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Math 221 – Exam III (50 minutes) – Friday April 19, 2002AnswersI. (10 points.) Fill in the boxes so as to complete the following statement:A definite integral can be approximated by a Riemann sum. More precisely,ifa = x0< x1< x2< ···xn−1< xn= b,∆xk=xk− xk−1,xk−1≤ ck≤ xk,andmax{∆x1, ∆x2, . . . , ∆xn} ≈ 0,thennXk=1f(ck) ∆xk≈Zbaf(x) dx.(This problem was on quiz 5.)Grader’s CommentsI think that though students may understand the concept of Riemann sum well,there is a gap between their understanding of it and their use of notations. I felt manyanswers came from the memorization of the answers from the quiz 5. Some answerssuggest confusion in understanding of the meaning of xk, ck, ∆, etc. In other words,they may understand what’s going on, but they don’t understand what xk, ck, and soon, represent in the definition.II. ( 20 points.) A continuous function f satisfiesf(1) = 3, f(1.6) = 6.7, f(2.8) = 11.2, f(3.3) = 9.9, f(4) = 3,f(x) is increasing for 1 ≤ x ≤ 2.8 and f(x) is decreasing for 2.8 ≤ x ≤ 4. Find aRiemann sum S such that3 < S <Z41f(x) dx.Sketch a possible graph and also draw the area represented by your Riemann sum.(You should leave the addition and multiplication undone so as to make your workeasier to grade.)Answer: Take the partitionx0= 1 < x1= 1.6 < x2= 2.8 < x3= 3.3 < x4= 41of the interval [1, 4] and in each interval [xk−1, xk] let ckbe the point where f(x) issmallest. Thus c1= x0= 1, c2= x1= 1.6, c3= x3= 3.3 and c4= x4= 4. TheRiemann sum isS =4Xk=1f(ck)(xk− xk−1)= f(1)(1.6 − 1) + f(1.6)(2.8 − 1.6) + f(3.3)(3.3 − 2.8) + f(4)(4 − 3.3)= 4(1.6 − 1) + 6.7(2.8 − 1.6) + 9.9(3.3 − 2.8) + 3(4 − 3.3)–6–4–20246810–1 1 2 3 4Grader’s CommentsMost students did well on this problem, having seen it before on a quiz. However,every once in a while i’d see such atrocities as rectangles fl oating above the x-axis, orworse, people not knowing how to compute areas of rectangles.III. (30 points.) Evaluate the integral:(1)Zx sin(2x2) dx.Answer: Let u = 2x2so du = 4x dx. ThenZx sin(2x2) dx =Zsin(u)du4=cos(u)4+ C =cos(2x2)4+ C.(This is problem 3 on page 187.)(2)Z20√4x + 1 dx.Answer: Let u = 4x + 1. Then du = 4 dx, u = 1 when x = 0, and u = 9 whenx = 2. ThenZ20√4x + 1 dx =Z90√udu4=u3/26¯¯¯¯90=276− 0.(This is problem 9 on page 210.)2Grader’s CommentsI graded 15 points each part. On the second integral I took 3 points off for doingthe substitution without changing the limits of integration, and 3 more if they actuallyuse them to evaluate the integral. I also took points for missing factors of integration.Some students gave the answer right away, I took 1 point for this, giving them thebenefit of doubt.IV. (20 points.) Evaluate the derivative:(1) F (x) =Zx1dtt. F0(x) =?Answer: By the Fundamental TheoremF0(x) =1x.(This is problem 19 on page 210.)(2) G (x) =Z2x1cos(t2) dt. G0(x) =?Answer: By the Fundamental Theorem and the Chain RuleG0(x) = cos((2x)2) · 2 = 2 cos(4x2).(This is problem 22 on page 210.)V. (30 points.) Find the volume generated when the area in the first quadrantbounded by the parabola y = 3x −x2and the line y = x is rotated about the x-axis.Set up a defi nite integral for the answer. Do not evaluate the integral. Do specify thelimits of integration. (Note: the x-axis is the line y = 0.)Answer: The curves intersect at (x, y) = (0, 0) and at (x, y) = (2, 2). Using themethod of washers with f1(x) = 3x − x2and f2(x) = x we getdV = π¡f1(x)2− f2(x)2¢dx = π¡(3x − x2)2− x2¢dxsoV =Z20π¡(3x − x2)2− x2¢dx.(This is problem 3 on page 245.)3Grader’s CommentsThis question was pretty well handled. Most students were aware that the problemrequired using washers, and almost eve ryone found the correct limits of integration.Attempted solutions that presented the correct limits of integration, but involved m ys-terious and unexplained integrands received only 15 points. The commonest errorin writing down the integrand arose from students confusing π(r2outer− r2inner) withπ(router− rinner)2.VI. (40 points.) A hemispherical bowl is obtained by rotating the semicirclex2+ (y − a)2= a2, y ≤ aabout the y-axis. It is filled with water toa depth of h, i.e. the water level is the liney = h.012345–4 –2 2 4(1) Find the volume of the water in the bowl as a function of h. (Set up a defi-nite integral for the answer. Do not evaluate the integral. Do specify the limits ofintegration.)Answer: The right half of the semicircle has equationx =pa2− (y − a)2=p2ay − y2.Using the method of disksdV = π(p2ay − y2)2dy = π(2ay − y2) dysoV =Zh0π(2ay − y2) dy.(2) Water runs into a hemispherical bowl of radius 5 ft at the rate of 0.2 ft3/sec. Howfast is the water level rising when the water is 4 ft deep? (Hint: Use the meth od ofrelated rates and the Fund amental Theorem.)Answer: By the previous formula with a = 5 we haveV =Zh0π(10y − y2) dy.By the Fundamental TheoremdVdh= π(10h − h2).We are given thatdVdt= 0.2.4By the chain ruledVdt=dVdh·dhdt,sodhdt¯¯¯¯h=4=dV/dtdV/dh¯¯¯¯h=4=0.2π(40 − 16).(This is problem 11 on page 239, but the wording has been changed to su ggest how toset up the problem. Here is the original wording: (1) A hemispherical bowl of radiusa contains water to a depth h. Find the volume of the water in the bowl. (2) Waterruns into a hemispherical bowl of radius 5 ft at the rate of 0.2 ft3/sec. How fast isthe water level rising when the water is 4 ft deep?)Grader’s CommentsThis question was a disaster. Perhaps I shouldn’t have changed the wording fromthe book. I thought I was making the problem easier by setting up the equations. AlsoI will ask a question like this on the final (with similar wording) where the integralcannot be done explicitly and the student must use the Fundamental Theorem. Hereare some of the comments I wrote on student papers.• h is a free variable, not the dummy variable (variable of integration).• Are you using shells or disks? (I can’t tell.)• Rotate about y-axis not the x-axis.• Use h, a in part (1), h = 4, a = 5 in part (2).• Incorrect limits of in tegration.• x should not appear in the integrand if y is the dummy variable.• Do not writeRwithout d.• Incorrect use of n otation: finite = infinitesim al.• The limits of integration depend on h.•pP2+ Q26= P +

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