 2 x| d x\).
Step-by-Step Solution
Verified Answer
The integral evaluates to 8.
1Step 1: Understand the Absolute Value Effect
The function inside the integral, \(\sin 2x\), oscillates between -1 and 1. The absolute value, \|\sin 2x|\, makes it positive, reflecting all negative values above the x-axis. Our task is thus to calculate the area under this curve in the interval [0, 4\pi].
2Step 2: Determine the Number of Full Oscillations
Since \(\sin 2x\) completes a full cycle (from 0 to \(\pi\)) in a period of \pi\ (dividing the regular period of \(2\pi\) by 2), check how many cycles occur from 0 to 4\pi\, which equals \frac{4\pi}{\pi} = 4\ cycles.
3Step 3: Calculate a Single Oscillation Area
Find the area for one complete oscillation of \|\sin 2x|\ from 0 to \pi\. This can be expressed as \int_0^{\pi}|\sin 2x|dx.\ Since \(\sin 2x\) is symmetric and positive in [0, \pi/2] and negative but flipped by the absolute value in [\pi/2, \pi], both halves contribute equally. First focus on 0 to \pi/2: \int_0^{\pi/2}\sin 2xddx. \ Using substitution, let \ u = 2x, \ du = 2dx \, this becomes: \ \= \frac{1}{2}\int_0^{\pi}\sin udu.This evaluates to: \ \left. -\frac{1}{2}\cos u \right|_0^{\pi} = 1.\ The total area per cycle: \ 2 \times 1 = 2.\ \
4Step 4: Multiply by the Number of Cycles
Since there are 4 complete oscillations in the interval from 0 to \(4\pi\), and each cycle contributes an area of 2, the total area is \ 4 imes 2 = 8.\ \
5Step 5: Final Integration Result
The integral of \(\int_0^{4\pi}|\sin 2x| dx \), accounting for the absolute value over the indicated interval, is equal to 8.
Key Concepts
Trigonometric FunctionsAbsolute ValueOscillationsSubstitution Method
Trigonometric Functions
When studying trigonometric functions, we often come across periodic behaviors and oscillations. In this exercise, the integral involves the function \( \sin 2x \). This function belongs to the trigonometric family, characterized by their wave-like appearance and periodicity.
The standard sine function \( \sin x \) has a period of \( 2\pi \), meaning it repeats its shape every \( 2\pi \) units. However, the function \( \sin 2x \) has a period of \( \pi \), because it oscillates twice as fast within the same interval, doubling the rate of its oscillation.
Understanding the period of trigonometric functions is crucial when integrating over extended intervals as it helps determine how many full cycles fit within that range.
The standard sine function \( \sin x \) has a period of \( 2\pi \), meaning it repeats its shape every \( 2\pi \) units. However, the function \( \sin 2x \) has a period of \( \pi \), because it oscillates twice as fast within the same interval, doubling the rate of its oscillation.
Understanding the period of trigonometric functions is crucial when integrating over extended intervals as it helps determine how many full cycles fit within that range.
Absolute Value
The use of absolute values in integrals changes the landscape of problem-solving significantly. Here, we are dealing with the absolute value of \( \sin 2x \). Absolute value, denoted as \( |\cdot| \), essentially makes all values positive by "flipping" any negative values above the x-axis.
This concept is particularly essential when dealing with oscillatory functions like sine or cosine. Since these functions oscillate above and below zero, taking the absolute value allows us to calculate the total area encompassed within one period or cycle, which is crucial in area calculation problems. As such, it simplifies our task by transforming the oscillation into a simple recurrent positive wave from which the area can be easily calculated.
This concept is particularly essential when dealing with oscillatory functions like sine or cosine. Since these functions oscillate above and below zero, taking the absolute value allows us to calculate the total area encompassed within one period or cycle, which is crucial in area calculation problems. As such, it simplifies our task by transforming the oscillation into a simple recurrent positive wave from which the area can be easily calculated.
Oscillations
Oscillations describe the repeating up and down movement of functions like sine and cosine. In the given problem, the function \( \sin 2x \) oscillates between -1 and 1, making it ideal for illustrating concepts of periodicity.
This oscillation completes every \( \pi \) units. For the interval from 0 to \( 4\pi \), there are exactly 4 full oscillations (or cycles). Each positive-half and negative-half of an oscillation is symmetric when the absolute value is taken into account. Hence, understanding oscillations helps in identifying repetitive patterns, which simplifies calculations over longer intervals by merely multiplying the area of one cycle by the number of total cycles.
This oscillation completes every \( \pi \) units. For the interval from 0 to \( 4\pi \), there are exactly 4 full oscillations (or cycles). Each positive-half and negative-half of an oscillation is symmetric when the absolute value is taken into account. Hence, understanding oscillations helps in identifying repetitive patterns, which simplifies calculations over longer intervals by merely multiplying the area of one cycle by the number of total cycles.
Substitution Method
The substitution method is a powerful tool in calculus for simplifying integrals. In this exercise, it is used to handle the integral \( \int_0^{\pi/2}\sin 2x \, dx \). Substitution reduces the complexity by converting the variable or function into a simpler form.
To perform substitution, we often let \( u = g(x) \) where \( g(x) \) is a part of the integrand, and then find \( du \) as the derivative of \( u \) with respect to \( x \). In our case, letting \( u = 2x \) simplifies the integral to involve just \( \sin u \), with \( du = 2 \, dx \), eventually leading to a simpler integration process.
This not only makes the integral easier to solve but also reduces potential mistakes in handling complex expressions.
To perform substitution, we often let \( u = g(x) \) where \( g(x) \) is a part of the integrand, and then find \( du \) as the derivative of \( u \) with respect to \( x \). In our case, letting \( u = 2x \) simplifies the integral to involve just \( \sin u \), with \( du = 2 \, dx \), eventually leading to a simpler integration process.
This not only makes the integral easier to solve but also reduces potential mistakes in handling complex expressions.
Other exercises in this chapter
Problem 47
Use periodicity to calculate \(\int_{0}^{4 \pi}|\cos x| d x\).
View solution Problem 47
, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{\pi / 6} \sin ^{3} \theta \cos \theta d \theta $$
View solution Problem 48
Use a graphing calculator to graph each integrand. Then use the Boundedness Property (Theorem \(C\) ) to find a lower bound and an upper bound for each definite
View solution Problem 48
, use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{\pi / 6} \frac{\sin \theta}{\cos ^{3} \theta} d \theta $$
View solution