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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 5.4.23
Chapter 5, Problem 5.4.23

Symmetry in integrals Use symmetry to evaluate the following integrals.
∫²₋₂ [(x³ ― 4x) / (x² + 1)] dx 

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Step 1: Recognize the symmetry in the integrand. The integral is over the interval [-2, 2], which suggests checking for even or odd symmetry in the function f(x) = (x³ - 4x) / (x² + 1).
Step 2: Test for odd symmetry. A function is odd if f(-x) = -f(x). Substitute -x into f(x): f(-x) = ((-x)³ - 4(-x)) / ((-x)² + 1) = (-x³ + 4x) / (x² + 1). Simplify to see if f(-x) = -f(x).
Step 3: Confirm that f(x) is odd. Since f(-x) = -f(x), the function is odd. For integrals of odd functions over symmetric intervals [-a, a], the result is always 0.
Step 4: Conclude that the integral ∫²₋₂ [(x³ - 4x) / (x² + 1)] dx evaluates to 0 due to the odd symmetry of the integrand.
Step 5: Use symmetry properties to save time and avoid direct computation, as the integral of an odd function over a symmetric interval is zero.

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Key Concepts

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

Symmetry in Functions

Symmetry in functions refers to the property where a function exhibits even or odd characteristics. An even function, f(x), satisfies f(x) = f(-x), while an odd function satisfies f(x) = -f(-x). Recognizing these properties can simplify the evaluation of integrals, particularly over symmetric intervals, as the integral of an odd function over a symmetric interval is zero.
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Definite Integrals

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫[a, b] f(x) dx, where 'a' and 'b' are the bounds of integration. Understanding how to evaluate definite integrals, especially using properties like symmetry, is crucial for solving problems involving area and accumulation.
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Properties of Integrals

Properties of integrals include various rules that can simplify the evaluation process. For instance, the integral of a sum can be split into the sum of integrals, and the integral of a constant multiplied by a function can be factored out. These properties, along with symmetry, allow for more efficient calculations and can lead to quicker solutions in integral evaluation.
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Related Practice
Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 ∫ sec 4w tan 4w dw

Textbook Question

Derivatives of integrals Simplify the following expressions.


d/dt ∫₀ᵗ d𝓍/(1 + 𝓍²) + ∫₁¹/ᵗ dx/(1 + 𝓍²)

Textbook Question

Average distance on a triangle Consider the right triangle with vertices (0,0) ,(0,b) , and (a,0) , where a > 0 and b > 0. Show that the average vertical distance from points on the 𝓍-axis to the hypotenuse is b/2 , for all a > 0 .

Textbook Question

Left and right Riemann sums Use the figures to calculate the left and right Riemann sums for f on the given interval and for the given value of n.

ƒ(𝓍) = x + 1 on [1,6] ; n = 5

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Textbook Question

Use the given substitution to evaluate the following indefinite integrals. Check your answer by differentiating.                                                                                              

                                                                                                                                                                                       

 ∫ 8𝓍 cos (4𝓍² + 3) d𝓍, u = 4𝓍² + 3

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Textbook Question

Approximating area from a graph Approximate the area of the region bounded by the graph (see figure) and the 𝓍-axis by dividing the interval [1, 7] into n = 6 subintervals. Use a left and right Riemann sum to obtain two different approximations.