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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 5.1.15a

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)

검증된 단계별 안내
1
Divide the interval [0, 4] into 4 subintervals: [0, 1], [1, 2], [2, 3], and [3, 4]. The length of each subinterval is Δt = 1 second.
Find the midpoint of each subinterval: For [0, 1], the midpoint is t = 0.5; for [1, 2], the midpoint is t = 1.5; for [2, 3], the midpoint is t = 2.5; and for [3, 4], the midpoint is t = 3.5.
Evaluate the velocity function v(t) = 3t² + 1 at each midpoint: v(0.5), v(1.5), v(2.5), and v(3.5).
Approximate the displacement on each subinterval by multiplying the velocity at the midpoint by the subinterval length Δt. For example, the displacement on [0, 1] is approximately v(0.5) * Δt.
Add the displacements from all subintervals to estimate the total displacement of the object on [0, 4].

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
4m

주요 개념

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

Velocity Function

The velocity function describes how the speed of an object changes over time. In this case, the function v(t) = 3t² + 1 indicates that the velocity increases quadratically as time progresses. Understanding this function is crucial for determining the object's speed at any given moment within the specified interval.
추천 영상:
가이드 코스
10:17
Using The Velocity Function

Midpoint Rule

The Midpoint Rule is a numerical method used to approximate the integral of a function. By evaluating the function at the midpoint of each subinterval, we can estimate the area under the curve, which in this context represents the displacement of the object. This method provides a more accurate approximation than using the endpoints of the intervals.
추천 영상:
5:50
Power Rules

Displacement

Displacement refers to the change in position of an object over a specific time interval. It can be calculated by integrating the velocity function over that interval. In this problem, estimating displacement involves summing the areas of rectangles formed by the velocity at midpoints of the subintervals, which approximates the total distance traveled by the object.
추천 영상:
가이드 코스
10:17
Using The Velocity Function
관련 실천
교과서 질문

{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.

(a) Write the midpoint Riemann sum in sigma notation for an arbitrary value of n.


∫₀⁴ (4𝓍― 𝓍²) d𝓍

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교과서 질문

Zero net area Consider the function ƒ(𝓍) = 𝓍² ― 4𝓍 .

(a) Graph ƒ on the interval 𝓍 ≥ 0.

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교과서 질문

Suppose ƒ is an odd function, ∫₀⁴ ƒ(𝓍) d𝓍 = 3 , and ∫₀⁸ ƒ(𝓍) d𝓍 = 9 .


(a) Evaluate ∫₋₈⁴ ƒ(𝓍) d𝓍 .

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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) = cos 𝓍 ; a = 0 , b = π/2 , c = π

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교과서 질문

{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.

(a) Write the midpoint Riemann sum in sigma notation for an arbitrary value of n.


∫₁⁴ 2√𝓍 d𝓍

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교과서 질문

Sigma notation Evaluate the following expressions.

(a)    10                                                                                                                                                                               

       ∑ κ                                                                                                                                                                          

       κ=1                         

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