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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 5.2.77a
Capitolo 5, Problema 5.2.77a

{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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Step 1: Understand the problem. A midpoint Riemann sum is a method to approximate the value of a definite integral by dividing the interval into subintervals, calculating the function value at the midpoint of each subinterval, and summing the areas of the rectangles formed. The goal is to express this sum in sigma notation for an arbitrary number of subintervals, n.
Step 2: Define the interval and subintervals. The integral ∫₀⁴ (4𝓍 - 𝓍²) d𝓍 is over the interval [0, 4]. Divide this interval into n subintervals of equal width, Δ𝓍 = (4 - 0)/n = 4/n.
Step 3: Determine the midpoints of the subintervals. The midpoints of the subintervals are given by 𝓍ᵢ = a + (i - 0.5)Δ𝓍, where a = 0 is the starting point of the interval, i is the index of the subinterval (ranging from 1 to n), and Δ𝓍 = 4/n.
Step 4: Write the function value at the midpoints. For each midpoint 𝓍ᵢ, evaluate the function f(𝓍) = 4𝓍 - 𝓍². Substitute 𝓍ᵢ into the function to get f(𝓍ᵢ) = 4(0 + (i - 0.5)(4/n)) - (0 + (i - 0.5)(4/n))².
Step 5: Write the midpoint Riemann sum in sigma notation. The sum is given by Sₙ = Σᵢ₌₁ⁿ f(𝓍ᵢ)Δ𝓍, where Δ𝓍 = 4/n and f(𝓍ᵢ) is the function value at the midpoint. Substitute Δ𝓍 and f(𝓍ᵢ) into the formula to express the sum in terms of n and i.

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Midpoint Riemann Sum

A Midpoint Riemann Sum is a method for approximating the value of a definite integral. It involves dividing the interval into 'n' subintervals, calculating the midpoint of each subinterval, and then evaluating the function at these midpoints. The sum of these function values, multiplied by the width of the subintervals, provides an estimate of the area under the curve.
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Left, Right, & Midpoint Riemann Sums

Sigma Notation

Sigma notation is a concise way to represent the sum of a sequence of terms. It uses the Greek letter sigma (Σ) to indicate summation, along with an index of summation that specifies the starting and ending values. In the context of Riemann sums, sigma notation is used to express the sum of function values at midpoints across all subintervals.
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Sigma Notation

Definite Integral

A definite integral represents the signed area under a curve defined by a function over a specific interval. It is denoted as ∫ from 'a' to 'b' of f(x) dx, where 'a' and 'b' are the limits of integration. The definite integral can be approximated using Riemann sums, which provide a numerical method to estimate the area when the exact integral is difficult to compute.
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Domanda del libro di testo

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

(a) Graph ƒ on the interval 𝓍 ≥ 0.

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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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{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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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(a) Consider the linear function ƒ(𝓍) = 2x + 5 and the region bounded by its graph and the x-axis on the interval [3,6]. Suppose the area of this region is approximated using midpoint Riemann sums. Then the approximations give the exact area of the region for any number of subintervals.

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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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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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