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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.R.1d
Chapter 5, Problem 5.R.1d

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.
(d) If ƒ is continuous on [a,b] and ∫ₐᵇ |ƒ(𝓍)| d𝓍 = 0 , then ƒ(𝓍) = 0 on [a,b] .

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Step 1: Begin by understanding the integral ∫ₐᵇ |ƒ(𝓍)| d𝓍 = 0. This represents the area under the curve of the absolute value of ƒ(𝓍) over the interval [a, b]. If this integral equals zero, it implies that the absolute value of ƒ(𝓍) contributes no area over the interval.
Step 2: Recall that the absolute value function |ƒ(𝓍)| is always non-negative. Therefore, for the integral of |ƒ(𝓍)| to equal zero, the function |ƒ(𝓍)| must be zero everywhere on [a, b].
Step 3: If |ƒ(𝓍)| = 0 for all 𝓍 in [a, b], then ƒ(𝓍) must also equal 0 for all 𝓍 in [a, b]. This is because the absolute value of a number is zero only when the number itself is zero.
Step 4: Conclude that the statement is true. If ƒ is continuous on [a, b] and ∫ₐᵇ |ƒ(𝓍)| d𝓍 = 0, then ƒ(𝓍) = 0 on [a, b]. The reasoning is based on the properties of the absolute value function and the definition of the integral.
Step 5: If needed, provide a counterexample to test the logic. For instance, consider a function ƒ(𝓍) that is non-zero at any point in [a, b]. The integral ∫ₐᵇ |ƒ(𝓍)| d𝓍 would not be zero, confirming the validity of the statement.

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Key Concepts

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

Continuity of Functions

A function is continuous on an interval if there are no breaks, jumps, or holes in its graph. This means that for every point in the interval, the function approaches the same value from both sides. Continuity is crucial for applying the Fundamental Theorem of Calculus, which connects differentiation and integration.
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Intro to Continuity

Definite Integral and Absolute Value

The definite integral of a function over an interval gives the net area under the curve of that function. When considering the integral of the absolute value of a function, ∫ₐᵇ |ƒ(𝓍)| d𝓍 = 0 implies that the function must be zero almost everywhere on that interval, as the absolute value cannot be negative, and the only way for the integral to equal zero is if the function itself is zero.
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Definition of the Definite Integral

Properties of Integrals

Integrals have specific properties that help in understanding the behavior of functions. One important property is that if the integral of a non-negative function over an interval is zero, then the function must be zero at every point in that interval. This property is essential for analyzing the statement in the question regarding the function ƒ and its integral.
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Properties of Functions
Related Practice
Textbook Question

Evaluating integrals Evaluate the following integrals.


∫₋₂² (3𝓍⁴―2𝓍 + 1) d𝓍

Textbook Question

Integration by Riemann sums Consider the integral ∫₁⁴ (3𝓍― 2) d𝓍.


(b) Use summation notation to express the right Riemann sum in terms of a positive integer n .

Textbook Question

Properties of integrals Suppose ∫₁⁴ ƒ(𝓍) d𝓍 = 6 , ∫₁⁴ g(𝓍) d𝓍 = 4 and ∫₃⁴ ƒ(𝓍) d𝓍 = 2 . Evaluate the following integrals or state that there is not enough information.


―∫₄¹ 2ƒ(𝓍) d𝓍

Textbook Question

Evaluating integrals Evaluate the following integrals.


∫π/₆^π/³ (sec² t + csc² t) dt

Textbook Question

Area functions and the Fundamental Theorem Consider the function

ƒ(t) = { t      if  ―2 ≤ t < 0

t²/2    if    0 ≤ t ≤ 2

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(b) Use the Fundamental Theorem to find an expression for F '(𝓍) for ―2 ≤ 𝓍 < 0.

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Textbook Question

Area functions and the Fundamental Theorem Consider the function

ƒ(t) = { t      if  ―2 ≤ t < 0

t²/2    if    0 ≤ t ≤ 2

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(e) Evaluate F ''(―1) and F ''(1). Interpret these values.