Skip to main content
Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.1.17
Capitolo 5, Problema 5.1.17

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.
v = 2t + 1(m/s), for 0 ≤ t ≤ 8 ; n = 2

Guida verificata passo dopo passo
1
Step 1: Understand the problem. The goal is to approximate the displacement of the object over the interval [0, 8] using the velocity function v = 2t + 1. The interval is subdivided into n = 2 subintervals, and the left endpoint of each subinterval is used to compute the height of the rectangles.
Step 2: Divide the interval [0, 8] into n = 2 subintervals. The width of each subinterval (Δt) is calculated as Δt = (8 - 0) / 2 = 4. So the subintervals are [0, 4] and [4, 8].
Step 3: Determine the left endpoints of each subinterval. For the subinterval [0, 4], the left endpoint is t = 0. For the subinterval [4, 8], the left endpoint is t = 4.
Step 4: Evaluate the velocity function v = 2t + 1 at each left endpoint. For t = 0, v(0) = 2(0) + 1 = 1. For t = 4, v(4) = 2(4) + 1 = 9.
Step 5: Approximate the displacement by summing the areas of the rectangles. The area of each rectangle is given by height × width. For the first rectangle, the area is v(0) × Δt = 1 × 4. For the second rectangle, the area is v(4) × Δt = 9 × 4. Add these areas together to approximate the total displacement.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Velocity and Displacement

Velocity is the rate of change of displacement with respect to time. In this context, the velocity function v(t) = 2t + 1 describes how the object's speed changes over time. Displacement can be approximated by integrating the velocity function over a specified interval, which gives the total change in position of the object.
Video consigliato:
Percorso guidato
10:17
Using The Velocity Function

Riemann Sums

Riemann sums are a method for approximating the integral of a function, which in this case represents the displacement. By dividing the interval into n subintervals and using the left endpoint of each subinterval to determine the height of rectangles, we can estimate the area under the velocity curve. This area corresponds to the total displacement over the interval.
Video consigliato:
Percorso guidato
06:11
Introduction to Riemann Sums

Subintervals and Left Endpoint Approximation

Subdividing the interval into n subintervals allows for a more manageable calculation of the area under the curve. The left endpoint approximation specifically uses the value of the function at the left end of each subinterval to calculate the height of the rectangles. This method provides a way to estimate the integral, which is particularly useful when the exact integral is difficult to compute.
Video consigliato:
Percorso guidato
07:59
Estimating the Area Under a Curve Using Left Endpoints
Pratica correlata
Domanda del libro di testo

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.

v = [1 / (2t + 1)] (m/s), for 0 ≤ t ≤ 8 ; n = 4

52
views
Domanda del libro di testo

Average value of the derivative Suppose ƒ ' is a continuous function for all real numbers. Show that the average value of the derivative on an interval [a, b] is ƒ⁻' = (ƒ(b) ―ƒ(a))/ (b―a) . Interpret this result in terms of secant lines.

128
views
Domanda del libro di testo

General results Evaluate the following integrals in which the function ƒ is unspecified. Note that ƒ⁽ᵖ⁾ is the pth derivative of ƒ and ƒᵖ is the pth power of ƒ. Assume ƒ and its derivatives are continuous for all real numbers. 

∫ (5 ƒ³ (𝓍) + 7ƒ² (𝓍) + ƒ (𝓍 )) ƒ'(𝓍) d𝓍

85
views
Domanda del libro di testo

The linear function ƒ(𝓍) = 3 ― 𝓍 is decreasing on the interval [0, 3]. Is its area function for ƒ (with left endpoint 0) increasing or decreasing on the interval [0, 3]? Draw a picture and explain. 

75
views
Domanda del libro di testo

Variations on the substitution method Evaluate the following integrals.                                                                                                        

                                                                                                                                                                    

 ∫ y²/(y + 1)⁴ dy

66
views
Domanda del libro di testo

Derivatives of integrals Simplify the following expressions.


d/d𝓍 ∫₀ˣ (√1 + t²) dt (Hint: ∫ˣ₋ₓ (√1 + t²) dt = ∫⁰₋ₓ (√1 + t²) dt + ∫ˣ₋ₓ (√1 + t²) dt ) .

60
views