ࡱ> ,. )*+z7E l8bjbjss @B5R 8 !|™[# [-[-[--N6NN lpNMNNN[--|:W:W:WN[-L-:WN:W:WHO+\#-{9QJ:y0™ёR~T~t##^~NN:WNNNNNUHNNN™NNNN~NNNNNNNNN : Section 7.1: Integration by Parts SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Find the indefinite integral using integration by parts technique Find the definite integral using integration by parts technique Vocabulary: Integration by Parts an integration technique that sometimes helps to simplify hard integrals Key Concept:  EMBED PowerPoint.Show.8  See pg. 475 for the basic integration rules that we should be familiar with so far. Integration by Parts The product rule states  EMBED Equation.3 . So,  EMBED Equation.3  OR  EMBED Equation.3  Therefore,  EMBED Equation.3 . Generally, we start with  EMBED Equation.3 . This is the rule called Integration by Parts. Examples:  EMBED Equation.3  we need to write this in the form u dv If we let u = x and dv = cos x dx, then du = dx and v = sin x So  EMBED Equation.3   EMBED Equation.3  You can check your answers by taking the derivative! Generally, we want to choose u so that taking its derivative makes a simpler function. Examples:  EMBED Equation.3   EMBED Equation.3   See example 2 on page 477. This shows  EMBED Equation.3  -  EMBED Equation.3  Repeated Integration by Parts Perform integration by parts until the integral you began with appears on the right or you get to a point that you can use a basic integration rule. Then add or subtract accordingly and then multiply or divide. Example:  EMBED Equation.3  More Examples:  EMBED Equation.3   EMBED Equation.3  Definite Integration by parts  EMBED Equation.3  combine this formula with the fundamental theorem of calculus, assume  EMBED Equation.3 and  EMBED Equation.3 are continuous, and you get  EMBED Equation.3  example:  EMBED Equation.3  Tabular view of repeated Integration by Parts If you have a polynomial as one of the two factors in a integration by parts problem, then the following is a short cut to solving the problem. Steps: Draw a 3 column table and label the first column Dif and the third column Int Put p(x) (the polynomial) in the first column and differentiate it until you obtain 0 Put f(x) (the other function) in the third column and integrate repeatedly until you reach the 0 in the first column Draw an arrow from the Dif column to the row below it in the Int column Label the arrows, starting with + and alternating with - From each arrow form the product of the expressions at the tail and the tip of the arrow and multiply that expression by +1 or -1 based on the sign on the arrow Add the results together to obtain the value of the integral Example: Find the integral of +" (2x - 7x + 3x  4) ex dx DifInt2x - 7x + 3x  4+ex6x - 14x + 3-ex12x - 14+ex12-ex0+ex +" (2x - 7x + 3x  4) ex dx = (2x - 7x + 3x  4) ex  (6x - 14x + 3) ex + (12x - 14) ex  (12) ex = ex [(2x - 7x + 3x 4) (6x - 14x + 3) + (12x - 14) (12)] = ex [(2x - 13x + 29x 33) Note that lots of work in setting up u= and dv= over and over again has been saved making this a great time saver as well as something that is less prone to making sign errors. Try it on the following problem: Find the integral of +" (6x + 3x - 5x  7) ex dx DifInt +" (6x + 3x - 5x  7) ex dx = Concept Summary: Integration by parts is analogous to the product rule of differentiation The goal of integration by parts is either to get an integral that is simpler or one that repeats Homework: pg 480 482: Day1: 3, 4, 7, 9, 36; Day 2: 1, 14, 19, 51 Read: Section 7.2 Section 7.2: Trigonometric Integrals SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Solve integrals involving trigonometric functions Vocabulary: None new Key Concepts:  EMBED