75. f(y) = 3y³ +5 Y ; F(1) = 3, y>0

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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(Question 75 and 79) Show work, thank you! Answers bellow 75 = y^3+5 In y+2, y>0 79 = g(x) = 7/8 x^8 -x^2/2 + 13/8
69-76. Particular antiderivatives For the following functions f, find the
antiderivative F that satisfies the given condition.
69. f(x) = x5 2x² + 1; F(0)
70. f(x) = 4√x+6; F(1) = 8
71. f(x) = 8x³+ sin x; F(0) = 2
-
72. f(t)= sec²t; F(π/4) = 1, π/2 <t</2
73. f(v) = sec v tan v; F(0) = 2, π/2 <V <π/2
74. f(u)
2e + 3; F(0) = 8
3y³ +5
=
-
= 1
75. f(y) =
Y
76. f(0) = 2 sin 0 - 4 cos 0; F(7) =
= 2
-; F(1) = 3, y > 0
Transcribed Image Text:69-76. Particular antiderivatives For the following functions f, find the antiderivative F that satisfies the given condition. 69. f(x) = x5 2x² + 1; F(0) 70. f(x) = 4√x+6; F(1) = 8 71. f(x) = 8x³+ sin x; F(0) = 2 - 72. f(t)= sec²t; F(π/4) = 1, π/2 <t</2 73. f(v) = sec v tan v; F(0) = 2, π/2 <V <π/2 74. f(u) 2e + 3; F(0) = 8 3y³ +5 = - = 1 75. f(y) = Y 76. f(0) = 2 sin 0 - 4 cos 0; F(7) = = 2 -; F(1) = 3, y > 0
77-86. Solving initial value problems Find the solution of the following initial
value problems.
77. f'(x) = 2x - 3; f(0) = 4.
78. g'(x) = 7x6 - 4x³ + 12; g(1) = 24
= 2
79. g'(x) = 7x (xº – ½ ); 9(1) =
80. h' (t) = 1 + 6 sin t; h( 77 )
81. f'(u) = 4(cos u — sin u); ƒ(7) = 0
3
82. p' (t) = 10et + 70; p(0) = 100
3
83. y' (t)=
84. u'(x) =
=
t
==
+ 6; y(1) = 8, t > 0
xe²x + 4e
xex
-3
-; u(1) = 0, x > 0
Transcribed Image Text:77-86. Solving initial value problems Find the solution of the following initial value problems. 77. f'(x) = 2x - 3; f(0) = 4. 78. g'(x) = 7x6 - 4x³ + 12; g(1) = 24 = 2 79. g'(x) = 7x (xº – ½ ); 9(1) = 80. h' (t) = 1 + 6 sin t; h( 77 ) 81. f'(u) = 4(cos u — sin u); ƒ(7) = 0 3 82. p' (t) = 10et + 70; p(0) = 100 3 83. y' (t)= 84. u'(x) = = t == + 6; y(1) = 8, t > 0 xe²x + 4e xex -3 -; u(1) = 0, x > 0
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