SolutionsAMS 161 Final Exams

# SolutionsAMS 161 Final Exams - Solutions to Past AMS 161...

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Solutions to Past AMS 161 Final Exams AMS 161 Final Exam Solution for Spring 2010 1. a) (e^9/3 - 9^2/2)-(1/3), b) -cos(e^(3x^3))/9, c) 4x - 4ln(x+1), d) (5/3)sqrt(2x+2)(x- 2) (done by parts and simplified) 2. a) e^4x[x/4 - 1/16], b) (x^8/8)[ln(x) - 1/8], c) -x^2*cos(2x^2)/4+sin(2x^2)/8 3. a) -1/[2(x-2)^2] undefined at x=2, b) 1/7, c) -1/(x-1) undefined at x=1 4. LH < MP < TI < TP < RH 5. a) int from 1 to 3 of pi[(2/y^2)^2], b) int from -1 to +1 of pi[(6-x^3)^2 - (2-x)^2] 6. a) int from 20 to 80 of 50 + 5(80-x), b) int from 0 to 12 of 62.5pi(10-10h/12)^2*h, c) int from 0 to 250 of 62.5(250-y)4ln(y+1) 7. a) ln(2) + x/2 - x^2/8 + x^3/24, b) ln(2) + x^2 - x^4/2 + x^3/6, c) xln(2) + (1- ln(2)/6)x^3 + (-1/2 + 1/6 + ln(2)/120)^x^5 8. an = 5^n/n!, limit of 5/(n+1) -> 0, rad of covergence is unbounded 9. a) y = Ae^[-(2/3)e^(-3x)], where A = 5e^(2/3), b) 6e^t - e^(4t) 10. a) T = 70 + 60e^(kt) where k = (1/8)ln(1/2), b) y = 4000(1-e^(-t/20) AMS 161 Final Exam Solutions for Fall 2009 1.{6 pts each} a) e^8/4 - 25/4, b) (1/6)sin(e^(3x^2)), c) 2x + 6ln(x-3), d) (done by

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SolutionsAMS 161 Final Exams - Solutions to Past AMS 161...

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