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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.1.5a
Capitolo 5, Problema 5.1.5a

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)                                                                                                             
                                                                                                                                                                                                
fig1

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Divide the interval [0,6] into three subintervals: [0,2], [2,4], and [4,6].
For each subinterval, determine the velocity at the right endpoint by evaluating the graph of v = f(t). For [0,2], the right endpoint is t=2; for [2,4], the right endpoint is t=4; and for [4,6], the right endpoint is t=6.
Approximate the displacement on each subinterval by multiplying the velocity at the right endpoint by the length of the subinterval. For example, for [0,2], the displacement is v(2) * (2-0). Repeat this for [2,4] and [4,6].
Add the displacements from all three subintervals to estimate the total displacement of the object on [0,6].
Ensure that the units of the final displacement are consistent (e.g., ft) since velocity is given in ft/s and time in seconds.

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Velocity Function

The velocity function, denoted as v = f(t), describes how the velocity of an object changes over time. In this context, it provides the speed of the object at any given moment t within the specified interval. Understanding this function is crucial for estimating displacement, as it directly influences how far the object travels during each subinterval.
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Using The Velocity Function

Subintervals

Subintervals are smaller segments into which a larger interval is divided for analysis. In this problem, the interval [0, 6] is divided into three subintervals: [0, 2], [2, 4], and [4, 6]. This division allows for the approximation of displacement by evaluating the velocity at the right endpoint of each subinterval, simplifying the calculation of the total distance traveled.
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Introduction to Riemann Sums

Riemann Sum

A Riemann sum is a method for approximating the total area under a curve, which in this case represents the displacement of the object. By using the right endpoint of each subinterval to determine the height of rectangles, the sum of the areas of these rectangles provides an estimate of the total displacement over the interval [0, 6]. This concept is fundamental in calculus for understanding integration and area calculations.
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Introduction to Riemann Sums
Pratica correlata
Domanda del libro di testo

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

(a) Graph ƒ on the interval 𝓍 ≥ 0.

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Free fall On October 14, 2012, Felix Baumgartner stepped off a balloon capsule at an altitude of almost 39 km above Earth’s surface and began his free fall. His velocity in m/s during the fall is given in the figure. It is claimed that Felix reached the speed of sound 34 seconds into his fall and that he continued to fall at supersonic speed for 30 seconds. (Source: http://www.redbullstratos.com)

(a) Divide the interval [34, 64] into n = 5 subintervals with the gridpoints x₀ = 34 , x₁ = 40 , x₂ = 46 , x₃ = 52 , x₄ = 58 , and x₅ = 64. Use left and right Riemann sums to estimate how far Felix fell while traveling at supersonic speed.

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Sigma notation Evaluate the following expressions.

(a)    10                                                                                                                                                                               

       ∑ κ                                                                                                                                                                          

       κ=1                         

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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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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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