Textbook Question
76. Different Substitutions
b. Show that ∫(1/√(x - x²)) dx = 2 sin⁻¹√x + C using substitution u = √x
Ch. 8 - Integration Techniques
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.9.88b
88. Incorrect Calculation
b. Evaluate ∫(from -1 to 1) dx/x or show that the integral does not exist.
Verified step by step guidance1
Identify the integral to evaluate: \(\int_{-1}^{1} \frac{dx}{x}\). Notice that the integrand \(\frac{1}{x}\) is undefined at \(x=0\), which lies within the interval of integration.
Since the function is not defined at \(x=0\), split the integral into two improper integrals at the point of discontinuity: \(\int_{-1}^{1} \frac{dx}{x} = \int_{-1}^{0} \frac{dx}{x} + \int_{0}^{1} \frac{dx}{x}\).
Rewrite each integral as a limit to handle the improper nature: \(\int_{-1}^{0} \frac{dx}{x} = \lim_{t \to 0^-} \int_{-1}^{t} \frac{dx}{x}\) and \(\int_{0}^{1} \frac{dx}{x} = \lim_{s \to 0^+} \int_{s}^{1} \frac{dx}{x}\).
Evaluate the antiderivative of \(\frac{1}{x}\), which is \(\ln|x|\), and apply it to each integral before taking the limits.
Examine the behavior of the limits as \(t \to 0^-\) and \(s \to 0^+\) to determine if the integrals converge or diverge. If either limit diverges, conclude that the original integral does not exist.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Improper Integrals
An improper integral occurs when the integrand is undefined or unbounded within the interval of integration. In this problem, the function 1/x is undefined at x = 0, which lies between -1 and 1, making the integral improper and requiring special evaluation methods.
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Improper Integrals: Infinite Intervals
Cauchy Principal Value
The Cauchy principal value is a method to assign a finite value to certain improper integrals that are otherwise divergent. For ∫ from -1 to 1 of 1/x dx, the principal value considers symmetric limits approaching zero from both sides, potentially yielding a finite result despite the singularity.
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06:37Average Value of a Function
Non-Existence of the Integral
If the limits of the integral do not converge to a finite number due to the singularity, the integral is said not to exist in the usual sense. Since 1/x has an infinite discontinuity at zero, the integral from -1 to 1 diverges unless interpreted as a principal value.
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03:07Cases Where Limits Do Not Exist
Related Practice
Textbook Question
Let L(c) be the length of the parabola f(x) = x² from x = 0 to x = c, where c ≥ 0 is a constant.
b. Is L concave up or concave down on [0, ∞)?
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Textbook Question
63. Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. If m is a positive integer, then ∫[0 to π] sin^m(x) dx = 0.
Textbook Question
The Eiffel Tower Property Let R be the region between the curves y = e^(-c·x) and y = -e^(-c·x) on the interval [a, ∞), where a ≥ 0 and c > 0.
The center of mass of R is located at (x̄, 0), where x̄ = [∫(a to ∞) x·e^(-c·x) dx] / [∫(a to ∞) e^(-c·x) dx]
(The profile of the Eiffel Tower is modeled by these two exponential curves; see the Guided Project ""The exponential Eiffel Tower"")
b. With a = 0 and c = 2, find the equations of the lines tangent to both curves at x = 0
Textbook Question
66–71. {Use of Tech} Estimating error Refer to Theorem 8.1 in the following exercises.
67. Let f(x) = √(x³ + 1).
b. Calculate f''(x).
Textbook Question
68. Different methods
b. Evaluate ∫(cot x csc² x) dx using the substitution u=cscx.