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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.R.109

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

โˆซโ‚€ยน ๐“โฟd๐“ + โˆซโ‚€ยน โฟโˆš(๐“d๐“) = 1

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1
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.
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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 โˆซโ‚€ยน x^n dx equals 1/(n+1), which decreases as n increases, reflecting how the area under x^n changes shape over [0,1].
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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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Related Practice
Textbook Question

Area versus net area Find (i) the net area and (ii) the area of the region bounded by the graph of ฦ’ and the ๐“-axis on the given interval. You may find it useful to sketch the region.

ฦ’(๐“) = ๐“โด โ€• ๐“ยฒ on [โ€•1, 1]

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Textbook Question

Evaluate the following derivatives.


d/d๐“ โˆซโ‚ƒแต‰หฃ cos tยฒ dt

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Evaluating integrals Evaluate the following integrals.


โˆซฯ€/โ‚†^ฯ€/ยณ (secยฒ t + cscยฒ t) dt

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Textbook Question

Evaluating integrals Evaluate the following integrals.


โˆซโ‚€ยน โˆš๐“ (โˆš๐“ + 1) d๐“

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Textbook Question

Area of regions Compute the area of the region bounded by the graph of ฦ’ and the ๐“-axis on the given interval. You may find it useful to sketch the region.                                              

                                                                                                                                                                                    

 ฦ’(๐“) = 2 sin ๐“/4 on [0, 2ฯ€]

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Textbook Question

Properties of integrals Suppose โˆซโ‚โด ฦ’(๐“) d๐“ = 6 , โˆซโ‚โด g(๐“) d๐“ = 4 and โˆซโ‚ƒโด ฦ’(๐“) d๐“ = 2 . Evaluate the following integrals or state that there is not enough information.


โˆซโ‚ยณ ฦ’(๐“)/g(๐“) d๐“

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