Calculus 2-A9 - Calculus II Calculus Early Transcendental...

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Calculus II 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). Written Assignment 9 Complete the following textbook exercises for sections 13.1 and 13.2, and submit them to your mentor for correction and grading. Show all calculations. SECTION 13.1 Exercises Do exercises 6, 10, 14, 18, 22, 28, 32, 36, 50, 52, 70, and 72 on pages 892–894 of the textbook. In Exercises 5– 16, find and simplify the function values. (A) ƒ(0,0)= 4 – (0) 2 – 4(0) 2 = 4 (B) ƒ(0,1)= 4 – (0) 2 – 4(1) 2 = 4 – 4 = 0 (C) ƒ(2,3)= 4 – (2) 2 – 4(3) 2 = 4 – 4 – 4(9) = -36 (D) ƒ(1,y)= 4 – (1) 2 – 4(y) 2 = 3 – 4y 2 (E) ƒ(x,0)= 4 – (x) 2 – 4(0) 2 = 4 – x 2 (F) ƒ(t,1)= 4 – (t) 2 – 4(1) 2 = 4 – t 2 – 4 = -t 2 (A) ƒ(0,5,4)= √(0+5+4) = √9 = 3 (B) ƒ(6,8, -3)= √(6+8+(-3)) = √11 (C) ƒ(4,6,2)= √(4+6+2) = √12 = 2√3 (D) ƒ(10,-4,-3)= √(10+(-4)+(-3)) = √3 (A) g(4,1) = ƒ 1 4 (1/t)dt = [ln|t|] 1 4 = ln1 - ln4 = 0 – ln4 = - ln4 (B) g(6,3) = ƒ 3 6 (1/t)dt = [ln|t|] 3 6 = ln3 - ln6
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= ln3 - ln2 – ln3 = - ln2 (C) g(2,5) = ƒ 5 2 (1/t)dt = [ln|t|] 5 2 = ln5 - ln2 = ln(5/2) (D) g(1/2,7) = ƒ 7 1/2 (1/t)dt = [ln|t|]
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Calculus 2-A9 - Calculus II Calculus Early Transcendental...

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