PowerPoint.Show.8  Type I: sinn x or cosn x, n is odd Keep one sin x or cos x or for dx Convert remainder with sin x + cos x = 1 Examples: +" sin3 x dx +" cos5 x dx Type II: sinn x or cosn x, n is even Use half angle formulas: sin x = (1 - cos 2x) cos x = (1 + cos 2x) Examples: +" sin x dx +" cos4 x dx Type III: sinm x " cosn x, n or m is odd From odd power, keep one sin x or cos x, for dx Use identities to substitute Example: +" sin3 x cos4 x dx Type IV: : sinm x " cosn x, n and m are even. Use half angle identities Example: +" sin x cos x dx Type V: tann x or cotn x From power pull out tan2 x or cot2 x and substitute cot2 x = csc2 x - 1 or tan2 x = sec2 x  1 Examples: +" cot4 x dx +" tan5 x dx Type VI: tanm x" secn x or cotm x " cscn x , where n is even Pull out sec2 x or csc2 x for dx Example:  EMBED Equation.3   EMBED PowerPoint.Show.8  Examples: +" sin x dx (check the double angle formula!) +" tan5 x dx (check your work from previous page!) Homework  Problems: pg 488-489, Day 1: 1, 2, 5, 9, 10 Day 2: 3, 7, 11, 14, 17 Read: Section 7.3 Section 7.3: Trigonometric Substitution SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Use trigonometric identities to simply certain hard integrals Vocabulary: None Key Concept:  EMBED PowerPoint.Show.8  Example 1: ( (4 - x Example 2: ( x (4-x2)-3/2 Example 3: ( 1/(x + 9) Example 4: ( (4 + x Example 5: ( 1/(x(9 + 9x) Example 6: ( ((x - 16)/x Application Problem: Find the area under the curve y = (16 - 4x between x = 0 and x = 2. Homework Problems: pg 494 495 Day 1: 1, 5, 9, 17 Day 2: 6, 13, 22, 33 Read: Section 7.4 Section 7.4: Integration of Rational Functions by Partial Fractions SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Solve integral problems using the technique of partial fractions Vocabulary: None new Key Concept: Type I Improper: (degree of numerator e" degree of denominator) Start with long division Examples:  EMBED Equation.3   EMBED Equation.3  Type II  Proper: decompose into partial fractions Example:  EMBED Equation.3   EMBED PowerPoint.Show.8   EMBED Equation.3   EMBED Equation.3  (If a factor is repeated, i.e.  EMBED Equation.3 , we must consider all possibilities and thus write  EMBED Equation.3 ) Type III Variations of Arctan:  EMBED Equation.3  Examples:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Type IV Variations of Arcsin:  EMBED Equation.3  EMBED Equation.3  Examples:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Homework Problems: pg 504-505, Day 1: 1, 2, 3, 7 Day 2: 4, 10, 19, 40 Read: Section 7.5 Section 7.5: Strategy for Integration SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Nothing at this time Vocabulary: none new Key Concept: Not Covered at this Time  EMBED PowerPoint.Show.8  Homework Problems: none Read: section 7.6 Section 7.6: Integration Using Tables and Computer Algebra Systems SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Solve problems involving density Vocabulary: None Key Concept: Not Covered at this Time Homework Problems: none Read: read 7.7 Section 7.7: Approximate Integration SOL: APC.16: The student will compute an approximate value for a definite integral. This will include numerical calculations using Riemann Sums and the Trapezoidal Rule. Objectives: Students will be able to: Approximate integrals using Riemann Sums and the Trapezoidal Rule Vocabulary: None Key Concept: Some elementary functions do not possess antiderivatives that are elementary functions, i.e.  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4 . Approximation techniques must be used. We have done approximations using rectangles, primarily with left endpoints, right endpoints and midpoints. Another technique for approximations is the trapezoidal rule.  