Textbook Question
Area functions for linear functions Consider the following functions Ζ and real numbers a (see figure).
(a) Find and graph the area function A (π) = β«βΛ£ Ζ(t) dt .
Ζ(t) = 2t + 5 , 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 5.2.57a
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
(a) β«βΒΉ (4πβ2πΒ³) dπ
Verified step by step guidance1
Step 1: Recognize that the given integral in part (a), β«βΒΉ (4π β 2πΒ³) dπ, can be split into two separate integrals using the linearity property of integrals: β«βΒΉ (4π β 2πΒ³) dπ = β«βΒΉ 4π dπ β β«βΒΉ 2πΒ³ dπ.
Step 2: Factor out constants from each integral. For the first term, β«βΒΉ 4π dπ becomes 4β«βΒΉ π dπ. For the second term, β«βΒΉ 2πΒ³ dπ becomes 2β«βΒΉ πΒ³ dπ.
Step 3: Use the given value of I = β«βΒΉ (πΒ³ β 2π) dπ = β3/4 to extract the value of β«βΒΉ πΒ³ dπ. Rewrite I as β«βΒΉ πΒ³ dπ β β«βΒΉ 2π dπ = β3/4. This equation can be solved to find β«βΒΉ πΒ³ dπ.
Step 4: Compute β«βΒΉ π dπ using the power rule for integration. The integral of π with respect to π is (πΒ²)/2. Evaluate this from 0 to 1 to find the value of β«βΒΉ π dπ.
Step 5: Substitute the values of β«βΒΉ πΒ³ dπ and β«βΒΉ π dπ into the expression for β«βΒΉ (4π β 2πΒ³) dπ = 4β«βΒΉ π dπ β 2β«βΒΉ πΒ³ dπ to compute the final result.
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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 ability to split integrals, are fundamental in calculus. Linearity allows us to factor constants out of integrals and combine integrals of the same limits. This means that if we have an integral of a sum, we can separate it into the sum of integrals, which simplifies calculations.
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Properties of Functions
Definite Integrals
Definite integrals represent the signed area under a curve between two points on the x-axis. The notation β«βα΅ f(x) dx indicates the integral of the function f(x) from a to b. The value of a definite integral can be interpreted as the accumulation of quantities, which is essential for evaluating integrals over specific intervals.
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05:43Definition of the Definite Integral
Integration Techniques
Various techniques exist for evaluating integrals, including substitution and integration by parts. In this context, recognizing patterns in the integrand can help simplify the integral. For example, if an integral can be expressed in terms of known integrals, such as the one provided (I), it can be evaluated more easily by leveraging previously calculated values.
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Related Practice
Textbook Question
Mass from density A thin 10-cm rod is made of an alloy whose density varies along its length according to the function shown in the figure. Assume density is measured in units of g/cm. In Chapter 6, we show that the mass of the rod is the area under the density curve.
(a) Find the mass of the left half of the rod (0 β€ x β€ 5) .
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Textbook Question
Approximating areas Estimate the area of the region bounded by the graph of Ζ(π) = xΒ² + 2 and the x-axis on [0, 2] in the following ways.
(a) Divide [0, 2] into n = 4 subintervals and approximate the area of the region using a left Riemann sum. Illustrate the solution geometrically.
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Textbook Question
Use Table 5.6 to evaluate the following indefinite integrals.
(a) β« eΒΉβ°Λ£ dπ
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Textbook Question
Area functions The graph of Ζ is shown in the figure. Let A(x) = β«βΛ£ Ζ(t) dt and F(x) = β«βΛ£ Ζ(t) dt be two area functions for Ζ. Evaluate the following area functions.
(a) A(2)
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Textbook Question
Area functions for the same linear function Let Ζ(t) = 2t β 2 and consider the two area functions A (π) = β«βΛ£ Ζ(t) dt and F(π) = β«βΛ£ Ζ(t) dt .
(a) Evaluate A (2) and A (3). Then use geometry to find an expression for A (π) , for π β₯ 1 .
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