Textbook Question
7–64. Integration review Evaluate the following integrals.
51. ∫ from -1 to 0 of x / (x² + 2x + 2) dx
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 8.9.13
7–58. Improper integrals Evaluate the following integrals or state that they diverge.
13. ∫ (from 0 to ∞) cos x dx
Verified step by step guidance1
Recognize that the integral \( \int_0^{\infty} \cos x \, dx \) is an improper integral because the upper limit of integration is infinite.
Rewrite the integral as a limit: \( \lim_{t \to \infty} \int_0^t \cos x \, dx \). This allows us to evaluate the integral over a finite interval first and then analyze the behavior as \( t \) approaches infinity.
Find the antiderivative of \( \cos x \), which is \( \sin x \). So, \( \int_0^t \cos x \, dx = \sin t - \sin 0 = \sin t \).
Evaluate the limit \( \lim_{t \to \infty} \sin t \). Since \( \sin t \) oscillates between -1 and 1 and does not approach a single value, this limit does not exist.
Conclude that because the limit does not exist, the improper integral \( \int_0^{\infty} \cos x \, dx \) diverges.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Improper Integrals
Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, limits are used to define the integral as a limit of definite integrals over finite intervals. Determining convergence or divergence is essential.
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Improper Integrals: Infinite Intervals
Convergence and Divergence of Integrals
An integral converges if its limit exists and is finite; otherwise, it diverges. For improper integrals over infinite intervals, convergence depends on the behavior of the integrand as the variable approaches infinity. Testing convergence often involves comparison or limit tests.
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05:44Divergence Test (nth Term Test)
Behavior of Oscillatory Functions in Integrals
Functions like cosine oscillate indefinitely without settling to a limit. When integrated over infinite intervals, their oscillations can cause the integral to fail to converge. Understanding how oscillatory behavior affects the integral's limit is crucial for evaluating such integrals.
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05:11Integrals of General Exponential Functions
Related Practice
Textbook Question
23-64. Integration Evaluate the following integrals.
23. ∫ [3 / ((x - 1)(x + 2))] dx
Textbook Question
7–84. Evaluate the following integrals.
30. ∫ from 5/2 to 5√3/2 [1 / (v² √(25 - v²))] dv
Textbook Question
7–84. Evaluate the following integrals.
22. ∫ [1 / ((x - a)(x - b))] dx, where a ≠ b
2
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Textbook Question
48. Integral of sec³x Use integration by parts to show that:
∫ sec³x dx = (1/2) secx tanx + (1/2) ∫ secx dx
Textbook Question
7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
40. ∫ (e³ᵗ / √(4 + e²ᵗ)) dt