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Calculus I group study Mod 5

Calculus I group study Mod 5 - A =(π 2 2 –-1 1 A =(π 2...

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Calculus I Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0- 618-60624-6). Calculus SG4 Module 5 5.2 # 40 lim n (2 i / n ) (2 / n ) x > ∞ l - 1 lim (4 / n 2 ) (∑ n ) x > ∞ l - 1 lim (4 / n 2 ) [( n + 1) n / 2] x > ∞ 2 lim ( n 2 + 1 / n 2 ] x > ∞ 2 lim ( 1 + (1/ n 2 )] x > ∞ 2 ( 1 + 0) = 2 5.4 # 44 y = x + sin x Area = A A = ƒ x 0 (x + sin x) dx A = [(x 2 / 2) –cos x] x 0 A = [(π 2 / 2) –cos π] – ( 0 – cos 0)
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Unformatted text preview: A = (π 2 / 2) –( -1) +1 A = (π 2 / 2) + 2 5.5 # 28 ƒ (1 / 2 √x)dx = 1/2 ƒ x -1/2 dx = ½ * 2√x + c = √x + c d / dx (√x + c) = (1 / 2 √x) 5.7 #36 ƒ (Sec t + tan t) dt = ƒ sec t dt + ƒ tan t dt Integrals ƒ Sec u du = 1n | Sec u + tan U | + C ƒ tan u du = -ln |Cos u | + C ƒ (Sec t + tan t ) dt = ln | Sec t + tan t | -ln | Cost | + C =ln (|Sec t + tan t|) / (|Cos t|) +C...
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