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# Study Guide for MATH 116 with Professor Mermin at KU

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This 7 page Study Guide was uploaded by an elite notetaker on Friday February 6, 2015. The Study Guide belongs to a course at Kansas taught by a professor in Fall. Since its upload, it has received 17 views.

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Date Created: 02/06/15
Math 116 Je Mermin7s sections Final Exam Review You will be given the following tables 0 The values of the functions sin cos tan on certain important angles 0 The derivatives of the trigonometric funtions o The integrals of the trigonometric functions 1 TrueFalse questions There will be ten truefalse questions taken verbatim from the quizzes All quizzes are fair game including those given before Test 1 2 Evaluate the following 11 3 2x3 3x2 ldx 10 Solution We compute 11 2 3 11 2x3 3x2 ldx 1954 gags 95 10 0 1 4 3 gt 11 95 95 95 2 l lt lt lt1gt31 lt0gt4lt0gt30 b e dz du du Solution Set u 7m so a 7r and dye 7 Then we have e dz eudl 7r leudu 7r 1 LL 5 C 7r1 0 H 2Hquot u c 9525de Solution We use integration by parts Set u x2 and dv exdx Then du 2955195 and v equot We have x25dx udv uv 7 vdu x251 7 2xexdxi To compute 2xexdx we use integration by parts again Set u 2x and dv exdx Then du Edge and v equot We have 2xexdx udv uv7 vdu 29551 72515195 2x51 7 255 C x25dx 95255 7 2xexdx x251 7 2x51 7 255 C 9525 7 2955 25 C Thus 21 U dx We have x 1 de m 5 3 x x Ci Book problems on this material 0 Chapter 61 pp 4077408 9750 Chapter 62 pp 4197420 1750 0 Chapter 64 p 439 17740 0 Chapter 65 p 449 1728 Chapter 6 review p 490 1732 0 Chapter 71 p499 1732 0 Chapter 7 review pl 534 176 3 Compute the following or state that they diverge No justi cation is necessary for correct answers on this problem However incorrect answers may receive partial credit if they are well justi ed a b C d e w E 5 quot n0 Solution This is a geometric series with rst term a 5 0 l and ratio r e li Thus since 5 1 i is between fl and 1 it 1 e ee1 a 177 71I converges to n l n n Solution This is equal to lim new 5 i Alternativel lu in a lar e number such as n 1010 and y p g g a get an m 2718281828 R lim new lt1 which is the de nition of n 3 hm ln2n 3 new n Solution Plugging in n 1000000 gives an m 000042 R 37L lim n 7 new n Solution Plugging in n 200 gives us an m 292 m i Perhaps more convincingly we have 7 37 i in This simpli cation allows us to plug in much larger numbers for n without causing over ow errors for example 11000000 m 2999958 m i xw igdx 11 95 Solution We have xw 5 fidx 75lnxlo 11 95 blim 75lnxl blim 75lnb7751nl lim 75lnb baoo OO so the integral Alternatively it by the power testi f 7 273 7 271 Solution This series is l l 39 1 ms 113202714 3 1 It telescopes l at V at l ma Book problems on this material 0 Chapter 74 p 532 15412 0 Chapter 7 review p 534 1530 0 Chapter 112 p 717 30414 0 Chapter 113 p 727 526 0 Chapter 11 review p 767 1118 11 ye 1 Evaluate dydx 10 y1 Solution We have 11 ye 11 1 dydx doc 10 y1 10 1 Express each of the following in terms of one or more de nite integrals or double integrals with explicit bounds Do not actually compute explicit answers a 3cos xdydx where R is the region bounded by x g x 13 R y0 and ysecx1 Solution The left and right boundaries of this region are x 1 and x g respectively and the top and bottom boundaries are y secx and y 0 Since the top boundary involves y as a function of x the yintegral must go on the inside Thus this is 1 ysec1 3 cos xdydx 1 y0 1E b sinx cos ydydx where R is the region bounded by x 7r R y7rx0andy27rl Solution The left and right boundaries of R are x 0 and x 7r respectively and the top and bottom boundaries are y 7r and y 27L Since all these bounds are constants ie the region is a rectangle we can put either on the outside Thus the 17r y27r integral is sinx cos ydydx which is the same y 7r 10 y27r 17r as sinx cos xdxdy y7r 10 C The area of the region R bounded by y 4 7 295 y 2 x 0 and x 1 Solution These lines form a triangle with vertices 0 2 04 and 1 2 If we want to use a single xintegral or a double integral with the xintegral on the outside we have the left boundary at x 0 and the right boundary at the rightmost point x 1 Then the top boundary is y 4 7 2x and the bottom boundary is y 2 11 so the area is 4 7 2x 7 dye which is the same as 10 11 y472c dydx 10 y2 If we want to use a single yintegral or a double integral with the yintegral on the outside we have the bottom boundary at y 2 and the top boundary at the topmost point y 4 Then the left boundary is x 0 and the right boundary is y 4 7 2x y4 or x 47 Thus the area is 4 7 7 dy which y 2 y4 14 is the same as dxdy y2 10 Book problems on this material 0 Chapter 66 pp 4617463 1742 0 Chapter 87 p 608 1725 0 Chapter 8 reView p 619 43746 6 Solve the following 3 Find all the relative maxima and minima if any of the function fxay 2xy75x2 72y24x4y74 Solution We compute all the rst and second partial derivatives of f f1 2y 7 1095 4 fy 2x 7 43 4 f 710 fry 2 fyy 4 The critical points are the simultaneous solutions to f1 0 and fy 0 Setting fy 0 and solving for 95 gives us 295 7 4y 74 or x 72 2y Plugging this into f1 0 gives us 23 71072 2y 4 0 or y Plugging this back into ac 72 23 gives us as Thus the only critical point is g g Applying the second derivative test at this point we have fmfyy7 2 40 7 4 36 gt 0 so the critical point is a maximum or a my 2 4 3 3 minimum Since f 710 lt 0 is a relative maximum b Find the maximum and minimum of the function fx y 3x y7 6 subject to the constraint x2 y2 4 Solution We create the auxiliary function Fx y 3x y 7 6 M952 y2 7 4 and take its partials FE 3 M295 Fy 1 2y F x2 y2 7 4 The maximum and minimum will occur at the simultaneous so lutions to F1 0 Fy 0 and F 0 We solve FE 0 and Fy 0 for getting 7 and 7i This gives us iiei x 3y Plugging this into F 0 we get 3y2y274 0 iiei 10y2 4 or y i Plugging these solutions into ac 3y we nd two critical points 3 and Magr i We have f3 10 76 and a r 710 g 7 6 Thus the maximum is 10 g 7 6 at 3 and the minimum is 710 g 7 6 at 73 7 Book problems on this material 0 Chapter 83 p 570 1720 0 Chapter 85 p 594 1716 0 Chapter 8 review p 618 29738

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