Textbook Question
Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus. Sketch the graph of the integrand and shade the region whose net area you have found.
β«ββ΅ (πΒ²β9) dπ
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 5.2.67
Use geometry and properties of integrals to evaluate
β«βΒΉ (2π + β(1βπΒ²) + 1) dπ
Verified step by step guidance1
Step 1: Break the integral into separate terms using the property of integrals: β«βα΅ [f(π) + g(π) + h(π)] dπ = β«βα΅ f(π) dπ + β«βα΅ g(π) dπ + β«βα΅ h(π) dπ. This gives β«βΒΉ (2π + β(1βπΒ²) + 1) dπ = β«βΒΉ 2π dπ + β«βΒΉ β(1βπΒ²) dπ + β«βΒΉ 1 dπ.
Step 2: Evaluate β«βΒΉ 2π dπ. Use the power rule for integration: β« πβΏ dπ = (πβΏβΊΒΉ)/(n+1) + C. Here, n = 1, so β« 2π dπ = 2 * (πΒ²/2) = πΒ². Apply the limits of integration from 0 to 1.
Step 3: Evaluate β«βΒΉ β(1βπΒ²) dπ. Recognize that β(1βπΒ²) represents the equation of a semicircle with radius 1 centered at the origin. The integral β«βΒΉ β(1βπΒ²) dπ calculates the area of one-quarter of the circle. Use the formula for the area of a circle, A = ΟrΒ², and divide by 4.
Step 4: Evaluate β«βΒΉ 1 dπ. The integral of a constant c over [a, b] is given by c(bβa). Here, c = 1, a = 0, and b = 1, so β«βΒΉ 1 dπ = 1 * (1β0).
Step 5: Combine the results from Steps 2, 3, and 4. Add the values obtained for β«βΒΉ 2π dπ, β«βΒΉ β(1βπΒ²) dπ, and β«βΒΉ 1 dπ to get the final result of the integral.
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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«_a^b f(x) dx, where 'a' and 'b' are the limits of integration. The value of a definite integral can be interpreted geometrically as the accumulation of the area between the function f(x) and the x-axis from x = a to x = b.
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Properties of Integrals
Properties of integrals, such as linearity, allow us to break down complex integrals into simpler parts. For example, β«(f(x) + g(x)) dx = β«f(x) dx + β«g(x) dx. This property is particularly useful when evaluating integrals that consist of multiple terms, as it enables the evaluation of each term separately.
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Geometric Interpretation of Functions
Understanding the geometric interpretation of functions is crucial for evaluating integrals. For instance, the term β(1 - xΒ²) represents a semicircle with radius 1. Recognizing the shapes formed by the functions involved can simplify the evaluation of the integral by allowing the use of geometric area formulas instead of purely algebraic methods.
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Related Practice
Textbook Question
Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.
β« πeΛ£Β² dπ
Textbook Question
Areas of regions Find the area of the following regions.
The region bounded by the graph of Ζ(π) = x /β(πΒ² β9) and the π-axis between and π = 4 and π= 5
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Textbook Question
Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.
β« 2π(πΒ² β 1)βΉβΉ dπ
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β«βαΆ Ζ(π) dπ
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Textbook Question
Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.
β« πΒ³ (πβ΄ + 16)βΆ dπ