What is the area of the first trapezoid?  EMBED Equation.3 , where  EMBED Equation.DSMT4   Now add each trapezoid together to get: Examples: Use the trapezoidal rule with n = 4 to approximate EMBED Equation.3 . Use the trapezoidal rule with n = 5 to approximate EMBED Equation.3 . Pond Problem: Use the trapezoidal rule to estimate the surface area of the pond. Suppose measurements are taken every 20 feet. The trapezoidal rule averages the results of the left endpoint and right endpoint rules. If f is an increasing function: left end d"  EMBED Equation.DSMT4 d" right end If f is a decreasing function: right end d"  EMBED Equation.DSMT4 d" left end If the graph of f is concave down, the trapezoidal rule underestimates the area; if it is concave up, it overestimates the area. How far off is our estimate? What is the error? Error Approximation & the Trapezoidal Rule:  Two types of problems 1) find the maximum error and 2) find  EMBED Equation.3 (the number of sub-intervals) given a specific error. Examples: Use the error formula to find the maximum possible error in approximating the integral, with n = 4.  EMBED Equation.3  Use the error formula to find n so that the error in the approximation of the definite integral is less than .00001.  EMBED Equation.3  Homework Problems: pg 527 529: probs: 7, 8, 9 Read: section 7.8 Section 7.8: Improper Integrals SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric functions. The special integration techniques of substitution (change of variables) and integration by parts will be included. Objectives: Students will be able to: Solve improper integrals Vocabulary: None Key Concept: Infinite Limits of Integration Type I: One infinite limit 1.  EMBED Equation.3  2.  EMBED Equation.3  You must rewrite these as limits:  EMBED Equation.3   EMBED Equation.3  If the limit exists, then the integral is said to converge to the limit value. If the limit fails to exist, then the integral is said to diverge. Type II: Two infinite limits you must rewrite this as the sum of 2 limits:  EMBED Equation.3  3.  EMBED Equation.3  4.  EMBED Equation.3  Infinite Discontinuities Type I: If  EMBED Equation.3  is continuous on  EMBED Equation.3  and is discontinuous at  EMBED Equation.3 , then  EMBED Equation.3 . If  EMBED Equation.3  is continuous on  EMBED Equation.3  and is discontinuous at  EMBED Equation.3 , then  EMBED Equation.3 . 1.  EMBED Equation.3  2.  EMBED Equation.3  Type II: If  EMBED Equation.3  is continuous on  EMBED Equation.3  except at  EMBED Equation.3 , where  EMBED Equation.3 , then  EMBED Equation.3 , provided both integrals converge. 3.  EMBED Equation.3  4.  EMBED Equation.3  Comparison Theorem: Suppose that f(x) and g(x) are continuous functions with f(x) e" g(x) for x e" a. If  EMBED Equation.3  converges, then  EMBED Equation.3  also converges. If  EMBED Equation.3  diverges, then  EMBED Equation.3  also diverges. See page 530. Homework  Problems: example problem 2 and 4 Read: review and study chapter 7 Chapter 7: Review SOLs: None Objectives: Students will be able to: Know material presented in Chapter 7 Vocabulary: None new Key Concept: The book review problems are on page 534. Homework Problems: pg 468-469: 2, 7, 13, 25, 27, 30 Read: Study for Chapter 7 Test Non-Calculator Multiple Choice 1.  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  2.  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  3.  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  4.  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  5.  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  EMBED Equation.DSMT4  6.  