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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 5.3.94a

Working with area functions Consider the function ƒ and the points a, b, and c.
(a) Find the area function A (𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem.
ƒ(𝓍) = sin 𝓍 ; a = 0 , b = π/2 , c = π

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Step 1: Understand the problem. You are tasked with finding the area function A(𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem of Calculus. The given function is ƒ(𝓍) = sin(𝓍), and the lower limit of integration is a = 0.
Step 2: Recall the Fundamental Theorem of Calculus. It states that if ƒ(t) is continuous on [a, x], then the derivative of the area function A(𝓍) with respect to 𝓍 is equal to ƒ(𝓍). In other words, A'(𝓍) = ƒ(𝓍).
Step 3: To find A(𝓍), integrate ƒ(t) = sin(t) with respect to t from the lower limit a = 0 to the upper limit x. The integral of sin(t) is -cos(t). Therefore, A(𝓍) = -cos(t) evaluated from t = 0 to t = x.
Step 4: Apply the limits of integration. Substitute the upper limit x and the lower limit 0 into the expression for A(𝓍). This gives A(𝓍) = [-cos(x)] - [-cos(0)]. Simplify the expression using the fact that cos(0) = 1.
Step 5: The area function A(𝓍) is now expressed in terms of x. You can use this function to evaluate the area for specific values of x, such as b = π/2 and c = π, by substituting these values into A(𝓍).

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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주요 개념

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

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration. It states that if a function is continuous on an interval, then the integral of its derivative over that interval can be computed using the values of the function at the endpoints. This theorem allows us to evaluate definite integrals and find area functions effectively.
추천 영상:
가이드 코스
06:11
Fundamental Theorem of Calculus Part 1

Definite Integral

A definite integral represents the signed area under a curve defined by a function over a specific interval [a, b]. It is calculated as the limit of Riemann sums and provides a numerical value that corresponds to the total accumulation of the function's values between the two bounds. In this context, it is used to find the area function A(x) from the function f(t).
추천 영상:
가이드 코스
05:43
Definition of the Definite Integral

Area Function

An area function A(x) is defined as the integral of a function f(t) from a constant lower limit a to a variable upper limit x. It represents the accumulated area under the curve of f(t) from a to x. In this problem, A(x) = ∫ₐˣ f(t) dt will help us determine the area under the curve of sin(x) from 0 to x, which is essential for understanding the behavior of the function over the specified interval.
추천 영상:
05:06
Finding Area When Bounds Are Not Given
관련 실천
교과서 질문

Displacement from a velocity graph Consider the velocity function for an object moving along a line (see figure).

(a) Describe the motion of the object over the interval [0,6].

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교과서 질문

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ, ƒ', and ƒ'' are continuous functions for all real numbers.                                                                                                                                                           

                                                                                                                                                                    

(a) ∫ ƒ(𝓍) ƒ'(𝓍) d𝓍 = ½ (ƒ(𝓍))² + C.

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교과서 질문

The velocity in ft/s of an object moving along a line is given by v = ƒ(t) on the interval 0 ≤ t ≤ 6 (see figure), where t is measured in seconds.


(a) Divide the interval [0,6] into n = 3 subintervals, [0,2] , [2,4] and [4,6]. On each subinterval, assume the object moves at a constant velocity equal to the value of v evaluated at the right endpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0,6] (see part (a) of the figure)                                                                                                             

                                                                                                                                                                                                

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

(a) Consider the linear function ƒ(𝓍) = 2x + 5 and the region bounded by its graph and the x-axis on the interval [3,6]. Suppose the area of this region is approximated using midpoint Riemann sums. Then the approximations give the exact area of the region for any number of subintervals.

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교과서 질문

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

(a) Find and graph the area function A(𝓍) = ∫ₐˣ ƒ(t) dt for ƒ.

ƒ(t) = 5 , a = 0

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교과서 질문

Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3t² + 1 on the interval 0 ≤ t ≤ 4, where t is measured in seconds.

(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)

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