Calculus II MATH 1920
pellissippi state community college
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Date Created: 11/01/15
INTEGRATION USING TABLES Objective Evaluate integrals using tables of integrals It is frequently necessary to use substitution along with integral tables It may be necessary to use a formula more than once orto use two or more formulas 2 Evaluate I dx x 4 The closest formula is17 J 2d 2 zltani1 c a u a a x212 8 2 1 2 X 4 X 4 2 2 x 12 2 8 o A IIn 17WItha2 dx 1 dx ppyg 39IOX24 39IOI x24j 2 X8Dltan 11 24tan 11 2 2 20 Evaluate 2 J39X dx J5 4x2 2 2 o The closest formula is 34 J39u du 2 u2 a sin 1 C m 3 Using substitution let u 2X 2 du 2dx and x E 5 4x2 2 5 u2 2 J 2 Using34witha25lj 2 duzlj U du 2 5 u2 8 5 u2 1 U 2 AU 5 u sm C 8 2 2 JE Integration Using Tables p 1 J Ahrens 20002006 APPLICATIONS OF INTEGRATION 65 Applications to Physics and Engineering Objective Calculate work and centers of mass l Force push or pull on an object A A constant force does not change in magnitude while it is being applied to an object B A variable force changes while it is being applied to an object eg because of friction or inertia C If an object moves along a straight line with position function st then the force F on the object in the same direction is de ned by Newton s Second Law of Motion d2s dt2 as the product of its mass m and its acceleration F m ll Work constant force F X distance d moved in direction ofthe force A W Fd work force X distance B US Customary system 1 Force is in lb 2 Displacement is in ft or in 3 Work is in ftlb or inlb 1 ftlb e 136 J C SI metric system 1 Force is in Newtons N 1 N 1 kgms2 2 Displacement is in m 3 Work is in Nm joule J III A 12 kg book is lifted off the oor and put on a desk 07 m high A F mg 2298 1176 N B W Fd 117607 s 82 J IV A 20lb weight is lifted 6 ft off the ground A Force 20 lb weight is a force B W Fd 206 120 ftlb V Example 1 force is variable in this example A If f is continuous on a b and an object is moved along the XaXis from X a to n b xbthenWlim fXAX fXdX 00in an B A force of x2 2x lb acts on a particle located at a distance X ft from the origin How much work is done in moving it from X 1 to X 3 3 3 WJ393X22Xdx Xx2 ftIb 1 3 1 3 C An analogous de nition can be stated for an interval on the yaxis V Hooke39s Law The force fX required to maintain a spring stretched X units beyond its natural length is given by fX kx where k is a positive constant called the spring constant Example 2 A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm Find the work done in stretching the spring from 15 cm to 18 cm 1 The distance stretched is 5 cm 05 m 2 f05 40 2 005k 40 2 k 800 3 fX 800x 4 Work done in stretching spring from 15 cm to 18 cm is 08 2 w 08800de seer J 156J 05 2 05 VI Suppose we wish to move a solid vertically eg pumping out uid or raising elevator Example 3 on p 476 see drawing A Divide the interval 2 10 into n subintervals with endpoints X0 X1 xn and choose x in the ith subinterval B Water is sliced into n layers C Each slice is approximately a circular cylinder 1 Thickness of slice Ax interesting notationl 2 Radius ofslice ri and L i 2 ri 310 x 10 Xi 1O 5 3 Volume of slice m 727l39i2AX 21 10 X132 4 Mass density X volume 4 25 5 Force must overcome gravity Fi mg 98160n10 xf x e 15707r10 xf x 6 Work to raise one slice mi e 1000 10 