Textbook Question
Area functions for linear functions Consider the following functions ƒ and real numbers a (see figure).
(a) Find and graph the area function A (𝓍) = ∫ₐˣ ƒ(t) dt .
ƒ(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 5.2.58a
Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1
(a) ∫₀^π/2 (2 sin θ ― cos θ) dθ
Verified step by step guidance1
Step 1: Recognize that the given integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 can be used to evaluate the new integral by leveraging the linearity property of integrals.
Step 2: Rewrite the new integral ∫₀^π/2 (2 sin θ ― cos θ) dθ as ∫₀^π/2 (―(cos θ ― 2 sin θ)) dθ by factoring out a negative sign.
Step 3: Use the property of integrals that states ∫ₐᵇ c·f(x) dx = c·∫ₐᵇ f(x) dx, where c is a constant. Apply this to factor out the negative sign, resulting in ―∫₀^π/2 (cos θ ― 2 sin θ) dθ.
Step 4: Substitute the value of the given integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 into the expression. This gives ―(―1).
Step 5: Simplify the expression to find the value of the new integral. The result will be the negation of the given integral's value.
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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 symmetry, allow us to manipulate and evaluate integrals more easily. For instance, the linearity property states that the integral of a sum is the sum of the integrals, and constants can be factored out. Understanding these properties is crucial for simplifying complex integrals and relating them to known values.
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Properties of Functions
Definite Integrals
A definite integral represents the signed area under a curve between two limits. In this case, the integral is evaluated from 0 to π/2, which means we are interested in the behavior of the function within this interval. Knowing how to compute definite integrals and interpret their results is essential for solving the given problem.
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05:43Definition of the Definite Integral
Substitution in Integrals
Substitution is a technique used to simplify integrals by changing the variable of integration. This method can help transform a complex integral into a more manageable form. In the context of the given problem, recognizing how to relate the integrals through substitution can lead to an easier evaluation based on the known value of the first integral I.
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Related Practice
Textbook Question
Suppose ƒ is an odd function, ∫₀⁴ ƒ(𝓍) d𝓍 = 3 , and ∫₀⁸ ƒ(𝓍) d𝓍 = 9 .
(a) Evaluate ∫₋₈⁴ ƒ(𝓍) d𝓍 .
Textbook Question
Working with area functions Consider the function ƒ and its graph.
(a) Estimate the zeros of the area function A(𝓍) = ∫₀ˣ ƒ(t) dt , for 0 ≤ 𝓍 ≤ 10 .
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Textbook Question
Area functions The graph of ƒ is shown in the figure. Let A(x) = ∫₋₂ˣ ƒ(t) dt and F(x) = ∫₄ˣ ƒ(t) dt be two area functions for ƒ. Evaluate the following area functions.
(a) A (―2)
Textbook Question
Sigma notation Express the following sums using sigma notation. (Answers are not unique.)
(a) 1 + 2 + 3 + 4 + 5
Textbook Question
Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3t² + 1 on the interval 0 ≤ t ≤ 4, where t is measured in seconds.
(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)