Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.R.109

Chapter 5, Problem 5.R.109

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

∫₀¹ 𝓍ⁿd𝓍 + ∫₀¹ ⁿ√(𝓍d𝓍) = 1

Verified step by step guidance

First, recognize that the problem involves two integrals over the interval from 0 to 1: \(\int_0^1 x^n \, dx\) and \(\int_0^1 \sqrt[n]{x} \, dx\), where \(n\) is a positive integer.

Recall the general formula for the integral of a power function: for any real number \(m > -1\), \(\int_0^1 x^m \, dx = \frac{1}{m+1}\).

Apply this formula to the first integral: since \(x^n\) is \(x\) raised to the power \(n\), we have \(\int_0^1 x^n \, dx = \frac{1}{n+1}\).

For the second integral, note that \(\sqrt[n]{x} = x^{1/n}\). Using the same formula, \(\int_0^1 x^{1/n} \, dx = \frac{1}{(1/n) + 1} = \frac{1}{\frac{n+1}{n}} = \frac{n}{n+1}\).

Add the two results together: \(\frac{1}{n+1} + \frac{n}{n+1} = \frac{1 + n}{n+1} = 1\). This shows why the sum of the two integrals equals 1 without evaluating the integrals explicitly.

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Key Concepts

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

Definite Integrals and Area Under the Curve

A definite integral from a to b represents the net area under the curve of a function between those limits. For positive functions, this area is positive and can be interpreted geometrically as the region bounded by the curve, the x-axis, and the vertical lines x = a and x = b.

Properties of Power Functions and Their Integrals

Power functions of the form x^n, where n is a positive integer, have integrals that can be computed using the power rule. The integral ∫₀¹ x^n dx equals 1/(n+1), which decreases as n increases, reflecting how the area under x^n changes shape over [0,1].

Complementary Functions and Geometric Interpretation

The functions x^n and the nth root of x (x^(1/n)) are inverses in a geometric sense on [0,1]. Their integrals sum to 1 because the areas under their curves complement each other, filling the unit square between x=0 and x=1 without overlap, illustrating a symmetry in their graphs.

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Textbook Question

Evaluating integrals Evaluate the following integrals.

∫₋₂² (3𝓍⁴―2𝓍 + 1) 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.

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Textbook Question

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

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Textbook Question

Area of regions Compute the area of the region bounded by the graph of ƒ and the 𝓍-axis on the given interval. You may find it useful to sketch the region.

ƒ(𝓍) = 2 sin 𝓍/4 on [0, 2π]

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Textbook Question

Symmetry properties Suppose ∫₀⁴ ƒ(𝓍) d𝓍 = 10 and ∫₀⁴ g(𝓍) d𝓍 = 20. Furthermore, suppose ƒ is an even function and g is an odd function. Evaluate the following integrals.

(a) ∫₋₄⁴ ƒ(𝓍) d𝓍

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Textbook Question

Symmetry properties Suppose ∫₀⁴ ƒ(𝓍) d𝓍 = 10 and ∫₀⁴ g(𝓍) d𝓍 = 20. Furthermore, suppose ƒ is an even function and g is an odd function. Evaluate the following integrals.

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Geometry of integrals Without evaluating the integrals, explain why the following statement is true for positive integers n: ∫₀¹ 𝓍ⁿd𝓍 + ∫₀¹ ⁿ√(𝓍d𝓍) = 1

Verified video answer for a similar problem:

Key Concepts

Definite Integrals and Area Under the Curve

Properties of Power Functions and Their Integrals

Complementary Functions and Geometric Interpretation

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

∫₀¹ 𝓍ⁿd𝓍 + ∫₀¹ ⁿ√(𝓍d𝓍) = 1