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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
5장, 문제 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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1
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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
가이드 코스
05:50
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.
추천 영상:
05:06
Finding Area When Bounds Are Not Given

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.
추천 영상:
가이드 코스
06:11
Fundamental Theorem of Calculus Part 1
관련 실천
교과서 질문

Does a right Riemann sum underestimate or overestimate the area of the region under the graph of a function that is positive and decreasing on an interval [a,b]? Explain.

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교과서 질문

Evaluate ∫₀² 3𝓍² d𝓍 and ∫₋₂² 3𝓍² d𝓍. 

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교과서 질문

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 ∫ 2 / (𝓍√4𝓍² ―1) d𝓍 , 𝓍 > ½ 

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교과서 질문

Use symmetry to explain why.

∫⁴₋₄ (5𝓍⁴ + 3𝓍³ + 2𝓍² + 𝓍 + 1) d𝓍 = 2 ∫₀⁴ (5𝓍⁴ + 2𝓍² + 𝓍 + 1) d𝓍 .

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교과서 질문

{Use of Tech} Sigma notation for Riemann sums Use sigma notation to write the following Riemann sums. Then evaluate each Riemann sum using Theorem 5.1 or a calculator.

The midpoint Riemann sum for f(x) = x³ on [3,11] with n = 32.

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교과서 질문

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

                                                                                                                                                                              

 ∫₁/₃^¹/√³ 4/(9𝓍² + 1) d𝓍

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