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by: Blake Lobato

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# Midterm 2 Taylor Polys and Improper Integrals with Solutions Math 31B

Marketplace > University of California - Los Angeles > Math > Math 31B > Midterm 2 Taylor Polys and Improper Integrals with Solutions
Blake Lobato
UCLA
GPA 3.5

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Midterm review and midterm solutions for Taylor Polynomials, arc length, error of polynomials, infinite integrals, and convergent/divergent integrals.
COURSE
Math
PROF.
Prof. Lafferty
TYPE
Study Guide
PAGES
5
WORDS
CONCEPTS
Calculus, 31B, Taylor, polynomials, Error, infinite, divergent, convergent
KARMA
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## Popular in Math

This 5 page Study Guide was uploaded by Blake Lobato on Wednesday February 24, 2016. The Study Guide belongs to Math 31B at University of California - Los Angeles taught by Prof. Lafferty in Winter 2016. Since its upload, it has received 50 views. For similar materials see Math in Math at University of California - Los Angeles.

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Date Created: 02/24/16
M ATH 31B | WINTER 2016 Practice Midterm #2 Solutions Version A Problem 1: Consider the half-circle S deﬁned by the equation x ¯(y ¡4) ˘ 1 and y ‚ 4. Find the area of the surface of revolution obtained by rotating S around the x-axis. Solution: First we solve for y, x ¯(y ¡4) ˘1 () (y ¡4) ˘ 1¡x 2 p () y ¡4 ˘ 1¡x 2 p 2 () y ˘ 1¡x ¯4 Note that we only considered the positive root in the above since y¡4‚0. Next, we note that dy ˘ p ¡x . dx 1¡x 2 Therefore, the surface area in question is given by Z s µ ¶ Z s 1 p 2 · ¡x 2 1 p 2 · x2 2… 1¡x ¯4 1¯ p 2 dx ˘ 2… 1¡x ¯4 1¯ 2dx ¡1 1¡x ¡1 1¡x Z1 p · 1 ˘ 2… 1¡x 2¯4 p dx ¡1 1¡x 2 Z1 4 ˘ 2… 1¯ p dx ¡1 1¡x 2 ‡ 1 ˘ 2… x ¯4sin¡1(x)ﬂ ¡1 ‡ … ‡ … ·· ˘ 2… 1¯4 ¡ ¡1¡4 2 2 ˘ 2… (¯4… ) 4…¯8… 2 1 Problem 2: Find an interval [a,b] containing 0 such that if x is in [a,b], the error of the 5th x ¡18 Taylor polynomial for f (x)˘e centered at 0 (i.e. c ˘0) is less than or equal to 10 . Solution: Let [a,b] be an interval containing 0. Then for any x 2[a,b] we know that 5¯1 6 x x jxj je ¡T (5)j • K ¢ (5¯1)! ˘ K ¢720 , where K is an upper bound on jf (5¯1(x)j over the interval [a,b]. Since (6) x f (x)˘e , we know that K ˘ e since this is the maximum value of f (6)(x) over the interval [a,b]. Putting this altogether we have b 6 x e ¢jxj je ¡T (5)j • 720 for all x 2[a,b]. Since we may choose [a,b], let’s choose b so that the above bound is easier to work with. In particular, we can take b ˘0. Doing so, the above bound becomes 6 x jaj je ¡T 5x)j • 720 . 6 Here we’re using the fact that since a •b ˘0, the largest that jxj can be over the interval [a,b] is jaj . Now, we want 6 jaj • 10 ¡18, 720 and solving for jaj we get r 720 jaj ˘ 6 . 10 18 Since a •0, we have r 6 720 a ˘¡ 18. 10 Hence, the interval " r # 6 720 ¡ 18,0 10 works. 