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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.2.58a
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

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1
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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Definition 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

Area functions for the same linear function Let ƒ(t) = t and consider the two area functions A(𝓍) = ∫₀ˣ ƒ(t) dt and F(𝓍) = ∫₂ˣ ƒ(t) dt .

(b) Evaluate F(4) and F(6). Then use geometry to find an expression for F (𝓍) , for 𝓍 ≥ 2. 

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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 .                                                                                                                               

                                                                                                                                                                               

 <IMAGE>                                                                                                                                                                                                           

                                                                                                                                                                                     

 ƒ(t) = 4t + 2 , a = 0

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Textbook Question

Suppose ƒ is an even function and ∫⁸₋₈ ƒ(𝓍) d𝓍 = 18

(b) Evaluate ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍 .

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Textbook Question

Bounds on an integral Suppose ƒ is continuous on [a, b] with ƒ''(𝓍) > 0 on the interval. It can be shown that (b―a) ƒ [(a + b) /2] ≤ ∫ₐᵇ ƒ(𝓍) d𝓍 ≤ (b―a) [ (ƒ(a) + ƒ(b)) /2]                                                         

                                                                                                                                                                               

(a) Assuming ƒ is nonnegative on [a, b], draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b. 

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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

Sigma notation Express the following sums using sigma notation. (Answers are not unique.)

(a) 1 + 2 + 3 + 4 + 5

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