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 5.1.15a

Chapter 5, Problem 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)

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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].

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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 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.

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.

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.

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Related Practice

Textbook Question

Area functions for linear 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 .

ƒ(t) = 2t + 5 , a = 0

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Textbook Question

Area functions for linear 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 .

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ƒ(t) = 4t + 2 , a = 0

Textbook Question

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

(a) Evaluate ∫₋₈⁴ ƒ(𝓍) d𝓍 .

Textbook Question

Working with area functions Consider the function ƒ and its graph.

(a) Estimate the zeros of the area function A(𝓍) = ∫₀ˣ ƒ(t) dt , for 0 ≤ 𝓍 ≤ 10 .

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Textbook Question

Sigma notation Express the following sums using sigma notation. (Answers are not unique.)

(a) 1 + 2 + 3 + 4 + 5

Textbook Question

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.

I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1

(a) ∫₀^π/2 (2 sin θ ― cos θ) dθ