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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 5.2.55b
Chapter 5, Problem 5.2.55b

Properties of integrals Consider two functions ƒ and g on [1,6] such that ∫₁⁶ƒ(𝓍) d𝓍 = 10 and ∫₁⁶g(𝓍) d𝓍 = 5, ∫₄⁶ƒ(𝓍) d𝓍 = 5 , and ∫₁⁴g(𝓍) d𝓍 = 2. Evaluate the following integrals.


(b) ∫₁⁶ (f(𝓍) ― g(𝓍)) d𝓍

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1
Step 1: Recall the property of integrals that states ∫ₐᵇ (ƒ(𝓍) ± g(𝓍)) d𝓍 = ∫ₐᵇ ƒ(𝓍) d𝓍 ± ∫ₐᵇ g(𝓍) d𝓍. This allows us to split the integral into two separate integrals.
Step 2: Apply the property to the given integral ∫₁⁶ (ƒ(𝓍) ― g(𝓍)) d𝓍. This becomes ∫₁⁶ ƒ(𝓍) d𝓍 ― ∫₁⁶ g(𝓍) d𝓍.
Step 3: Substitute the given values for the integrals: ∫₁⁶ ƒ(𝓍) d𝓍 = 10 and ∫₁⁶ g(𝓍) d𝓍 = 5.
Step 4: Perform the subtraction operation symbolically: 10 ― 5.
Step 5: The result of the subtraction gives the value of the integral ∫₁⁶ (ƒ(𝓍) ― g(𝓍)) d𝓍. This completes the evaluation process.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Properties of Definite Integrals

Definite integrals have several key properties, including linearity, which states that the integral of a sum of functions is the sum of their integrals. This means that for any two functions f and g, ∫(f + g) dx = ∫f dx + ∫g dx. Additionally, the integral of a constant multiplied by a function can be factored out: ∫k * f dx = k * ∫f dx. Understanding these properties is essential for evaluating integrals involving combinations of functions.
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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then ∫ₐᵇ f(x) dx = F(b) - F(a). This theorem allows us to evaluate definite integrals by finding the antiderivative of the integrand. It is crucial for solving problems involving definite integrals, as it provides a method to compute the area under a curve.
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Substitution in Integrals

Substitution is a technique used in integration to simplify the process of finding integrals. It involves changing the variable of integration to make the integral easier to evaluate. For example, if we let u = g(x), then the integral ∫f(g(x))g'(x)dx can be transformed into ∫f(u)du. This method is particularly useful when dealing with composite functions or when the integrand can be expressed in a simpler form.
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Related Practice
Textbook Question

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. 

I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4

(b) ∫₁⁰ (2𝓍―𝓍³) d𝓍

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Textbook Question

Use Table 5.6 to evaluate the following indefinite integrals.                                                                                                               

                                                                                                                                                                  

 (b) ∫ sec 5𝓍 tan 5𝓍 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

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


f(x) = sin 2x on [0,3π/4]


(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

Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.

(b) ∫₃⁶ (―3g(𝓍)) d𝓍

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Textbook Question

Working with area functions Consider the function ƒ and the points a, b, and c.

(b) Graph ƒ and A.

ƒ(𝓍) = 1/𝓍 ; a = 1 , b = 4 , c = 6

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