Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.1.3a

Chapter 5, Problem 5.1.3a

The velocity in ft/s of an object moving along a line is given by v = ƒ(t) on the interval 0 ≤ t ≤ 8 (see figure), where t is measured in seconds.
a) Divide the interval [0,8] into n = 2 subintervals, [0,4] and [4,8]. On each subinterval, assume the object moves at a constant velocity equal to the value of v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0,8] (see part (a) of the figure)

Verified step by step guidance

Step 1: Divide the interval [0,8] into two subintervals: [0,4] and [4,8]. These subintervals are determined based on the problem's requirement to use n = 2 subintervals.

Step 2: Identify the midpoints of each subinterval. For [0,4], the midpoint is t = 2, and for [4,8], the midpoint is t = 6.

Step 3: Evaluate the velocity function v = ƒ(t) at the midpoints. From the graph, v(2) ≈ 40 ft/s and v(6) ≈ 60 ft/s.

Step 4: Calculate the displacement for each subinterval using the formula displacement = velocity × time. For [0,4], displacement ≈ v(2) × (4 - 0) = 40 × 4. For [4,8], displacement ≈ v(6) × (8 - 4) = 60 × 4.

Step 5: Add the displacements from both subintervals to estimate the total displacement of the object on [0,8]. Total displacement ≈ displacement from [0,4] + displacement from [4,8].

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Key Concepts

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

Velocity Function

The velocity function, denoted as v = ƒ(t), describes how the velocity of an object changes over time. In this context, it provides the instantaneous speed of the object at any given time t within the interval [0, 8]. Understanding this function is crucial for estimating displacement, as it directly influences how far the object travels during each time segment.

Midpoint Rule

The Midpoint Rule is a numerical method used to approximate the area under a curve, which in this case represents displacement. By evaluating the velocity at the midpoint of each subinterval, we can assume the object moves at a constant velocity during that interval. This method simplifies calculations and provides a reasonable estimate of total displacement over the entire interval.

Displacement

Displacement refers to the total distance an object moves in a specific direction over a given time period. In this problem, it is calculated by summing the products of the velocity (evaluated at midpoints) and the duration of each subinterval. Understanding displacement is essential for interpreting the motion of the object and applying the results of the Midpoint Rule effectively.

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Using The Velocity Function

Related Practice

Textbook Question

Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.

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Textbook Question

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

(a) Write the left and right Riemann sums in sigma notation for an arbitrary value of n.

∫₀¹ cos ⁻¹ 𝓍 d𝓍

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Textbook Question

Properties of integrals Consider two functions ƒ and g on [1,6] such that ∫₁⁶ƒ(𝓍) d𝓍 = 10 and ∫₁⁶g(𝓍) d𝓍 = 5, ∫₄⁶ƒ(𝓍) d𝓍 = 5 , and ∫₁⁴g(𝓍) d𝓍 = 2. Evaluate the following integrals.

(a) ∫₁⁴ 3f(𝓍) d𝓍

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Textbook Question

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) = ― 12𝓍 (𝓍―1) (𝓍― 2) ; a = 0 , b = 1 , c = 2

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Textbook Question

Average value with a parameter Consider the function ƒ(𝓍) = a𝓍 (1―𝓍) on the interval [0, 1], where a is a positive real number.

(a) Find the average value of ƒ as a function of a .

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Textbook Question

Planetary orbits The planets orbit the Sun in elliptical orbits with the Sun at one focus (see Section 12.4 for more on ellipses). The equation of an ellipse whose dimensions are 2a in the 𝓍-direction and 2b in the y-direction is (𝓍²/a²) + (y² /b²) = 1.

(a) Let d² denote the square of the distance from a planet to the center of the ellipse at (0, 0). Integrate over the interval [ ―a, a] to show that the average value of d² is (a² + 2b²) /3 .

137

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