CALC II SCI ENG
CALC II SCI ENG MATH 2260
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This 6 page Class Notes was uploaded by Joanne Bergnaum on Saturday September 12, 2015. The Class Notes belongs to MATH 2260 at University of Georgia taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/202086/math-2260-university-of-georgia in Mathematics (M) at University of Georgia.
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Date Created: 09/12/15
Practice problems Math 2260 March 17 2009 Notes One of the skills you need to practice is recognizing which method to use The practice problems on page 499 provide a random assort me nt 1 A charged particle moves on the real line It is attracted by a charge located at it 72 and repelled by a charge located at it 1 so that the force acting on it is of the form A B F 7 7 3 x 2 x 71gt where A and B are constants It is determined experimentally that the force is given by 2 x2 20 it 14 ac 2 ac 7 1 i Find A and B Check your work Solution This is a simpli ed version of the standard partial fractions problem because you are told in advance that the Fla and Til terms aren t needed Get a common denominator BAx24B72Ax4BA 96 22 96 i 02 You get three equations the solution of which is A 72 and B 4 Ordinarily with three equations you would require three unhnowns but the coe icients were carefully chosen here to make it work out ii Find the work done by F in moving the particle from it 0 to ac 71 Solution The work is the integral of the force 71 72 4 W 7 7d 71 0 ltx2gt2ltx71gt2 3 2 A solid is formed by revolving the region in the first quadrant bounded above 7r by y sec2x below by y 0 with 0 S at S 1 around the y axis Find the volume of the solid Solution Using the cylindrical shell method the volume is ff 27m sec2xdac Inte grate by parts with u ac and do sec2xdac Then 1 tanx The inde nite integral works out to 27rac tanx 7 tanacdac 27rac tanx 7 ln l secxl O l When we plug in the endpoints we get 7139 Volume 27d tan4 7 ln l sedg 7 0 27d 7 ln 3 Here is how to integrate sec3ac i Write fsec3xdac as fsecac sec2acdac and integrate by parts Solution With it secac and do sec2acdac we have u tanx and du secac tanxdac Make sure you know how to integrate and di erentiate the si1 basic trig functions This gives sec3dac secx tanx 7 secx tan2xdac ii After integrating by parts use the substitution tan2x sec2 7 1 in the new integral Solution sec3dac secx tanx 7 sec3ac 7 secacdac iii Solve for fsec3acdac Solution 2sec3acd secx tanxsecacdac secx tanln l sectanl0 SeC3d secac tanx ln2l secac tanacl O 4 i Set up the integral to find the length of the curve y g from x 0 to x 1 Solution Length fol 11y2dac fol xl 2dac ii Convert the integral in part i to a trigonometric integral by making an appro priate trigonometric substitution Solution Draw a triangle Let x tan6 Then die sec26d6 When it 0 60 Whenicl6 Weget 1 sec36d6 0 iii Perform the integral See problem 3 Solution Using problem 5 we get the following answer for the length of the curve sec6 tan6l ln l sec6 tan6l 2 10 ln2 1 5 Approximate ff idac with n 4 using i Midpoint Rule Solution Let x The midpoint approximation is g f 54 f 74 f 94 f ll4 108975 ii apezoid Rule Solution f1 2f32 2f 2 2f52 f3 m 111667 iii Simpson7s Rule Solution T22 f 1 4f32 2f2 4f52 f 3 11 The following table shows the error estimates for the midpoint trapezoidal and Simpson approximations m Error at most where O is an upper bound for Midpoint C lf ll apezoidal 223 0 WWW Simpson 53150 WWW Which value of n would you pick to compute ln3 to 5 decimal places using iv Midpoint Rule Solution fio 70 On the interval 13 we get an upper bound 0 1 So we want 22 i lt 00001 2n Thus n gt 20000 So take n 20001 V apezoid Rule Solution f ac On the interval 13 we get an upper bound 0 2 So we want 23 12112 This works out to n gt 3651 So take n 366 a dramatic improvement 2 lt 00001 Vi Simpsorfs Rule Solution f4ac 73 On the interval 13 we get an upper bound 0 24 So we want 5 2 24 00001 180114 lt This works out to n gt 255 So take n 26 even more dramatic 6 Compute the following integrals i fo arctand Solution Integrate by parts and use polynomial division 1 3 1 2 1 2 7x arctan 77x 71n1x 0 3 6 6 ii fies arctandx Solution Integrate by parts and use polynomial division 1 1 1 1 1 4 arctan 7 E 3 1 at 7 Z arctan 0 iii fac1nx2dx Solution Integrate by parts with u 1nx2 and du audio Then you ll have to integrate by parts a second time with u 1nic and du xdx 1 1 1 21nx27 21nxlx20 1 1V f Wdac Solution This is not a partial fractions problem The integrand is already in the desired form It s a trig substitution problem After letting it tan6 you get sec2 6 d6 6 sin 6 cos 9 cos26d6 From the triangle draw it one gets 1 5 arctan V f sin5acdac 5 Solution Pull o a power of sinx Then make the substitution u cosx and write sin4x in terms of cosac The answer is 7 cos5 g cos3 7 cos 0 Vi fe c sin2acdac Solution Lihe problem 5 this is a problem where you reproduce the original integral You will have to integrate by parts twice using do emdac both times Answer 2 1 igem cos2ac gemsin2 0 vii fm52dac Solution Here you need to do polynomial division 253 7 10 755p 250 gm 3 ltx5gt2 To integrate the fraction substitute u it 5 Answer 125 5751n50 12 7 710 29c 26 3m24m5 V111 f Wdr Solution Partial Fractions Answer 2in31n1x2arctanac0 7 Integrating the inverse function Suppose x and gx are functions such that fgac ac Suppose is an antiderivative of i Show that icgac 7 is an antiderivative of This is nice because it says that if you know how to integrate a function you know how to integrate its inverse Solution The easiest way to do this is to take the derivative 96996 Fg veg96 996 FQWDQQE But Fac at and fgac ac so we just get gac as required ii Use part i to integrate arctanx Then show that this gives the same result you would get by integrating arctanac by parts Math 2260 43859 Spring 2009 Instructor Neil Lyall7 Boyd 602A7 lyall mathugaedu Course Web Page wwwmathugaedulyall226OSpring09 TakeHome Math 2250 Review Due in class on Thursday 15th of January 1 Find the most general antiderivative of a 35 W 7 m b M siny 7 cos 2y c 712 322 7 524 2 Evaluate the following inde nite integrals a sin2 z dz b x W d c m d 03 Solve the following differential equations subject to the given restraints initialboundary conditions In other words7 nd x given that a fm 490 371 9 0 H90 4 31 955717 U 0 c f z 3em 5sinz7 f0 17 and f 0 2 d f 2 m gt 0 f1 0 and f2 0 F Recall that if f is continuous on 11 then f is also integrable on 11 and the de nition of the de nite integral simpli es to b fxdz lim Rn a mace where Rnbaifltakbagt k1 a Use this formula to evaluate the integral gm 7 3m dz 10 Check your answer to part a by instead evaluating the integral using the Fundamental Theorem of Calculus Part II
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