xf2Ax 1607r10 xf x vv e Fin e1570nxi1o xf2Ax n k k 7 W lim Z1570nxi 10 xi ZAX awm 2101570nx1o X2dx 1570n 100x 20x2 X3dx e 34x 106J Vll Moments and centers of mass Jgt Goal is to nd point P on which a thin plate of any given shape balances horizontally center of mass or center of gravity Weight is determined by gravity weight on moon 16 weight on Earth w C Mass is constant D F ma Newton39s second law of motion 1 Slug unit of mass in US Customary system a 32 ftsec2 2 Kilogram unit of mass in SI metric system a 981 msec2 D We assume that a mass is concentrated at a point a pointmass 1 Let two masses be attached to a rod of negligible mass on opposite sides of the fulcrum balance point a Pointmasses m1 and m2 are placed at distances x1 and x2from the fulcrum P b The rod balances if m1x1 m2x2 i Law ofthe Lever ii m1x1 is the moment of mass m with respect to the origin iii m2X2 is the moment of mass m2 with respect to the origin 2 Suppose the rod lies along the Xaxis Let Xcoordinates of m1 m2 and P be x1 x2 and Y i m1X 39 X1 m2 X 39 X2 quot X m1X1m2X2 m1m2 E Center of mass Y is obtained by adding the moments ofthe masses and dividing n 2mm i1 m by the total mass m m1 m2 2 Y VIII If S is a system of pointmasses in a coordinate plane A Moment of m about Xaxis m1yi n Moment ofS about Xaxis MX 2mm i1 B Moment of m1 about yaXis m1xi n Moment ofS about yaXis My ZmX E Center of mass is i y where i yand y m IX Find the moments and center of mass of the system of objects that have masses 3 4 and 8 at points 1 1 2 1 and 3 2 respectively A My31428329 B MX 31418215 o m34815 D izMyzgandy Miz 1 m 15 X Find the centroid ofa lamina at plate with uniform density n A Let the region be bounded by the graphs of y 6 X2 and y 3 2X 1 For simplicity let p 1 it would cancel out in the calculations anywayl 2 Draw graph of region APPLICATIONS OF INTEGRATION 62 Volumes Objective Find volume of solids of revolution De nition ofvolume S is a solid which lies between X a and X b A B Crosssectional area of S in plane PX through X and perpendicular to XaXis is AX C A is a continuous function D v lim imxfmx J39bAXdX awm a II A solid ofrevolution is formed by revolving a region in a plane about a line in the plane called the aXis of revolution lll Disk method Find the volume of the solid of revolution generated by revolving the region formed byyX3 2X2X4XOX3andthe 39939 XaXis about the X aXis lquot A Consider a regular partition on 0 3 with AX 12 ll B Rotate area about X aXis C Each rectangle generates a circular disk D Volume ofa disk nr2h E Radius ofa disk fX F Height thickness of a disk Ax G Volume ofa disk 7rradius2 thickness 2 H Volume ofa disk n x3 2X2 X 4 AX vLnlflt gt2f l f aimMerci oi lat 21 m2 will vL e 2835 vR e 2591 vM e 27107 INTEGRATION USING PARTIAL FRACTIONS Objective Evaluate integrals using partial fraction decomposition Any rational function a ratio of polynomials can be integrated by expressing it as asum of simpler fractions 39 Case 1 Degree of numerator is greater than or equal to degree of denominator perform the long division first 3 Integrate Ixxjjx dx using partial fraction decomposition Since the denominator is linear you could also use synthetic division 711073 0 71 12 17172 2 3 2 X73X 2 L XX I X1dx imx x 2X1dx 3 2 2x2nx1 x3x2712x1 Integrate J39 2 12 x x 7 dx using partial fraction decomposition