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by: Evert Christiansen


Evert Christiansen
Texas A&M
GPA 3.92


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This 9 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 151 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 39 views. For similar materials see /class/226044/math-151-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
M151B Practice Problems for Exam 3 Calculators will not be allowed on the exam On the exam you will be given the following identities n nn1 n 27nn12n1 37 71n12 gk 2 gk 6 gk lt H 1 Write down a general expression for the sequence with terms 357 911 E77 7187 327 2 Find all xed points for the recursion 1 1 an1 7 an7 and use the method of cobwebbing to determine which limit will be achieved from the starting value 10 71 3 Find all xed points for the recursion t1 telimb and determine whether each is asymptotically stable or unstable 4 The discrete logistic population model is N Nt1 N RNt1i Take R 1 and K 10 and show that one drawback of this model is that it can start with a positive population No gt 0 and return a negative population N1 5 Use a geometric argument to evaluate the integral 1 71 6 Use the method of Riemann sums to evaluate 4 1 7 zzdz 1 2 v1 sinxdm 0 as the limit of a Riemann sum Be sure to de ne all quantities that appear in your expression 7 Express the integral 8 Determine whether or not the Fundamental Theorem of Calculus can be applied to the function x 03x31 s f 2m 1lt 2 on the interval 0 2 If so nd the anti derivative and use it to compute 9 Compute 10 Evaluate the following inde nite integrals 10a cos2 71dz x 7d x21 11 Evaluate the following de nite integrals 11a w zsinz2dx 0 10b 11b 5 xlns idz 1 s 12 Evaluate the following inde nite integral sins z cos x m V1 sin2 z 13 Find the total area between the curves y x2 and y 2 7 z for z E 0 2 14 Suppose that for each value x a certain solid has cross sectional area AW and density The density of an object is its mass per unit volume p Find a formula for the total mass of such an object on the interval 1 b 15 Find the volume of the solid obtained by rotating about the s axis the area between the graph of f and the s axis for z E 01 16 Find the volume obtained by rotating the region between y 2 and y V5 for z E 0 4 about the x axis 17 Find the volume obtained by rotating the region bounded by the curves yz z and m y 5 about the y axis 18 Suppose the base of a certain solid is the region in the xy plane bounded by y 4 and y x2 Find the volume of the solid created if every cross section is a square perpendicular to the x axis 19 Compute the average value of the function n7z7 for z E 13 20 Determine the length of the graph of f x 1 for z E 04 Solutions 1 First we get the sign right with 71 1 n 12 and we observe that the numerator is 271 1 for n 12 The easiest way to understand the denominator is to factor out the common factor 2 a useful trick in general We nd 2n1 7 n1 7 an771 W n712 2 First the xed point equation is 1 2 a77a77a 71a77a so that the xed points are 3 1 2 3 1 3 7 7 7 7 O 077 4a2a a42a a 2 For the cobwebbing we can plot fa ia 7a2 1671 by noticing that its a parabola opening downward with x intercepts a a 0 and a and therefore has a maximum value t at the midpoint of i6 7 i We nd that for 10 71 see the gure A 32 lim an 0 Hoe 3 First7 in order to nd the xed points we solve z zelim 17 617m 07 from which we have the xed points z 01 In order to evaluate the stability of these points7 we set f96 RH and compute fs 617m 61 w71 61 w1 7 We have fO 6 i lfO gt1 z 0 is unstable f 1 0 i lf 0l lt 1 i z 1 is asymptotically stable 4 First7 for R 1 and K 10 the model becomes N M17MM17175 We see that if M is large the second term will be negative7 and as a convenient value we can take N0 50 We nd N1 50 5017 5 50 7 200 7150 5 The graph of the function fx looks like a V on 7117 and the area under the curve consists of two triangles with equal areas Each triangle has baselength 1 and height 17 and so the area of each is 1 We conclude 2 1 lsldm 1 71 6 In this case Ax 17 i g and we use right endpoints m 1 kAx We have i 3k237 k k2 3 An717171 16n9n2n 18k 27k 18 27 2 F n3 Wzk zk 7 k1 k1 18nn1 27nn12n1 n2 2 n3 6 H M w H We conclude lim An 79 7 9 718 4 7 The Riemann sum is 2 7L v1 sin mdz lim x1 sin ckAzk 0 1 HPH70 k where P is a partition of the interval 07 2 with points P 207 quotWall is the norm of P maxk Ask An M 7 1 and ck E zk17xk for each k 1771 8 This function is continuous on the interval 07 2 and so FTC applies In order to compute the anti derivative7 we rst observe that for z E 01 we have m 2 F yea3 0 as expected For x E 12 we must keep in mind that we have m 1 m 2 FWF0 fydyOydy127ydy2z That is7 This can easily be veri ed by a geometric argument 9 According to Leibniz7 rule7 we have d 2471 2 2 2 2 i d7 6 dx 6 1 22 7 e cos m7 s1n 2 cos x 10a Use the substitution u 2x 7 17 so that 