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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 5.2.32d
Capitolo 5, Problema 5.2.32d

{Use of Tech} Approximating definite integrals Complete the following steps for the given integral and the given value of n. 
(d) Determine which Riemann sum (left or right) underestimates the value of the definite integral and which overestimates the value of the definite integral..


∫₀² (𝓍²―2) d𝓍 ; n = 4

Guida verificata passo dopo passo
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Step 1: Understand the problem. The goal is to approximate the definite integral ∫₀² (𝓍² - 2) d𝓍 using Riemann sums with n = 4 subintervals. Additionally, determine which Riemann sum (left or right) underestimates or overestimates the integral.
Step 2: Divide the interval [0, 2] into n = 4 subintervals. The width of each subinterval, Δ𝓍, is calculated as Δ𝓍 = (b - a) / n, where a = 0 and b = 2. This gives Δ𝓍 = 2 / 4 = 0.5.
Step 3: For the left Riemann sum, use the left endpoints of each subinterval to evaluate the function f(𝓍) = 𝓍² - 2. The left endpoints are 𝓍₀ = 0, 𝓍₁ = 0.5, 𝓍₂ = 1, and 𝓍₃ = 1.5. Compute the sum as L = Δ𝓍 * [f(𝓍₀) + f(𝓍₁) + f(𝓍₂) + f(𝓍₃)].
Step 4: For the right Riemann sum, use the right endpoints of each subinterval to evaluate the function f(𝓍) = 𝓍² - 2. The right endpoints are 𝓍₁ = 0.5, 𝓍₂ = 1, 𝓍₃ = 1.5, and 𝓍₄ = 2. Compute the sum as R = Δ𝓍 * [f(𝓍₁) + f(𝓍₂) + f(𝓍₃) + f(𝓍₄)].
Step 5: Analyze the behavior of the function f(𝓍) = 𝓍² - 2 over the interval [0, 2]. Since the function is increasing on this interval, the left Riemann sum will underestimate the integral (as it uses lower values of the function), and the right Riemann sum will overestimate the integral (as it uses higher values of the function).

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Riemann Sums

Riemann sums are a method for approximating the value of a definite integral by dividing the area under a curve into smaller rectangles. The sum of the areas of these rectangles provides an estimate of the integral's value. Depending on whether the left or right endpoints of the subintervals are used, the Riemann sum can either overestimate or underestimate the actual area, which is crucial for understanding the behavior of the integral.
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Introduction to Riemann Sums

Definite Integrals

A definite integral represents the signed area under a curve between two specified limits. It is denoted as ∫ₐᵇ f(x) dx, where 'a' and 'b' are the lower and upper limits, respectively. The value of a definite integral can be interpreted as the accumulation of quantities, such as area, over an interval, making it essential for applications in physics, engineering, and economics.
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Definition of the Definite Integral

Underestimation and Overestimation

In the context of Riemann sums, underestimation occurs when the sum of the areas of the rectangles is less than the actual area under the curve, while overestimation occurs when the sum exceeds the actual area. For a function that is increasing on the interval, the left Riemann sum will underestimate the integral, and the right Riemann sum will overestimate it. Conversely, for a decreasing function, the opposite is true, highlighting the importance of the function's behavior in determining the accuracy of the approximation.
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Left, Right, & Midpoint Riemann Sums Example 1
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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.


(d) ∫₄⁶ (g(𝓍) ― f(𝓍) d𝓍

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Midpoint Riemann sums Complete the following steps for the given function, interval, and value of n.


ƒ(𝓍) = 2x + 1 on [0,4] ; n = 4


d) Calculate the midpoint Riemann sum.

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Properties of integrals Use only the fact that ∫₀⁴ 3𝓍 (4 ―𝓍) d𝓍 = 32, and the definitions and properties of integrals, to evaluate the following integrals, if possible.

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Midpoint Riemann sums Complete the following steps for the given function, interval, and value of n.


{Use of Tech} ƒ(𝓍) = √x on [1,3] ; n = 4


(d) Calculate the midpoint Riemann sum.

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Use Table 5.6 to evaluate the following definite integrals.                                                                                                                    

 (d) ∫₀^π/¹⁶ sec ² 4𝓍 d𝓍

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Midpoint Riemann sums Complete the following steps for the given function, interval, and value of n.


ƒ(𝓍) = 1/x on [1,6] ; n = 5


(d) Calculate the midpoint Riemann sum.

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