Integration Using Partial Fractions p 1 JAhrens 20012006 Case 2 The denominator is a product of distinct linear factors where no factor is repeated Background from addition of fractions 2 1 72X271X71 L5 X71 x27 X 1XXt2 X2X72 How do we undo this process painfullyl Integrate dx using partial fraction decomposition X x 7 2 L5 L5 L L X2X12 X71x2 7 x71 x2 NOTE Degree of each numerator is 1 less than degree of its denominator x5Ax2Bx71 Multiply both sides by LCD x 5 Ax2ABx7B Distribute X AX Bx and 5 2A 7 B Equate coefficents of like powers 1 A B 5 2A 7 B 3 A 2 B 71 Solve for coeffICIents X 5 2 1 ldel 7xzjdX2lnlx1lnx2IC 2 Case 2 Integrate I dx using partial fraction decomposition X3 3X2 7 2x X22gtlt71 x22x71 A B C 2x33x272gtlt x2x71x2 2x71 x2 x2 2X71 A2X71X 2 BXX2 CX2X71 Multiply both sides by LCD x2 2x71 2A B2Cx2 3AZB7 Cx7 2A Equate coefficents of like powers 2AB2C1 1 1 1 3A 237 C 2 3 A B C 7 Solve for coefficients 2 5 10 72A 71 I gtlt22gtlt71dXIiJr 1 7 1 dx 2x3 3x272x 2X 52 1 10X2 1 1 1 Elnlxl ln2x717Enlx2 C Integration Using Partial Fractions p 2 J Ahrens 20012006 Integrate I 2 7 x 22x 1dx x 39 Case 3 The denominator is a product of linear factors some of which are repeated 39 Integrate 4 2 W dx using partial fraction decomposition x 7 x 7 x 1 4 2 Dividefirst mwwzm h 3 ix de X x 7x 7x 7x27x1 Let Qx x3 7 x2 7 x 1 Then Q1 0 implies that x 1 is a factor of Qx x3 7 x2 7 gtlt 1 by the Remainder Theorem Dividing shows that x3 7 x27 gtlt1 x71gtlt271 x71gtlt71gtlt1 Partial fraction decomposition XL L L x71gtlt71gtlt17x71x12gtlt1 4x Agtlt71gtlt1Bgtlt1 Cgtlt712 4x ACx2B7ZCx7ABC Multiply both sides by LCD Equate coefficients of like A C 0 A 7C Solve for cofficients B72C 4 B42C 7ABC 0 C4ZCC0 C 71 ThenA1andB2 Integration Using Partial Fractions p 3 JAhrens 20012006 4 2 IWdxzjx1dx x7x7x1 X l x71 th 2 x71 x1 2 XTXlYIIX717 X717nx16 or xln 1 2 76x24x3 8 39 Case 3 Background 4X t 3 x3 X x34x 3 We know there must have been a fraction with denominator of x3 but there may also have been fractions with denominators of x2 andor x dx using partial fraction decomposition 39 Inte rate 9 l x34x 3 6 A E 3 x34x3 4Xt3 X x x3 Complete this problem The denominator contains irreducible quadratic factors none of which is repeated degree of numerator 2 degree of denominator 39 Case 4 4x2 7 3x 2 2 dx usIng partIal fraction decomposItIon 4x 7 4x 3 Divide if degree of numerator degree of denominator 27 I4x2 3x2dX I1 2x 1 de 4x 74x3 4x 74x3 39 Integrate Integration Using Partial Fractions p 4 JAhrens 20012006 4X2 7 4X 3 is irreducible because iTs discriminanT is negoTive Therefore iT cannoT be focTored CompleTe The square of The polynomial 2 4x274x34x27x4ij371 22x7 22x7122 u1 LeTu2X1 Then du2dx ondx 1 u1 71 X 4X 74x3 2x71 2 2 u 2 1u71 1 u 1 1 X dux du7 du 439l u22 439l u22 439l u22 1 2 1 1 1u I X nu 2 7 Tan C Table 17WITha J 8 4 J5 J5 X in4gtlt2 7 4X 3 7 LTan 1271 C 8 4f 2 J5 Case 5 The denominaTor conToins a repeaTed irreducible quadroTic focTor 17gtlt2X27X3 InTegroTe XX2 02 de using porTial fracTion decomposiTion 17x2X27X3 A BXC DxE XX2 12 X X2 1 x2 12 17X2X27X3AX212BXCXX21DXEX SoluTion A1B 1C 1D1andEO 2 3 I 17X22X 72X de gtlt21Jr 2x ZJdX XX 1 X x1 x 1 1 X M X21 g XIdx leTux21 X 1 X 1 nx 7 lngtlt2 17 Tan 1X7 C 2 2X 1 Integration Using Partial Fractions p 5 J Ahrens 20012006
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