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Find the length of the curve 3= + e), Setup the integral. dy 2 ° ² - [ √ - - - _ ) « - / √ - + [ } @ ² - - -]=[]}{²+ra ( L= 1+ dx = 1+ (e" e dx 0 0 0 dx 0≤x≤2. ) dx 2 2 2 dy 2 ° 1 - [²√ · + ( + ) » - L √ · · (² «^² - 5F³*=£*½œ²=« »« L= 1+ dx = 1+ (e 0 dx (et - e ) dx dx 0 0 0 2 2 dy 2 ° ^ = [ {√ - + [ # F « = [ {√{1+ ( ² «^-e » Fa=[{²+0%@ L= 1+ dx (e .e dx (e+ e-x) dx dx 0 0 2 dy ° + - / √ + + ( )* ^ - £ √/ - ( ² ~ ~ ~ ~ ~ » ] = {{²+ L= 1+ dx = 1+-(e dx (ex ex) dx dx 0 0 0 dx 2 ·2 ° - - £; {√ · · ( ) « - £; {√ + + – ² «^² - - - »)]*=£}r=0%« S². (et - e dy L= 1+ dx = 1+= dx 0 0 dx

Find the length of the curve 3= + e), Setup the integral. dy 2 ° ² - [ √ - - - _ ) « - / √ - + [ } @ ² - - -]=[]}{²+ra ( L= 1+ dx = 1+ (e" e dx 0 0 0 dx 0≤x≤2. ) dx 2 2 2 dy 2 ° 1 - [²√ · + ( + ) » - L √ · · (² «^² - 5F³*=£*½œ²=« »« L= 1+ dx = 1+ (e 0 dx (et - e ) dx dx 0 0 0 2 2 dy 2 ° ^ = [ {√ - + [ # F « = [ {√{1+ ( ² «^-e » Fa=[{²+0%@ L= 1+ dx (e .e dx (e+ e-x) dx dx 0 0 2 dy ° + - / √ + + ( )* ^ - £ √/ - ( ² ~ ~ ~ ~ ~ » ] = {{²+ L= 1+ dx = 1+-(e dx (ex ex) dx dx 0 0 0 dx 2 ·2 ° - - £; {√ · · ( ) « - £; {√ + + – ² «^² - - - »)]*=£}r=0%« S². (et - e dy L= 1+ dx = 1+= dx 0 0 dx

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find the length of the curve y = 1/² (e* + e²³¹) ₂ dx Setup the integral. dy 2 1 ^² = £ *^{√ ² + ( ^^ )² * = £^_^{√ ¹ + { } ( ² = x + ]«<=[[{{² =>@ dx 1+ dx S (et + e¯x) dx dx 0 dy 2 1 1 ° ² = £*{√ 4+ ( ^ ^ )³ ^ = £^*^{√ ¹ + ( < * = x + }&=£²½«²=e*%& ) L= 1+ dx 1+ (et - -x) dx dx 0 0 2 dy 2 ° i = £ * {√ ₁ + (^~^)*^*^ = £^*^{√ ₁ + ( + 0 ² x » »] = [² => L= 1+ dx 1+ dx (et+ e-x) dx dx 0 0 0 0≤ x ≤ 2. 2 dy 2 1 ° ² - £*{√ ¹ +( ^^) «^ - £ ^ √{ ¹ +{ } ( ^ - e ~» ] =S²½ (²=e^r« + + L= 1+ (2)² dx = dx dx 0 2 2 dy 2 ° ² = £*{√ ₁ + ( 2 )³ ^ = £ * {√ ¹ +[ ^_~ »][««£}<=c»« L= 1+ dx / ( (et. - e¯x) dx dx 0 0 dx
Transcribed Image Text:Find the length of the curve y = 1/² (e* + e²³¹) ₂ dx Setup the integral. dy 2 1 ^² = £ *^{√ ² + ( ^^ )² * = £^_^{√ ¹ + { } ( ² = x + ]«<=[[{{² =>@ dx 1+ dx S (et + e¯x) dx dx 0 dy 2 1 1 ° ² = £*{√ 4+ ( ^ ^ )³ ^ = £^*^{√ ¹ + ( < * = x + }&=£²½«²=e*%& ) L= 1+ dx 1+ (et - -x) dx dx 0 0 2 dy 2 ° i = £ * {√ ₁ + (^~^)*^*^ = £^*^{√ ₁ + ( + 0 ² x » »] = [² => L= 1+ dx 1+ dx (et+ e-x) dx dx 0 0 0 0≤ x ≤ 2. 2 dy 2 1 ° ² - £*{√ ¹ +( ^^) «^ - £ ^ √{ ¹ +{ } ( ^ - e ~» ] =S²½ (²=e^r« + + L= 1+ (2)² dx = dx dx 0 2 2 dy 2 ° ² = £*{√ ₁ + ( 2 )³ ^ = £ * {√ ¹ +[ ^_~ »][««£}<=c»« L= 1+ dx / ( (et. - e¯x) dx dx 0 0 dx
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