2 Problem 3: Determine whether each of the following integrals converge or diverge. If the integral converges, determine what it converges to. Be sure to justify your answer. Z 0 (a) 2 dx ¡1 Solution: Z 0 Z 0 x x 2 dx ˘ lt!¡1 2 dx ¡1 t ﬂ 1 xﬂ0 ˘ lim 2 ﬂ t!¡1 ln(2) t 1 0 1 t 1 ˘ 2 ¡ lim 2 ˘ ln(2) t!¡1 ln(2) ln(2) The integral converges to 1/ln(2). Z 1 (b) cos(…x)dx ¡1 Solution: Z 1 Z 0 Z 1 cos(…x)dx ˘ cos(…x)dx ¯ cos(…x)dx ¡1 ¡1 Z 0 Z 0 s ˘ lim cos(…x)dx ¯ lim cos(…x)dx t!¡1 t s!1 0 1 ﬂ0 1 ﬂs ˘ lim sin(…x)ﬂ¯ lim sin(…x)ﬂ t!¡1 … t s!1 … 0 1 1 ˘ lim sin(…t)¯ lim sin(…s) t!¡1 … s!1 … Since neither of the above limits exist, the integral diverges. Z 3 (c) p 1 dx 0 2 9¡x Solution: The integrand is discontinuous at the upper limit of integration, x ˘ 3. Therefore, Z 3 Z t 1 1 p 2 dx ˘ lt!3¡ p 2dx 0 9¡x 0Z 9¡x 1 t 1 ˘ lim ¡ p dx t!3 3 0 1¡(x/3) 2 ﬂt ˘ lim sin¡1(x/3)ﬂ t!3¡ 0 … ˘ lim sin ¡1(t/3) ˘ . t!3¡ 2 The integral converges to …/2. 3 Problem 4: Consider the inﬁnite rotational solid generated by rotating the area under the curve f (x)˘1/x on [1,1) around the x-axis. (a) Set up (but do not attempt to evaluate!) an improper integral that gives the surface area of this solid. Solution: We know that the surface area is given by s Z 1 µ ¶2 Z 1 r 2… 1 1¯ ¡1 dx ˘ 2… 1 1¯ 1 dx 1 x x2 1 x x4 (b) Decide whether the integral you found in part (a) converges or diverges. Solution: Noting that r 1 1 1 • 1¯ x x x4 for all x 2[1,1), and Z1 1 dx 1 x diverges to 1, we know that the integral giving the surface area diverges by the Comparison Theorem. Problem 5: Prove from the deﬁnition that 1 n!1 p ˘ 0. n (Don’t worry too much about the structure of the proof, just focus on determining M for an arbitrary †¨0.) 2 Solution: Let †¨0 and set M ˘1/† . Then for all n ¨ M we have ﬂ ﬂ ﬂ 1 ¡0 ˘ 1 ˙ 1 ˘ †. ﬂp n ﬂ p n p1/†2 p Hence, the sequence {1/ n} converges to 0. 4 M ATH 31B | W INTER 2016 Practice Midterm #2 Version A Problem 1: Consider the half-circle S deﬁned by the equation x ¯(y ¡4) ˘ 1 and y ‚ 4. Find the area of the surface of revolution obtained by rotating S around the x-axis. th Problem 2: Find an interval [a,b] containing 0 such that if x is in [a,b], the error of the 5 Taylor polynomial for f (x)˘e centered at 0 (i.e. c ˘0) is less than or equa. to 10 Problem 3: Determine whether each of the following integrals converge or diverge. If the integral converges, determine what it converges to. Be sure to justify your answer. ∫ 0 (a) 2 dx ¡1 ∫ 1 (b) cos(▯x)dx ¡1 ∫ 3 1 (c) p 2dx 0 9¡x Problem 4: Consider the inﬁnite rotational solid generated by rotating the area under the curve f (x)˘1/x on [1,1) around the x-axis. (a) Set up (but do not attempt to evaluate!) an improper integral that gives the surface area of this solid. (b) Decide whether the integral you found in part (a) converges or diverges. Problem 5: Prove from the deﬁnition that 1 n!1 p ˘ 0. n (Don’t worry too much about the structure of the proof, just focus on determining M for an arbitrary ϵ¨0.) 1

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