Textbook Question
Area functions for linear functions Consider the following functions Ζ and real numbers a (see figure).
b) Verify that A'(π) = Ζ(π).
Ζ(t) = 4t + 2 , a = 0
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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 5.2.57b
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
(b) β«ββ° (2πβπΒ³) dπ
Verified step by step guidance1
Step 1: Recognize that the integral given in part (b) is the reverse of the integral provided in part (I). Specifically, β«ββ° (2π β πΒ³) dπ is the same as β«βΒΉ (πΒ³ β 2π) dπ, but with the limits of integration swapped.
Step 2: Use the property of integrals that states swapping 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 given integral. Since β«βΒΉ (πΒ³ β 2π) dπ = β3/4, swapping the limits gives β«ββ° (πΒ³ β 2π) dπ = 3/4.
Step 4: Notice that the integrand in part (b) is written as (2π β πΒ³), which is the negative of (πΒ³ β 2π). Therefore, β«ββ° (2π β πΒ³) dπ = -β«ββ° (πΒ³ β 2π) dπ.
Step 5: Substitute the value of β«ββ° (πΒ³ β 2π) dπ from Step 3 into the equation from Step 4. This gives β«ββ° (2π β πΒ³) dπ = -3/4.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Properties of Integrals
The properties of integrals, such as linearity and the additive property, allow us to manipulate and evaluate integrals more easily. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. Understanding these properties is essential for simplifying complex integrals and relating them to known values.
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Definite Integrals
A definite integral represents the signed area under a curve between two limits. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. Knowing how to evaluate definite integrals is crucial for solving problems that involve finding the total accumulation of quantities over an interval.
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Substitution Method
The substitution method is a technique used to simplify the evaluation of integrals by changing variables. This method involves substituting a part of the integral with a new variable, which can make the integral easier to solve. Mastery of this technique is important for tackling integrals that are not straightforward or that involve complex expressions.
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Related Practice
Textbook Question
Suppose Ζ is an even function and β«βΈββ Ζ(π) dπ = 18
(b) Evaluate β«βββΈ πΖ(π) dπ .
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Textbook Question
{Use of Tech} Functions defined by integrals Consider the function g, which is given in terms of a definite integral with a variable upper limit.
b) Calculate g'(π)
g(π) = β«βΛ£ sinΒ² t dt
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Textbook Question
Area functions for linear functions Consider the following functions Ζ and real numbers a (see figure).
(b) Verify that A'(π) = Ζ(π).
Ζ(t) = 3t + 1 , a = 2
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Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
(b) A left Riemann sum always overestimates the area of a region bounded by a positive increasing function and the x-axis on an interval [a,b].
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Textbook Question
{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.
(b) Evaluate each sum using a calculator with n = 20, 50, and 100. Use these values to estimate the value of the integral.
β«ββ΄ (4πβ πΒ²) dπ
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