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  EMBED Equation.DSMT4  C.    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S  0RN ^ ` E X*  0TN ^  E Z*  0N ^ ` E Z*H  0޽h ? 3380___PPT10.qp~2 Default Design$   $(  r  S (N `v  N r  S Nt ` N H  0޽h ? 3380___PPT10.qp~2r0WQe}ArialDefault DesignIntegration by Parts  Fonts UsedDesign Template Slide Titles FMicrosoft Equation 3.0 DS Equation Equation.39qX~I\nI ddxuCompObjfObjInfoEquation Native t_1104560142F(C9OC9v()=2uv+u2vI FMicrosoft Equation 3.0 DS Equation Equation.39qڄlIoI ddxuv()  +" =2uv  +" +u2v  +"Ole CompObjfObjInfoEquation Native _1104560228FOC9KE9Ole  CompObj!fObjInfo# FMicrosoft Equation 3.0 DS Equation Equation.39q`HII uv()=2uv  +" +u2v  +" FMicrosoft Equation 3.0 DS EqEquation Native $|_1104560291'FrE9rE9Ole &CompObj 'fuation Equation.39qPIyI u2v  +" =uv"2uv  +" FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo!)Equation Native *l_1104560349$FrE9E9Ole ,CompObj#%-fObjInfo&/Equation Native 0l_1104560480",)FE9E9PII udv  +" =uv"vdu  +" FMicrosoft Equation 3.0 DS Equation Equation.39q4mIyI xcosxdx  +"Ole 2CompObj(*3fObjInfo+5Equation Native 6P_1104560981.FE9E9Ole 8CompObj-/9fObjInfo0; FMicrosoft Equation 3.0 DS Equation Equation.39qxmIyI xcosxdx  +" =xsinx"sinxdx  +"Equation Native <_1104561024O3FE9E9Ole ?CompObj24@f FMicrosoft Equation 3.0 DS Equation Equation.39q@lIoI =xsinx+cosx+C FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo5BEquation Native C\_11045613108FE9E9Ole ECompObj79FfObjInfo:HEquation Native IT_11045613266J=FE9E98mIyI xe x dx  +" FMicrosoft Equation 3.0 DS Equation Equation.39q0lIoI xlnxdxOle KCompObj<>LfObjInfo?NEquation Native OL  +" FMicrosoft Equation 3.0 DS Equation Equation.39q,LnIHzI lnxdx  +"_1134753990BFE9E9Ole QCompObjACRfObjInfoDTEquation Native UH_1134754021@YGFE9E9Ole WCompObjFHXf FMicrosoft Equation 3.0 DS Equation Equation.39q8IppI u=lnx,dv=dx FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoIZEquation Native [T_1104561501TLFF9F9Ole ]CompObjKM^fObjInfoN`Equation Native a`_1104562401;QFF9F9D4xInI e x sinxdx  +" FMicrosoft Equation 3.0 DS Equation Equation.39qHIHI e x cOle cCompObjPRdfObjInfoSfEquation Native gdos2xdx  +" FMicrosoft Equation 3.0 DS Equation Equation.39qDII x 2 sinxdx  +"_1104561396VFF96F9Ole iCompObjUWjfObjInfoXlEquation Native m`_1134754250[F6F96F9Ole oCompObjZ\pf FMicrosoft Equation 3.0 DS Equation Equation.39qLnIHzI fx()2gx()dx=fx()gx()"gx()2fx()dx  +"  +"ObjInfo]rEquation Native s_1134754409`F6F96F9Ole wCompObj_axfObjInfobzEquation Native {0_1134754418eF6F9]F9 FMicrosoft Equation 3.0 DS Equation Equation.39qLnIHzI 2f FMicrosoft Equation 3.0 DS Equation Equation.39qOle |CompObjdf}fObjInfogEquation Native 0IppI 2g FMicrosoft Equation 3.0 DS Equation Equation.39qIܳI fx()2gx()dx ab +" =fx()gx() ab "_1134754449cmjF]F9 H9Ole CompObjikfObjInfolTSEquation Native _1134754616oF H91H9Ole CompObjnpfgx()2fx() ab +" dx FMicrosoft Equation 3.0 DS Equation Equation.39qhIoI xcsc 2 xdx 42 +"ObjInfoqEquation Native _1264959113vdO)H9p)L9Ole  dO)"Microsoft PowerPoint PresentationMSPresentationPowerPoint.Show.89q %_;  Chris HeadleeChris HeadleeOh+'0  XEPRINTsulCompObj{ObjInfotxPictures      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRtsVWXYZ[\]^_`abcdefghijklmnopqruvxwy4|}~l1'% EMFl\KhCF, EMF+@F\PEMF+"@ @ $@ 0@?!@ @ F(GDICFXLEMF+*@$?? @$yD;D!b $$==% % V0.x>.x>% % $$AA" FEMF+@ FGDICF(GDIC1FGDICF(GDICUFGDICF(GDIC1IFGDICRp@"Arial0 E0 "TiK0W0Hz08 #0 !0xXm50x (L( dv%    T#:KKKAJA2"LStrategies for Hard Trig Integrals           % ( F(GDICNF(GDIC;FGDICRp@"Arial0U 䌥0ȧTiK0W0䌥*ȧz08 0 !0 xXm50x=( dv%    T`bKKKAJAzLTsin % ( Rp@"ArialW0䌥L|Q|Hm|U ø0 !0z08m50x ( dv!