Textbook Question
Find the average value of Ζ(π) = eΒ²Λ£ on [0, ln 2] .
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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.R.18
Properties of integrals Suppose β«ββ΄ Ζ(π) dπ = 6 , β«ββ΄ g(π) dπ = 4 and β«ββ΄ Ζ(π) dπ = 2 . Evaluate the following integrals or state that there is not enough information.
ββ«βΒΉ 2Ζ(π) dπ
Verified step by step guidance1
Step 1: Recognize that the integral β«βΒΉ 2Ζ(π) dπ involves reversing the limits of integration. When the limits are reversed, the integral changes sign. Thus, β«βΒΉ 2Ζ(π) dπ = -β«ββ΄ 2Ζ(π) dπ.
Step 2: Use the property of integrals that allows constants to be factored out. Specifically, β«ββ΄ 2Ζ(π) dπ = 2β«ββ΄ Ζ(π) dπ.
Step 3: Substitute the given value of β«ββ΄ Ζ(π) dπ = 6 into the equation from Step 2. This gives β«ββ΄ 2Ζ(π) dπ = 2 Γ 6.
Step 4: Combine the results from Step 1 and Step 3 to express the integral as -β«ββ΄ 2Ζ(π) dπ = -(2 Γ 6).
Step 5: Conclude that the integral β«βΒΉ 2Ζ(π) dπ can be evaluated using the steps above, but the final numerical result is not calculated here as per the instructions.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Properties of Definite Integrals
Definite integrals have several key properties, including linearity and the ability to reverse limits. The linearity property states that β«[a,b] (c * f(x)) dx = c * β«[a,b] f(x) dx for any constant c. Additionally, reversing the limits of integration changes the sign: β«[b,a] f(x) dx = -β«[a,b] f(x) dx. Understanding these properties is essential for evaluating integrals efficiently.
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Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then β«[a,b] f(x) dx = F(b) - F(a). This theorem allows us to evaluate definite integrals by finding the antiderivative, which is crucial for solving integral problems and understanding the relationship between the two operations.
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Substitution in Integrals
Substitution is a technique used in integration to simplify the process of evaluating integrals. It involves changing the variable of integration to make the integral easier to solve. For example, if we let u = g(x), then the integral β« f(g(x)) g'(x) dx can be transformed into β« f(u) du, which may be simpler to evaluate. This concept is particularly useful when dealing with composite functions.
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Related Practice
Textbook Question
Evaluating integrals Evaluate the following integrals.
β« (9πβΈβ7πβΆ) dπ
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Textbook Question
Area functions and the Fundamental Theorem Consider the function
Ζ(t) = { t if β2 β€ t < 0
tΒ²/2 if 0 β€ t β€ 2
and its graph shown below. Let F(π) = β«ββΛ£ Ζ(t) dt and G(π) = β«ββΛ£ Ζ(t) dt.
(e) Evaluate F ''(β1) and F ''(1). Interpret these values.
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Textbook Question
Evaluating integrals Evaluate the following integrals.
β«βΒ² (2π + 1)Β³ dπ
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Textbook Question
Limits with integrals Evaluate the following limits.
lim β«βΛ£ eα΅Β² dt
πβ2 ---------------
π β 2
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Textbook Question
Integration by Riemann sums Consider the integral β«ββ΄ (3πβ 2) dπ.
(a) Evaluate the right Riemann sum for the integral with n = 3 .
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