page 1 of 8Math 1B: First ExamTuesday, 7 July 2009Instructor: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Name: ANSWERSProblem Number 1a 1b 1c 1d 2a 2b 3 4a 4b 4c TotalScoreMaximum 10 10 10 10 10 10 15 5 5 15 100Please do not begin this test until 2:10 p.m. You may work on the exam until 4 p.m.Please do not leave during the last 15 minutes of the exam time.You must always justify your answers: show your work, show it neatly, and when in doubt,use words (and pictures!) to explain your reasoning. Please box your final answers.Calculators are not allowed. Please sign the following honor code:I, the student whose name and signature appear on this midterm, have completed the examby myself, without any help during the exam from other people, or from sources other thanmy allowed one-page hand-written cheat sheet. Moreover, I have not provided any aid toother students in the class during the exam. I understand that cheating prevents me fromlearning and hurts other students by creating an atmosphere of distrust. I consider myselfto be an honorable person, and I have not cheated on this exam in any way. I promise totake an active part in seeing to it that others also do not cheat.Signature:Name: ANSWERS Math 1B First Exam: 7 July 2009 page 2 of 81. (40 pts – 4 questions, 10 pts each) Evaluate the following integrals.(a)Zdxx√x2+ 4We substitute x = 2 tan θ. Then:Zdxx√x2+ 4=Z2 sec2θ dθ2 tan θ√4 tan2θ + 4=Z2 sec2θ dθ2 tan θ 2 sec θ=12Zsec θtan θdθ=12Z1/ cos θsin θ/ cos θdθ=12Zcsc θ dθ=12ln |csc θ − cot θ|+ C=12ln√x2+ 4x−2x+ C=12ln√x2+ 4 − 2x+ C(b)Zx + 1x2− 6x + 5dxWe factor the denominator and expand into partial fractions:Zx + 1x2− 6x + 5dx =Zx + 1(x − 1)(x − 5)dx=Z−1/2x − 1+3/2x − 5dx= −12ln(x − 1) +32ln(x − 5) + CName: ANSWERS Math 1B First Exam: 7 July 2009 page 3 of 8(c)Zπ/30x sec x tan x dxWe integrate by parts with u = x and dv = sec x tan x dx:Zπ/30x sec x tan x dx = x sec xπ/30−Zπ/30tan x dx=x sec x π/30−ln |sec x + tan x| π/30=π3secπ3− 0 − lnsecπ3+ tanπ3+ ln |1 + 0|=π3· 2 − ln2 +√3=23π − ln2 +√3(d)Z(cos x + sin x)2dxWe expand and use trig identities:Z(cos x + sin x)2dx =Zcos2x + 2 cos x sin x + sin2xdx=Z(1 + 2 cos x sin x) dx=Z(1 + sin 2x) dx= x −12cos 2x + CName: ANSWERS Math 1B First Exam: 7 July 2009 page 4 of 82. (20 pts – 2 questions, 10 pts each) Determine whether each of the following improper integralsis convergent or divergent. You do not need to prove your answer, and you do not need toevaluate the integrals (although you are welcome to do either or both). For example, if I askedyou aboutR∞1(x2+ x)−1dx, you could evaluate the integral (partial fractions), or you coulduse the comparison test to prove that it converges, but it’s enough to write: “x2+ x ≈ x2asx → ∞, so we expect the integral to converge, sinceR∞1x−2dx converges by the p-test.”(a)Z10cos x√xdx. Hint: do not try to evaluate.The only potential divergence is at x = 0. When x ≈ 0, we have cos x ≈ 1, and soZsmall0cos x√xdx ≈Zsmall01√xdxwhich converges by the P-test with p = 1/2. Thus, we expect the integral to converge .(b)Z∞2dxx ln x. Hint: u-substitution.We substitute u = ln x. Then du = dx/x and ln ∞ = ∞, so the integral is:Z∞2dxx ln x=Z∞ln 2duuwhich diverges by the P-test. Thus the original integral diverges .Name: ANSWERS Math 1B First Exam: 7 July 2009 page 5 of 83. (15 pts) Let’s say you want to know the numerical value of ln 3 =R31x−1dx, and you decideto approximate the integral using the trapezoid rule. How many subintervals do you need totake in order to assure that your answer is accurate to within an error of 0.00001 = 10−5?We recall the error formula for the Trapezoid Rule:E ≤K(b − a)312 n2where K is any number such that K ≥ |f00(x)| for all x ∈ [a, b], where [a, b] = [1, 3] andf(x) = x−1. Then f00(x) = 2x−3, which is a decreasing function, so it is maximized whenx = 1, whence K = 2 works. Thus, we have:E ≤2 · 2312 n2=43n−2We want this number to be less than 10−5. Thus, we solve:43n−2≤ 10−534n2≥ 105n2≥43105n ≥r43105=p40/3 × 102We round 40/3 = 1313up to 16 (a perfect square), and conclude that n = 4 × 102= 400works.Name: ANSWERS Math 1B First Exam: 7 July 2009 page 6 of 84. Each of the following questions required the answers to the previous ones. If you answer oneincorrectly, I will grade the later problems as if your answer was correct (so that you will notbe penalized twice). If you cannot answer one question, you may ask me for the answer, andI will mark your test as such.(a) (5 pts) Let a be an arbitrary positive real number. Sketch the curve {y =√x : 0 ≤ x ≤a}. Write an expression for the arclength as an integral in terms of x.We sketch the curve y =√x:−11 2 3 4 5 6 7 8 912340−1Recalling the arclength formula, the length of the curve for 0 ≤ x ≤ a is:` =Za0r1 +14xdxsince if f(x) =√x then f0(x) = 1/2√x.(b) (5 pts) Explain why your expression in part (a) is an improper integral.The integral is improper at x = 0. The integrand blows up there, since limx&014x= ∞.We remark that the integral nevertheless converges (which we prove by evaluating itin the next question). Indeed: 1 +14x≤1xfor x ≤ 1/3, and soR1/30q1 +14xdx ≤R1/30x−1/2dx which converges by the P-test.Name: ANSWERS Math 1B First Exam: 7 July 2009 page 7 of 8(c) (15 pts) Evaluate your integral from part (a). Hint: substitute x =14tan2θ. Never leaveany answer in a form like sin(arccos b) — use a triangle to find a corresponding algebraicexpression.We are interested in the integralZa0r1 +14xdx. We substitute x =14tan2θ, whencedx =12tan θ sec2θ dθ. Thus:Za0r1 +14xdx =Zx=ax=0r1 +1tan2θ12tan θ sec2θ dθ=Zx=ax=0rtan2θ + 1tan2θ12tan θ sec2θ dθ=12Zx=ax=0rsec2θtan2θtan θ sec2θ dθ=12Zx=ax=0sec θtan θtan θ sec2θ dθ=12Zx=ax=0sec3θ dθ=1212sec θ tan θ +12ln |sec θ + tan θ|x=ax=0where the last line is from the cheat sheet. When x = 0, we have θ = 0, and so tan θ = 0and sec θ = 1. When x = a, we have tan θ = 2√x = 2√a, and sec θ =√1 + tan2θ =√1 + 4a. Thus, we see that:1212sec θ tan θ +12ln |sec θ + tan θ|x=ax=0=14[sec θ tan θ + ln |sec θ + tan θ|]x=ax=0=14√1 + 4a 2√a + ln√1 + 4a + 2√a =12√a√1 + 4a +14ln√1 + 4a + 2√aName: ANSWERS Math 1B
View Full Document
Unlocking...