Textbook Question
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = e^(-5x)
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.6.65
Evaluate the integrals in Exercises 53–76.
65. ∫3dr/√(1-4(r-1)²)
Verified step by step guidance1
Recognize that the integral has the form \( \int \frac{3 \, dr}{\sqrt{1 - 4(r-1)^2}} \), which resembles the standard integral \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\left(\frac{x}{a}\right) + C \).
Make a substitution to simplify the expression inside the square root. Let \( u = r - 1 \), so that \( du = dr \). The integral becomes \( \int \frac{3 \, du}{\sqrt{1 - 4u^2}} \).
Rewrite the expression under the square root as \( 1 - (2u)^2 \), which suggests another substitution or recognizing \( a = 1 \) and \( x = 2u \).
To match the standard form, factor out the constant inside the square root: \( \sqrt{1 - 4u^2} = \sqrt{1 - (2u)^2} \). Consider substituting \( w = 2u \), so \( dw = 2 du \) or \( du = \frac{dw}{2} \).
Rewrite the integral in terms of \( w \) and use the standard arcsine integral formula: \( \int \frac{dw}{\sqrt{1 - w^2}} = \arcsin(w) + C \). After integrating, substitute back \( w = 2u \) and \( u = r - 1 \) to express the answer in terms of \( r \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration of Functions Involving Square Roots
This concept involves integrating functions that contain square roots, often requiring substitution or recognizing standard integral forms. Understanding how to manipulate the integrand to match known integral formulas is essential for solving these problems.
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Trigonometric Substitution
Trigonometric substitution is a technique used to simplify integrals involving expressions like √(a² - x²), √(x² - a²), or √(a² + x²). By substituting x with a trigonometric function, the integral becomes easier to evaluate using trigonometric identities.
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Definite and Indefinite Integrals
Understanding the difference between definite and indefinite integrals is crucial. Indefinite integrals represent families of functions plus a constant, while definite integrals compute the net area under a curve between limits. This problem involves an indefinite integral, focusing on finding the antiderivative.
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Related Practice
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1
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