MATH 152 Midterm: MATH 152 TAMU 2012c Exam 2b Solutions
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Exam ii version b solutions: c as n , the sequence behaves like. Since bn is decreasing, it is bounded above and bounded below by 0. Finally, by b3 = applying l"hospital"s rule shows that bn = |an| . 4n2 : c 1 sin(n3) 1, so . 1 n2 0, an 0 by the squeeze n2: e since the rational expression is improper, use long division to obtain (cid:18)x + x + 4. (cid:18)x x2 + 1(cid:19) dx = x2 + 1(cid:19) dx = ln|x2+1|+ x2 x2 + 1. 0 q36 cos2(3t) + 36 sin2(3t) dt = /3. 0 = 2 . (note the curve is of a circle of: d. = sec2 y, so the length of the curve is s = 0 p1 + (sec2 y)2 dy = /4. 0 p1 + sec4 y dy: b let x = 3 sec . Then dx = 3 sec tan d .
