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CalculusCalculus10th Edition

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Chapter 8, Problem 68RE
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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

Solutions

Chapter
Section
Problem 1RE
Problem 2RE
Problem 3RE
Problem 4RE
Problem 5RE
Problem 6RE
Problem 7RE
Problem 8RE
Problem 9RE
Problem 10RE
Problem 11RE
Problem 12RE
Problem 13RE
Problem 14RE
Problem 15RE
Problem 16RE
Problem 17RE
Problem 18RE
Problem 19RE
Problem 20RE
Problem 21RE
Problem 22RE
Problem 23RE
Problem 24RE
Problem 25RE
Problem 26RE
Problem 27RE
Problem 28RE
Problem 29RE
Problem 30RE
Problem 31RE
Problem 32RE
Problem 33RE
Problem 34RE
Problem 35RE
Problem 36RE
Problem 37RE
Problem 38RE
Problem 39RE
Problem 40RE
Problem 41RE
Problem 42RE
Problem 43RE
Problem 44RE
Problem 45RE
Problem 46RE
Problem 47RE
Problem 48RE
Problem 49RE
Problem 50RE
Problem 51RE
Problem 52RE
Problem 53RE
Problem 54RE
Problem 55RE
Problem 56RE
Problem 57RE
Problem 58RE
Problem 59RE
Problem 60RE
Problem 61RE
Problem 62RE
Problem 63RE
Problem 64RE
Problem 65RE
Problem 66RE
Problem 67RE
Problem 68RE
Problem 69RE
Problem 70RE
Problem 71RE
Problem 72RE
Problem 73RE
Problem 74RE
Problem 75RE
Problem 76RE
Problem 77RE
Problem 78RE
Problem 79RE
Problem 80RE
Problem 81RE
Problem 82RE
Problem 83RE
Problem 84RE
Problem 85RE
Problem 86RE
Problem 87RE
Problem 88RE
Problem 89RE
Problem 90RE
Problem 91RE
Problem 1PS
Problem 2PS
Problem 3PS
Problem 4PS
Problem 5PS
Problem 6PS
Problem 7PS
Problem 8PS
Problem 9PS
Problem 10PS
Problem 11PS
Problem 12PS
Problem 13PS
Problem 14PS
Problem 15PS
Problem 16PS
Problem 17PS
Problem 18PS
Problem 19PS
Problem 20PS
Problem 21PS
Problem 22PS
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Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

Area In Exercises 75 and 76, find the area of the given region using any method.

y = 1 25 − x 2

Chapter 8, Problem 68RE, Area In Exercises 75 and 76, find the area of the given region using any method. y=125x2

To determine

To calculate: The area of the region covered by definite integration ∫04125−x2dx.

Explanation

Given: We are given with expression ∫04125−x2dx

Formula used:

∫dua2−u2=12aln|a+ua−u|+C

Calculation:

Let

I=∫04125−x2dx

∫

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Sect-8.1 P-1ESect-8.1 P-2ESect-8.1 P-3ESect-8.1 P-4ESect-8.1 P-5ESect-8.1 P-6ESect-8.1 P-7ESect-8.1 P-8ESect-8.1 P-9ESect-8.1 P-10E
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P-15RECh-8 P-16RECh-8 P-17RECh-8 P-18RECh-8 P-19RECh-8 P-20RECh-8 P-21RECh-8 P-22RECh-8 P-23RECh-8 P-24RECh-8 P-25RECh-8 P-26RECh-8 P-27RECh-8 P-28RECh-8 P-29RECh-8 P-30RECh-8 P-31RECh-8 P-32RECh-8 P-33RECh-8 P-34RECh-8 P-35RECh-8 P-36RECh-8 P-37RECh-8 P-38RECh-8 P-39RECh-8 P-40RECh-8 P-41RECh-8 P-42RECh-8 P-43RECh-8 P-44RECh-8 P-45RECh-8 P-46RECh-8 P-47RECh-8 P-48RECh-8 P-49RECh-8 P-50RECh-8 P-51RECh-8 P-52RECh-8 P-53RECh-8 P-54RECh-8 P-55RECh-8 P-56RECh-8 P-57RECh-8 P-58RECh-8 P-59RECh-8 P-60RECh-8 P-61RECh-8 P-62RECh-8 P-63RECh-8 P-64RECh-8 P-65RECh-8 P-66RECh-8 P-67RECh-8 P-68RECh-8 P-69RECh-8 P-70RECh-8 P-71RECh-8 P-72RECh-8 P-73RECh-8 P-74RECh-8 P-75RECh-8 P-76RECh-8 P-77RECh-8 P-78RECh-8 P-79RECh-8 P-80RECh-8 P-81RECh-8 P-82RECh-8 P-83RECh-8 P-84RECh-8 P-85RECh-8 P-86RECh-8 P-87RECh-8 P-88RECh-8 P-89RECh-8 P-90RECh-8 P-91RECh-8 P-1PSCh-8 P-2PSCh-8 P-3PSCh-8 P-4PSCh-8 P-5PSCh-8 P-6PSCh-8 P-7PSCh-8 P-8PSCh-8 P-9PSCh-8 P-10PSCh-8 P-11PSCh-8 P-12PSCh-8 P-13PSCh-8 P-14PSCh-8 P-15PSCh-8 P-16PSCh-8 P-17PSCh-8 P-18PSCh-8 P-19PSCh-8 P-20PSCh-8 P-21PSCh-8 P-22PS

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