Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.4.38

Chapter 5, Problem 5.4.38

Average height of a wave The surface of a water wave is described by y = 5 (1 + cos 𝓍) , for ― π ≤ 𝓍 ≤ π, where y = 0 corresponds to a trough of the wave (see figure). Find the average height of the wave above the trough on [ ―π , π] .

Verified step by step guidance

Step 1: Understand the problem. The goal is to find the average height of the wave above the trough over the interval [−π, π]. The wave is described by the function y = 5(1 + cos(x)), where y = 0 corresponds to the trough.

Step 2: Recall the formula for the average value of a function f(x) over an interval [a, b]: Average value = (1 / (b - a)) * ∫[a to b] f(x) dx. Here, f(x) = 5(1 + cos(x)), a = −π, and b = π.

Step 3: Set up the integral for the average value. Substitute the given function and interval into the formula: Average value = (1 / (π - (−π))) * ∫[−π to π] 5(1 + cos(x)) dx.

Step 4: Simplify the expression. The denominator becomes 2π, so Average value = (1 / 2π) * ∫[−π to π] 5(1 + cos(x)) dx. Factor out the constant 5 from the integral: Average value = (5 / 2π) * ∫[−π to π] (1 + cos(x)) dx.

Step 5: Break the integral into two parts: ∫[−π to π] (1 + cos(x)) dx = ∫[−π to π] 1 dx + ∫[−π to π] cos(x) dx. Evaluate each integral separately. The integral of 1 over [−π, π] is straightforward, and the integral of cos(x) over [−π, π] can be solved using trigonometric properties.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Average Value of a Function

The average value of a continuous function over an interval is calculated using the formula: (1/(b-a)) * ∫[a to b] f(x) dx. This concept is essential for determining the average height of the wave, as it allows us to find the mean value of the function y = 5(1 + cos x) over the specified interval from -π to π.

Integration

Integration is a fundamental concept in calculus that involves finding the area under a curve represented by a function. In this context, we will use definite integration to calculate the total area under the wave function y = 5(1 + cos x) over the interval [-π, π], which is necessary for computing the average height of the wave.

Cosine Function Properties

The cosine function is periodic and oscillates between -1 and 1. In the given wave function y = 5(1 + cos x), the term (1 + cos x) shifts the graph vertically, ensuring that the wave's height is always non-negative. Understanding the behavior of the cosine function is crucial for analyzing the wave's shape and determining its average height above the trough.

Recommended video:

06:21

Properties of Functions

Related Practice

Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.

∫ 𝓍eˣ² d𝓍

views

Textbook Question

A midpoint Riemann sum Approximate the area of the region bounded by the graph of ƒ(𝓍) = 100 ― x² and the x-axis on [0, 10] with n = 5 subintervals. Use the midpoint of each subinterval to determine the height of each rectangle (see figure).

111

views

Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.

∫ 𝓍³ (𝓍⁴ + 16)⁶ d𝓍

views

Textbook Question

Areas of regions Find the area of the region bounded by the graph of ƒ and the 𝓍-axis on the given interval.

ƒ(𝓍) = 𝓍³ ― 1 on [―1, 2]

102

views

Textbook Question

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.

{Use of Tech} v = 4 √(t +1) (mi/hr) . for 0 ≤ t ≤ 15 ; n = 5

views

Textbook Question

Derivatives of integrals Simplify the following expressions.

d/dz ∫¹⁰ₛᵢₙ ₂ dt /(t⁴ + 1)

views