7 2 The integral becomes 1 1 1 7 cosudu7sinuC7sin2z71C 2 2 2 10b Use the substitution u 2 17 so that 377 2x The integral becomes g 2 u d 1 d 1 1 u7 7UilnlulC lnlz21lC7 where since 2 1 is always positive the absolute value can be dropped 11a Use the substitution u 27 so that 77 2x The integral becomes 7 du 1 7 1 7 xs1nu77 sinudu77cosu 1 0 2x 2 0 2 0 5 11b Use the substitution u lnx so that 77 The integral becomes 5 1 xdu u du 1 0 Alternatively Problems 10 and 11 can be solved with fast substitution mm mim 12 We make the substitution u 1 sin2 x with du 2 sinzcos xdz and we nd sins z cos x du 1 sin2 z 7 du 2 sin z cos x 2 At this point we observe that sin2 z u 7 1 so we have 1 71 1 1 32 12 1 E u du u127u 12duilg 7l 1sin2z3271sin2x12C 13 Plotting these two curves together we can see that they intersect at z 1 and that for z lt 1 y 2 7 z is larger while for z gt 1 y 2 is larger The area between the curves is 1 2 A277x2dx 22727zdx 0 1 x2 x3 1 3 2 W E gt10lt 2 3gtl 2 1 3 14 As when developing our formula for the volume of such an object we consider a par tition of the interval 11 P 021 xn On the general subinterval k1xk we approximately have a cylinder with base area Ack and constant density pck where ck is any value ck E k1k The mass of such a cylinder is pckV pckAckAxk and so if we sum up these masses we have Mn pckAckAxk 71 k 1 Taking now a limit as the partition size goes to 0 we nd n b M lim ZpckAckAxk pzAzd HPH70 k 15 Since the object is being created by rotation the cross section at each point z is a circle with radius The area of the cross section at point z is Aw 7Tfz2 7m Recalling that our volume formula is V 1b Azdx we compute 1 1 1 V 7Td 7T72 7 0 2 0 16 In this case7 we use the method of washers7 for which we have 4 V W x2 7 g2dz 0 4 2 4 W 22 7 fdx 7T4 7 7 0 2 0 8739139 17 First7 we nd the points at which these curves intersect by solving 2 7952 07951 0 04 7 x 7 7 x x 7 7 x 2 4 4 7 The points of intersection are 07 0 and 47 2 If we rotate the region between these curves about the y axis the line z 2y describes the outer radius while the parabola z yz describes the inner radius The volume is 15 2 2 4 32 32 V W 212 7 1122dy W 412 7 y4dy 7 ngy3 7 5113 W17 7 7 0 0 3 5 7647T 71539 18 One side of each square runs from the curve y x2 to the line y 47 and so the sidelength is 4 7 2 The cross sectional area is consequently AW 7 4 7 962 and the volume is by symmetry 2 2 5 8 2 512 V247z22d21678z2z4dz216x77x3 7 0 0 3 5 0 15 19 We compute 1 3 1 11 3 19 11 7 d7721 1771377721 3 JCan 219H95 z 2l2nlll1 2l2nl 2l2l nf 20 First7 observe that 31 7 H9072 The formula for arclength is b L 1f z2dz7 so we have 4 9 L 1 izdm 0 4 We carry out this integral with substitution7 setting u 1 Ex so that if i g The integral mm u becomes 10 4 4 10 8 L u idu 77 410341 1 9 9 5 1 27 Fall 2007 Math 151 Exam 1 Supplement Mon 010ct 2007 Art Belmonte An alternative way to do problems XlAl4b and XlB15b is to use advanced theory from later in the course Many people tried to do this most left out important pieces Here is the gist Recall the piecewise de nitition of f le72l ifxlt2 fx x7327l if25xg4 2x74 ifxgt4 Note that forx lt 2 we have f x 72 x 7 2 4 7 2x while for 2 lt x lt 4we have f x x2 7 6x 8 Thus f is continuous on say 1 2 U 2 4 since polynomials are continuous 9 Now limxaf f x limxaf 72 72 9 Moreover limxaz f x limxaz 2x 7 6 72 9 Thus limxaz f x 72 9 Since limxaz fx 0 f2 we see thatfis continuous atx 2 Therefore f is continuous on say 1 4 an open interval containing x 2 9 With these four pieces the Assertion see below allows us to conclude that f 2 72 People made tacit use of the Assertion While it is true and even geometrically plausible it is not entirely obvious Without the continuity hypothesis it is not true So please read the Assertion and its proof carefully Or dispense with it altogether and do the problem directly as outlined in original exam solutions Assertion Let f be continuous on a b with nite derivative everywhere in a b exceptpossibly at p Iflimxap f x exists and is equal to A then f 7 must also exist andbe equal to A Proof 9 First observe that a lt p lt 1 since 7 is in the open interval a b o Letu E a p ie u is in the interval a 7 Then by hypothesis f is continuous on u p and differentiable on u 7 By the Mean Value Theorem MVT there is a number m E u 7 such that 7fp7fu7fuifp p7u 7 fwm 7p Hence as u 7 p we have m 7 p and thus fufp p u fw7A since limxap f x A by hypothesis In other words hmfw7 m u7p u 7 p 9 Now repeat this argument for o E p 1 Again by hypothesis f is continuous on p o and differentiable on p o By the MVT there is a numberz E p a such that f 0 i f P m777777 0 i 7 Hence as o 7 17 we have 2 7 7 and thus fm7 mfA 17 since limxap f x A by hypothesis In other words D 7 0 i 7 v7pJr A since the 9 Therefore f p xh39m w i 7 corresponding onesided limits exist and are both equal to A


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