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ex = 1+ sin x = x COS X = - 1 - 38 3! х 1! + x2 + N8 x5 5! 2! - + 3! 28 7! x4 + 2! 4! (a) Write the Maclaurin series for 4! = 7 + = ∞ Σ(-1)". n=0 ปร n!' n=0 x²+1 (2n+1)!' + = 6! Σ(-1)" x2n (2n)!' n=0 x6 f(x) = cos x + sin x, (b) for ex - COS X - sin x 9(x) = x2 ex - f(x) x2 (c) and for | 9(x)dx = [ ex COS X - sin x -dx. x2

ex = 1+ sin x = x COS X = - 1 - 38 3! х 1! + x2 + N8 x5 5! 2! - + 3! 28 7! x4 + 2! 4! (a) Write the Maclaurin series for 4! = 7 + = ∞ Σ(-1)". n=0 ปร n!' n=0 x²+1 (2n+1)!' + = 6! Σ(-1)" x2n (2n)!' n=0 x6 f(x) = cos x + sin x, (b) for ex - COS X - sin x 9(x) = x2 ex - f(x) x2 (c) and for | 9(x)dx = [ ex COS X - sin x -dx. x2

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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ex = 1+ sin x = x COS X = - 1 - 38 3! х 1! + x2 + N8 x5 5! 2! - + 3! 28 7! x4 + 2! 4! (a) Write the Maclaurin series for 4! = 7 + = ∞ Σ(-1)". n=0 ปร n!' n=0 x²+1 (2n+1)!' + = 6! Σ(-1)" x2n (2n)!' n=0 x6 f(x) = cos x + sin x, (b) for ex - COS X - sin x 9(x) = x2 ex - f(x) x2 (c) and for | 9(x)dx = [ ex COS X - sin x -dx. x2
Transcribed Image Text:ex = 1+ sin x = x COS X = - 1 - 38 3! х 1! + x2 + N8 x5 5! 2! - + 3! 28 7! x4 + 2! 4! (a) Write the Maclaurin series for 4! = 7 + = ∞ Σ(-1)". n=0 ปร n!' n=0 x²+1 (2n+1)!' + = 6! Σ(-1)" x2n (2n)!' n=0 x6 f(x) = cos x + sin x, (b) for ex - COS X - sin x 9(x) = x2 ex - f(x) x2 (c) and for | 9(x)dx = [ ex COS X - sin x -dx. x2
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