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Evaluate using the substitution method. √ dz 2x4 (6 +x5) 4 2 1 [(26²du) - 15 (6+23)² + = = + C +x5)³ 4 f( 245 th du .4 Ⓒ/("10"du). 3u 0 ( ³465²-¹ du) = 2 15 - . (6 +x5)-³ + C 3 2x5 5 ²² (6x + ²) ² x5 5 5 3u - 3 0 / ( ³45-² - du) = 0 . (6 + x³)¯³ + C dr + C 2 4 ²/5 - (6 +2³) ¹ + © +x5) C 15

Evaluate using the substitution method. √ dz 2x4 (6 +x5) 4 2 1 [(26²du) - 15 (6+23)² + = = + C +x5)³ 4 f( 245 th du .4 Ⓒ/("10"du). 3u 0 ( ³465²-¹ du) = 2 15 - . (6 +x5)-³ + C 3 2x5 5 ²² (6x + ²) ² x5 5 5 3u - 3 0 / ( ³45-² - du) = 0 . (6 + x³)¯³ + C dr + C 2 4 ²/5 - (6 +2³) ¹ + © +x5) C 15

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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Evaluate using the substitution method. 2x4 √ ( 6 + +25) ₁ dx 4 2 1 ○ f(2a²du) = -1/5 - (0 - 20³)² (6 +x5)³ + C -4 2u 0 / ( 245 + du) = - 2²/5 · (6 +2²³) - ² + 0 3 . +x5) 15 of("² = ² du) = + (50 + 5) 2x5 5 U 10 5 6x + C 3u - 3 [(345-4 (³45=²^ du) = 0 - (6 + 3²) ² + C 3u 4 ⒸS(³465-¹ du) - - 1²/25 · (6 + 3²) ² + c =
Transcribed Image Text:Evaluate using the substitution method. 2x4 √ ( 6 + +25) ₁ dx 4 2 1 ○ f(2a²du) = -1/5 - (0 - 20³)² (6 +x5)³ + C -4 2u 0 / ( 245 + du) = - 2²/5 · (6 +2²³) - ² + 0 3 . +x5) 15 of("² = ² du) = + (50 + 5) 2x5 5 U 10 5 6x + C 3u - 3 [(345-4 (³45=²^ du) = 0 - (6 + 3²) ² + C 3u 4 ⒸS(³465-¹ du) - - 1²/25 · (6 + 3²) ² + c =
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