CAS MA 124 Study Guide - Midterm Guide: Antiderivative, Improper Integral, Indeterminate Form

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turquoisegnu128

10 Oct 2018

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Boston University

Department

Mathematics & Statistics

Course

CAS MA 124

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All

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Ma 124 calculus ii c1, solutions to first midterm exam. Prof. nikola popovic, february 16, 2006, 08:00am - 09:20am. Determine whether the statements below are true or false. Z b a a a g(x) dx(cid:19): f(x) = 0, then z 1 a is convergent. f(x) dx (c) all continuous functions have antiderivatives. Take e. g. f(x) = x, g(x) = x, and [a; b] = [0; 1]. 0 x (cid:1) x dx = z 1. 0 x2 dx = but (cid:18)z b a f(x) dx(cid:19) (cid:1)(cid:18)z b a x dx(cid:19) (cid:1)(cid:18)z 1 g(x) dx(cid:19) = (cid:18)z 1 x and a = 1; then, f(x) is continuous on [1; 1), limx!1 x dx(cid:19) = Take e. g. f(x) = 1 and f(x) is decreasing on [1; 1) (since f 0(x) = (cid:0) 1 x2 < 0), but. However, even in that case, we know that g exists. ) F (x) = z x px et t dt:

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5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.

Need help with this study guide for my calculus 2 midterm. Please solve problems and show the solutions step by step for each problem. .

CAS MA 124 Study Guide - Midterm Guide: Antiderivative, Improper Integral, Indeterminate Form

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