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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.2.58b

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
(b) ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ

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1
Step 1: Recognize that the given integral (b) ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ can be expressed in terms of the original integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1.
Step 2: Factor out the constant multiplier from the integrand in (b). Using the property of integrals, ∫ₐᵇ k * f(x) dx = k * ∫ₐᵇ f(x) dx, rewrite the integral as ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ = 4 * ∫₀^π/2 (cos θ ― 2 sin θ) dθ.
Step 3: Substitute the value of the original integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 into the expression derived in Step 2.
Step 4: Multiply the constant factor (4) by the value of the original integral (―1) to simplify the expression.
Step 5: The result of the multiplication gives the value of the integral (b).

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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, allow us to manipulate integrals in useful ways. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. This is crucial for evaluating integrals efficiently, especially when they can be expressed in terms of known integrals.
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Definite Integrals

Definite integrals represent the signed area under a curve between two limits. In this case, the limits are from 0 to π/2. Understanding how to compute definite integrals and their properties is essential for evaluating integrals like the one given in the question, as it provides the numerical value of the area.
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Definition of the Definite Integral

Substitution in Integrals

Substitution is a technique used to simplify the evaluation of integrals by changing the variable of integration. This method can transform a complex integral into a simpler form, making it easier to compute. Recognizing when and how to apply substitution is key to solving integrals effectively.
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Related Practice
Textbook Question

{Use of Tech} Approximating net area The following functions are positive and negative on the given interval.

ƒ(𝓍) = tan⁻¹ (3x - 1) on [0,1]

(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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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 ƒ, ƒ', and ƒ'' are continuous functions for all real numbers.                                                                                                                                                           

                                                                                                                                                                    

(b) ∫ (ƒ(𝓍))ⁿ ƒ'(𝓍) d𝓍 = 1/(n + 1) (ƒ(𝓍))ⁿ⁺¹ + C , n ≠ ―1 .

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(c) For an increasing or decreasing nonconstant function on an interval [a,b] and a given value of n, the value of the midpoint Riemann sum always lies between the values of the left and right Riemann sums.

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Textbook Question

Substitutions Suppose ƒ is an even function with ∫₀⁸ ƒ(𝓍) d𝓍 = 9 . Evaluate each integral.                                                                                                       

(b) ∫²₋₂ 𝓍²ƒ(𝓍³) d𝓍

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Textbook Question

Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.



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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.                                                                          

                                                                                                                                                                                     (b) Suppose ƒ is a negative increasing function, for 𝓍 > 0 . Then the area function A(𝓍) = ∫₀ˣ ƒ(t) dt is a decreasing function of 𝓍 .

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