Textbook Question
Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result.
β«ββ΄ (8β2π) 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 5.3.31
Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus
β«ββΈ 8πΒΉ/Β³ dπ
Verified step by step guidance1
Step 1: Recognize that the integral β«ββΈ 8πΒΉ/Β³ dπ is a definite integral, and we will use the Fundamental Theorem of Calculus to evaluate it. The Fundamental Theorem states that if F'(π) = f(π), then β«βα΅ f(π) dπ = F(b) - F(a).
Step 2: Identify the function to integrate, which is f(π) = 8πΒΉ/Β³. To find the antiderivative, recall the power rule for integration: β«πβΏ dπ = (πβΏβΊΒΉ)/(n+1) + C, where n β -1.
Step 3: Apply the power rule to the term πΒΉ/Β³. The antiderivative of πΒΉ/Β³ is (πβ΄/Β³)/(4/3) = (3/4)πβ΄/Β³. Multiply this by the constant 8 to get the antiderivative of the entire function: F(π) = 8 * (3/4)πβ΄/Β³ = 6πβ΄/Β³.
Step 4: Use the Fundamental Theorem of Calculus to evaluate the definite integral. Substitute the upper limit (π = 8) and the lower limit (π = 1) into the antiderivative F(π). This gives F(8) - F(1), where F(π) = 6πβ΄/Β³.
Step 5: Write the expression for the result: F(8) - F(1) = 6(8β΄/Β³) - 6(1β΄/Β³). Simplify each term separately to find the final value of the definite integral.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation with integration, stating that if a function is continuous on an interval [a, b], then the integral of its derivative over that interval equals the difference in the values of the function at the endpoints. This theorem allows us to evaluate definite integrals by finding an antiderivative of the integrand.
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Fundamental Theorem of Calculus Part 1
Definite Integral
A definite integral represents the signed area under a curve defined by a function over a specific interval [a, b]. It is calculated using the limits of integration, which specify the interval, and provides a numerical value that reflects the accumulation of quantities, such as area, volume, or total change, over that interval.
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Antiderivative
An antiderivative of a function is another function whose derivative is the original function. In the context of the Fundamental Theorem of Calculus, finding the antiderivative is essential for evaluating definite integrals, as it allows us to compute the integral by substituting the limits of integration into the antiderivative and calculating the difference.
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Related Practice
Textbook Question
The linear function Ζ(π) = 3 β π is decreasing on the interval [0, 3]. Is its area function for Ζ (with left endpoint 0) increasing or decreasing on the interval [0, 3]? Draw a picture and explain.
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Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume Ζ and Ζ' are continuous functions for all real numbers.
(c) β«βα΅ Ζ'(π) dπ = Ζ(b) βΖ(a) .
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Textbook Question
Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. Explain why your result is consistent with the figure.
β«βΒΉ (πΒ² β 2π + 3) dπ
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Textbook Question
Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus
β«βββ»ΒΉ πβ»Β³ dπ
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Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume Ζ and Ζ' are continuous functions for all real numbers.
(d) If Ζ is continuous on [a,b] and β«βα΅ |Ζ(π)| dπ = 0 , then Ζ(π) = 0 on [a,b] .
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