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
(b) β«ββ° (2πβπΒ³) 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 5.2.51b
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.
(b) β«ββ΄ π(π β 4) d(π)
Verified step by step guidance1
Step 1: Begin by analyzing the given integral β«ββ΄ π(π β 4) dπ. Notice that the integrand π(π β 4) can be rewritten as a product of terms. Expand the expression π(π β 4) to simplify it into a polynomial form.
Step 2: Expand the integrand: π(π β 4) = πΒ² β 4π. This simplifies the integral to β«ββ΄ (πΒ² β 4π) dπ.
Step 3: Use the linearity property of integrals to split the integral into two separate integrals: β«ββ΄ (πΒ² β 4π) dπ = β«ββ΄ πΒ² dπ β β«ββ΄ 4π dπ.
Step 4: Factor out constants where applicable. For the second term, factor out the constant 4: β«ββ΄ πΒ² dπ β 4β«ββ΄ π dπ.
Step 5: Evaluate each integral using the definitions and properties of integrals. Recall that β«ββ΄ 3π(4 β π) dπ = 32 is given, and use this information to relate the results if necessary. Apply the power rule for integration to compute β«ββ΄ πΒ² dπ and β«ββ΄ π dπ.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral represents the signed area under a curve between two points on the x-axis. 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 [a, b]. Understanding how to evaluate definite integrals is crucial for solving problems involving areas and accumulated quantities.
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Properties of Integrals
The properties of integrals, such as linearity, additivity, and the ability to change variables, are essential for simplifying and evaluating integrals. For instance, the linearity property states that β«(c * f(x)) dx = c * β«f(x) dx for a constant 'c'. Additionally, the additivity property allows us to split integrals over adjacent intervals, which can be useful in evaluating more complex integrals by breaking them down into simpler parts.
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Integration by Substitution
Integration by substitution is a technique used to simplify the process of evaluating integrals by changing the variable of integration. This method involves substituting a new variable for a function of the original variable, which can make the integral easier to solve. It is particularly useful when dealing with composite functions or when the integrand can be expressed in a simpler form through substitution.
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