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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 5.3.2

Suppose F is an antiderivative of ƒ and A is an area function of ƒ. What is the relationship between F and A?

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Understand the definitions: An antiderivative F of a function ƒ is a function such that the derivative of F is equal to ƒ, i.e., F'=ƒ. An area function A of ƒ represents the accumulated area under the curve of ƒ from a fixed point to a variable point x.
Recall the Fundamental Theorem of Calculus: It states that if A(x) is the area function of ƒ, then A'(x) = ƒ(x). This means the derivative of the area function is the original function ƒ.
Recognize the connection: Since F is an antiderivative of ƒ, and A'(x) = ƒ(x), it follows that A(x) and F(x) differ by a constant. Specifically, A(x)=F(x)+C, where C is a constant.
Interpret the constant C: The constant C depends on the choice of the lower limit of integration in the area function A(x). If the lower limit is changed, the value of C will adjust accordingly.
Summarize the relationship: The area function A(x) is essentially an antiderivative of ƒ, but it includes a constant term that depends on the lower limit of integration. Both F and A are closely related through this constant adjustment.

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Key Concepts

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

Antiderivative

An antiderivative of a function f is another function F such that the derivative of F is equal to f, i.e., F' = f. This means that F represents a family of functions whose slopes at any point correspond to the values of f. Antiderivatives are essential in calculus for solving problems related to integration and finding areas under curves.
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Antiderivatives

Area Function

An area function A associated with a function f typically represents the accumulated area under the curve of f from a specific point to a variable endpoint. Mathematically, it is defined as A(x) = ∫[a to x] f(t) dt, where a is a constant. The area function is crucial for understanding how the total area changes as the endpoint varies, linking it to the concept of integration.
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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 the integral of f from a to b can be computed as F(b) - F(a). This theorem establishes that the area function A is directly related to the antiderivative F, as A(x) = F(x) - F(a), illustrating the deep relationship between these concepts.
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Related Practice
Textbook Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.                                                                                                                         

                                                                                                                                                                              

 ∫π/₄^π/² (cos 𝓍) / (sin² 𝓍) d𝓍

Textbook Question

Evaluate ∫₀² 3𝓍² d𝓍 and ∫₋₂² 3𝓍² d𝓍. 

Textbook Question

Area functions from graphs The graph of ƒ is given in the figure. A(𝓍) = ∫₀ˣ ƒ(t) dt and evaluate A(2), A(5), A(8), and A(12).


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Textbook Question

Integrals with sin² 𝓍 and cos² 𝓍 Evaluate the following integrals.                                                                                                             

                                                                                                                                                                    

 ∫₋π^π cos² 𝓍 d𝓍