Textbook Question
Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = β«βΒΉ (πΒ³ β 2π) dπ = β3/4
(a) β«βΒΉ (4πβ2πΒ³) dπ
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Ch. 5 - Integration
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 5, Problem 5.2.51a
Properties of integrals Use only the fact that β«ββ΄ 3π (4 βπ) dπ = 32, and the definitions and properties of integrals, to evaluate the following integrals, if possible.
(a) β«ββ° 3π(4 β π) d(π)
Verified step by step guidance1
Step 1: Recognize the integral given in part (a) is the same as the integral provided in the problem, except the limits of integration are reversed. The integral provided is β«ββ΄ 3π(4 β π) dπ = 32.
Step 2: Recall the property of definite integrals: reversing the limits of integration changes the sign of the integral. Mathematically, β«βα΅ f(π) dπ = -β«α΅β f(π) dπ.
Step 3: Apply this property to the integral in part (a). Since the limits are reversed (from 4 to 0 instead of 0 to 4), the integral becomes -β«ββ΄ 3π(4 β π) dπ.
Step 4: Substitute the value of the original integral, which is given as 32. Therefore, the integral in part (a) becomes -32.
Step 5: Conclude that the integral β«ββ° 3π(4 β π) dπ evaluates to -32 based on the properties of integrals and the given information.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integral
A definite integral represents the signed area under a curve defined by a function over a specific interval. It is denoted as β«βα΅ f(x) dx, where 'a' and 'b' are the limits of integration. The value of a definite integral can be interpreted as the accumulation of quantities, such as area, over the interval from 'a' to 'b'.
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Properties of Integrals
The properties of integrals include linearity, which allows for the integration of sums and scalar multiples, and the reversal of limits, which states that β«βα΅ f(x) dx = -β«α΅β f(x) dx. These properties enable the evaluation of integrals by transforming them into simpler forms or by changing the limits of integration.
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Substitution in Integrals
Substitution is a technique used in integration to simplify the integrand by changing variables. This method involves selecting a new variable that simplifies the integral, allowing for easier computation. For example, if u = g(x), then dx can be expressed in terms of du, transforming the integral into a more manageable form.
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Related Practice
Textbook Question
Area functions for linear functions Consider the following functions Ζ and real numbers a (see figure).
(a) Find and graph the area function A (π) = β«βΛ£ Ζ(t) dt .
Ζ(t) = 2t + 5 , a = 0
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Textbook Question
Area functions for linear functions Consider the following functions Ζ and real numbers a (see figure).
(a) Find and graph the area function A (π) = β«βΛ£ Ζ(t) dt .
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Ζ(t) = 4t + 2 , a = 0
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Textbook Question
Substitutions Suppose Ζ is an even function with β«ββΈ Ζ(π) dπ = 9 . Evaluate each integral.
(a) β«ΒΉββ πΖ(πΒ²) dπ
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Textbook Question
Area functions The graph of Ζ is shown in the figure. Let A(x) = β«ββΛ£ Ζ(t) dt and F(x) = β«βΛ£ Ζ(t) dt be two area functions for Ζ. Evaluate the following area functions.
(a) A (β2)
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Textbook Question
Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = β«β^Ο/2 (cos ΞΈ β 2 sin ΞΈ) dΞΈ = β1
(a) β«β^Ο/2 (2 sin ΞΈ β cos ΞΈ) dΞΈ
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