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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.R.11
Chapter 5, Problem 5.R.11

Limit definition of the definite integral Use the limit definition of the definite integral with right Riemann sums and a regular partition to evaluate the following definite integrals. Use the Fundamental Theorem of Calculus to check your answer. 


∫₀² (𝓍²―4) d𝓍

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Step 1: Recall the limit definition of the definite integral. For a function f(x) over the interval [a, b], the definite integral can be approximated using right Riemann sums: ∫ₐᵇ f(x) dx = lim(n→∞) Σᵢ₌₁ⁿ f(xᵢ)Δx, where Δx = (b - a)/n and xᵢ = a + iΔx.
Step 2: Identify the given function and interval. Here, f(x) = x² - 4, a = 0, and b = 2. Calculate Δx = (2 - 0)/n = 2/n, and the right endpoints xᵢ = 0 + iΔx = 2i/n.
Step 3: Substitute f(xᵢ) and Δx into the Riemann sum formula. The sum becomes Σᵢ₌₁ⁿ [(2i/n)² - 4](2/n). Expand and simplify the terms inside the summation.
Step 4: Use summation formulas to evaluate the sum. For example, Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6 and Σᵢ₌₁ⁿ 1 = n. Substitute these formulas into the expression and simplify.
Step 5: Take the limit as n → ∞. Simplify the resulting expression to find the value of the definite integral. Finally, use the Fundamental Theorem of Calculus to verify the result by directly evaluating ∫₀² (x² - 4) dx.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limit Definition of the Definite Integral

The limit definition of the definite integral involves approximating the area under a curve using Riemann sums. As the number of subintervals increases and their width approaches zero, the sum of the areas of rectangles formed under the curve converges to the exact area, which is represented by the integral. This concept is foundational for understanding how integrals are derived and calculated.
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Definition of the Definite Integral

Riemann Sums

Riemann sums are a method for approximating the total area under a curve by dividing the interval into smaller subintervals and summing the areas of rectangles formed. In the context of the limit definition, right Riemann sums use the right endpoint of each subinterval to determine the height of the rectangles. This approach is crucial for evaluating definite integrals and understanding their geometric interpretation.
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Introduction to Riemann Sums

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that if a function is continuous on an interval, then the integral of its derivative over that interval equals the difference in the values of the original function at the endpoints. This theorem provides a powerful tool for evaluating definite integrals and verifying results obtained through the limit definition and Riemann sums.
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Related Practice
Textbook Question

Evaluating integrals Evaluate the following integrals.


∫₀^²π cos² 𝓍/6 d𝓍

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Textbook Question

(b) Find the average value of ƒ shown in the figure on the interval [2,6] and then find the point(s) c in (2, 6) guaranteed to exist by the Mean Value Theorem for Integrals. 

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

(c) ∫ₐᵇ ƒ'(𝓍) d𝓍 = ƒ(b) ―ƒ(a) .

Textbook Question

Evaluating integrals Evaluate the following integrals.                                                                                                                                         

                                                                                                                                                                   

 ∫ (9𝓍⁸―7𝓍⁶) d𝓍

Textbook Question

Function defined by an integral Let ƒ(𝓍) = ∫₀ˣ (t ― 1)¹⁵ (t―2)⁹ dt .

(c) For what values of 𝓍 does ƒ have local minima? Local maxima?

Textbook Question

Geometry of integrals Without evaluating the integrals, explain why the following statement is true for positive integers n:

∫₀¹ 𝓍ⁿd𝓍 + ∫₀¹ ⁿ√(𝓍d𝓍) = 1