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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.3.88b

Working with area functions Consider the function ƒ and its graph.
(b) Estimate the points (if any) at which A has a local maximum or minimum.


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Guida verificata passo dopo passo
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Step 1: Understand the problem. The graph provided represents the function ƒ(t), and the task is to estimate the points at which the area function A(t) has a local maximum or minimum. The area function A(t) is defined as the integral of ƒ(t) from a fixed point to t.
Step 2: Recall that the derivative of the area function A(t) is equal to ƒ(t). This means that A'(t) = ƒ(t). To find local maxima or minima of A(t), we need to analyze where ƒ(t) changes sign, as this indicates critical points.
Step 3: Observe the graph of ƒ(t). The function ƒ(t) is a straight line with a positive slope throughout the interval shown (from t = 0 to t = 10). Since ƒ(t) does not change sign (it remains positive), there are no points where A(t) has a local maximum or minimum.
Step 4: Consider the behavior of A(t). Since ƒ(t) is always positive, the area function A(t) is continuously increasing over the interval. This means A(t) does not have any local maxima or minima within the interval.
Step 5: Conclude that the area function A(t) has no local maximum or minimum points in the interval shown because the graph of ƒ(t) does not change sign and remains positive throughout.

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Local Maximum and Minimum

A local maximum is a point where a function's value is higher than the values of the function at nearby points, while a local minimum is where the function's value is lower than those nearby. These points are critical for understanding the behavior of functions and can be found using the first derivative test, which involves analyzing where the derivative changes sign.
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The Second Derivative Test: Finding Local Extrema

Area Function

An area function typically represents the accumulation of quantities, such as the area under a curve. In calculus, it is often defined as the integral of a function over a specified interval. Understanding area functions is crucial for solving problems related to optimization and finding local extrema, as they relate to the fundamental theorem of calculus.
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Finding Area When Bounds Are Not Given

Graph Interpretation

Interpreting the graph of a function involves analyzing its shape, slopes, and intercepts to understand its behavior. In this context, the graph of ƒ(t) is linear, indicating a constant rate of change. This information is essential for estimating local maxima and minima, as the absence of curvature suggests that there are no local extrema in the given interval.
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Percorso guidato
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Graphing The Derivative
Pratica correlata
Domanda del libro di testo

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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Domanda del libro di testo

Working with area functions Consider the function ƒ and its graph.

(b) Estimate the points (if any) at which A has a local maximum or minimum.


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Domanda del libro di testo

{Use of Tech} Approximating definite integrals with a calculator Consider the following definite integrals.

(b) Evaluate each sum using a calculator with n = 20, 50, and 100. Use these values to estimate the value of the integral.


∫₀¹ cos ⁻¹ 𝓍 d𝓍

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Domanda del libro di testo

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

(b) If ƒ is a linear function on the interval [a,b] , then a midpoint Riemann sums give the exact value of ∫ₐᵇ ƒ(𝓍) d𝓍, for any positive integer n. 

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Domanda del libro di testo

Area functions for constant functions Consider the following functions ƒ and real numbers a (see figure).

(b) Verify that .A'(𝓍) = ƒ(𝓍)

                                                                                                                                                            

ƒ(t) = 5 , a = -5

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Domanda del libro di testo

Generalizing the Mean Value Theorem for Integrals Suppose ƒ and g are continuous on [a, b] and let h(𝓍) = (𝓍―b) ∫ₐˣ ƒ(t) dt + (𝓍―a) ∫ₓᵇg(t)dt.                                                                                                                                                                                                                                                                                                                                

(b) Show that there is a number c in (a, b) such that ∫ₐᶜ ƒ(t) dt = ƒ(c) (b ― c)                                                                                                              

                                                                                                                                                                                

(Source: The College Mathematics Journal, 33, 5, Nov 2002)

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