Math 150B S22 Midterm 2 Solutions

California State University Northridge MATH 150B: Calculus II, Spring 2022 Midterm Exam 2 March 14, 2022. Duration: 65 Minutes. Instructor: Dr. Jing Li Student Name: Student number: Instructions: • Take your time to read the entire paper before you begin to write, and read each question carefully. Remember that certain questions are worth more points than others. Make anote of the questions that you feel confident you can do, and then do those first: you do nothave to proceed through the paper in the order given. • This exam contains 4 problems on 7 pages (including this title page) for a total of 110 points. • You have 65 minutes to complete this exam. • This is a closed book exam, and no notes of any kind are allowed. No calculators are allowed. The use of cell phones, pagers or any text storage or communication device is not permitted. • The correct answer requires justification written legibly and logically: you must convince me that you know why your solution is correct. Answer these questions in the space pro-vided. Show all work that is used to derive your answers. Use the backs of pages if neces-sary. • Where it is possible to check your work, do so. • Good Luck! Page 2 3 4 5 6 7 Total Points 20 20 20 10 20 20 110 Your Grades  

Math 150B Midterm 2 Student Name: Student ID: 2 Question 1. [20 points] Evaluate the following integrals. a. (10 points) Z 16 x 2 ( x ° 6)( x + 2) 2 d x b. (10 points) Z 2 x 2 + x + 1 ( x ° 1)( x 2 + 1) d x Solution Using partial fraction we get Ifan fo tf En AMYIBY.fi tcn t Now We have 16 3 A 4 44 Bix 6 x 2 01 6 pick x 6 we get 16.62 A16 42 A 9 pick x 2 We get 16 27 01 2 6 c 8 pick x 0 We get 0 9.2 131 6 12 1 8 1 6 13 7 therefore I Egged x 91 6 Eg Eye dx 9h lx 61 7h lay 8 way t c h fixblunt tf t c Solution Using partial fraction we get IIII YE AMIIEI.ge Now wehave 2 4 1 AND Bx c ex1 A NID BR Bx tox C At B It l B tox CA c Equating the like powerterms on both sides of the above equation we get Bt 0 1 Atc 3 24 4 A 2 sub A 2 into we get 13 0 sub A 2 into we get c l therefore SEF.fi dx f1ftf dx zenlx il tank t c

Math 150B Midterm 2 Student Name: Student ID: 3 Question 2. [20 points] Evaluate the following improper integral or state that they diverge. You must write the integrals as limits. a. (10 points) Z 1 °1 e 3 x 1 + e 6 x d x b. (10 points) Z 3 0 1 p 9 ° x 2 d x Solution Wefirst write III dx IEfxdxtf.ie xdx aliI feIdxtbl5yfb dx Tofind ananti derivative for Efx We use substitution u e then du 3 e'dx e'ax 3 du Sff dx Stfu du tf du Itai ut c t tant le t c therefore 1 54 41 541 t tantalite tannery Is s Haiti tangy tiny's fan le tan D Solution We first note that x 3 IF o then we rewrite Stand x bing.S.ba dx 153 S.ba xdx 153 sine ing sin B sin s y sin lb o big sin b sin t sin I I

Math 150B Midterm 2 Student Name: Student ID: 4 Question 3. [30 points] Evaluate each series or state that it diverges. a. (10 points) 1 X k = 0 2 µ 43 ∂ ° k b. (10 points) 1 X k = 1 µ 1 k + 6 ° 1 k + 7 ∂ Solution We note that 214 214 then 0455 Eng This is a convergent geometric series since its ratio r y and irk l Note that the first term a 2 when ko Therefore the sum is 0445 É R Ir Ey 4 8 Solution We note that the givenseries is a telescoping series because the second part of each term cancels with thefirstpart of the succeeding term We then check its first n terms sum Sn E L that E t t int int f I Therefore Eh ta ness hi t

Math 150B Midterm 2 Student Name: Student ID: 5 c. (10 points) 1 X k = 0 √ 3 µ 25 ∂ k ° 2 µ 57 ∂ k ! Solution We examine the two series É 3 5 and É 2157 individually ID For E 315 the first term of the series is a 3 5 3 the ratio r f and w al therefore É 315 77 5 For Et the first term of the series is a 215 2 the ratio r 57 and in a l therefore EEE I Er Eq 7 Thus É 315 E 4 8 315 EEE I 5 7 2

Math 150B Midterm 2 Student Name: Student ID: 6 Question 4. [40 points] Determine if the series converge or diverge. a. (10 points) 1 X k = 1 2 k 2 + 1 3 k 2 ° 2 k + 1 b. (10 points) 1 X k = 2 1 k (ln k ) 2 Solution We note that ak YEI and Isak 1 4 4 5 0 By the Divergence test we conclude that the giver series diverges Solution By the Integral Test We let fix xp Note that fix is continuous positive and decreasing for x 2 Because Sifu dx SE dy dx blimp ftp.dx To find an anti derivative for Hp we let u thx and thendu dx hence I xp dx fated tut c ut t c therefore Sig dx bliss pdx bist tux l big tub the ut We conclude that the given series converges

Math 150B Midterm 2 Student Name: Student ID: 7 c. (10 points) 1 X k = 1 1 k p k + 3 d. (10 points) 1 X k = 3 1 5 k ° 3 k Solution Use the comparison Test note that for kal at Because É is a p series and p 3 2 21 it converges we conclude that the given series converges Solution Use the Limit comparison Test with the series whos be Note that fight List II É Is It 1 and the series E E t converges because it is a geometric series with Fts therefore the given series also converges