\P|p|m|Z|m|dv%    TTc~KKKAJAzLP % ( Rp@"Arial L|Q|Hm|U ĸm50x (z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TTc~KKKAJAzLPx % ( Rp@"Arial L|Q|Hm|U Ÿ! \P|p|m|z08(dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    Tlb]KKKAJAzLX+ cos % ( Rp@"Arial \L|Q|Hm|U Ƹ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TT^beKKKAJA^zLP % ( Rp@"Arial \L|Q|Hm|U Ǹ! \P|p|m|z08̋dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TTmc|~KKKAJAmzLPx % ( Rp@"Arial \L|Q|Hm|U ȸ! \P|p|m|z08Xdv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T`bKKKAJAzLT= 1 % ( F(GDIC;FGDICRp@"Arial0 \0 TiK0W0$z08 0 !0xXm50x (L( dv%    TjA:^KKKAJAjYLlFrom odd power,    % ( Rp@"Arial0U X0 TiK0W0XBz08l 0 !0 xXm50x`=(d dv%    T_|KKKAJAw Lhkeep one sin  % ( Rp@"ArialW0XL|Q|Hm|U ޸0 !0z08m50x (d dv!\P|p|m|Z|m|dv%    TT`{KKKAJAwLPx % ( Rp@"Arial L|Q|Hm|U ߸m50x (dz08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T`_|KKKAJAwLTor   % ( Rp@"Arial L|Q|Hm|U ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T`_|KKKAJAwLTcos % ( Rp@"Arial \L|Q|Hm|U ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TT!`0{KKKAJA!wLPx % ( Rp@"Arial \L|Q|Hm|U ! \P|p|m|z08@dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    Tp/_d|KKKAJA/wLX, for   % ( Rp@"Arial \L|Q|Hm|U ! \P|p|m|z08̋dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TXe`{KKKAJAewLPdx % ( Rp@"Arial0 q0짥TiK0W0<짥z08 0 !0xXm50x(L(dv%    T/mKKKAJA/LUse identities to substitute % ( F(GDIC;FGDICRp@"Arial$=0xж=z08=0xж=$=0=820 dv%    T`.SQpKKKAJA.kLTsin % ( Rp@"Arial$=@| Q|Hm|U =0xж=$=0z08=820 dv!PD|p|m|Z|m|dv%    TTRU_gKKKAJARdLPm % ( Rp@"Arialж=@| Q|Hm|U 820 z08dv! 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PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TT]lKKKAJA]LPx % ( Rp@"Arial0 @| Q|Hm| ܸ! PD|p|m|z08(dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    T`rKKKAJArLTor   % ( Rp@"Arial P@| Q|Hm| ݸ! PD|p|m|z08dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    T`KKKAJALTcos % ( Rp@"Arial P@| Q|Hm| ޸! PD|p|m|z08@dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TTKKKAJA LPn  % ( Rp@"Arial P@| Q|Hm| ߸! PD|p|m|z08Xdv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial P@| Q|Hm| ! PD|p|m|z08䌥dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TXKKKAJALP,  % ( Rp@"Arial0U ]0  TiK0W0z08 0 !0xXm50x`(tPdv%    TT(65KKKAJA(1LPn % ( Rp@"ArialW0L|Q|Hm|U ^0 !0z08m50x (tPdv!\P|p|m|Z|m|dv%    Tx>6KKKAJA>1L\is even % ( F(GDICFGDICRp@"Arial0 䌥0ȧTiK0W0䌥*ȧz08 0 !0 xXm50x=( dv%    T`KKKAJALTsin % ( Rp@"ArialW0䌥L|Q|Hm| 0 !0z08m50x ( dv!\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@"Arial L|Q|Hm| m50x (z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial L|Q|Hm| ! \P|p|m|z08(dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    Tl]KKKAJALX+ cos % ( Rp@"Arial \L|Q|Hm| ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TT^eKKKAJA^LP % ( Rp@"Arial \L|Q|Hm| ! \P|p|m|z08̋dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TTm|KKKAJAmLPx % ( Rp@"Arial \L|Q|Hm| ! \P|p|m|z08Xdv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T`KKKAJALT= 1 % ( F(GDICFGDICRp@"Arial0 䌥0 "TiK0W0䌥Hz08 #0 !0 xXm50x=(d dv%    T KKKAJA  LhKeep one sin  % ( Rp@"ArialW0䌥L|Q|Hm| 0 !0z08m50x (d dv!\P|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial L|Q|Hm| m50x (dz08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T`KKKAJALTor   % ( Rp@"Arial L|Q|Hm| ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T`KKKAJALTcos % ( Rp@"Arial \L|Q|Hm| ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TT%KKKAJALPx % ( Rp@"Arial \L|Q|Hm| ! \P|p|m|z08@dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    Tx+rKKKAJA+L\or for    % ( Rp@"Arial \L|Q|Hm| ! \P|p|m|z08̋dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TXsKKKAJAsLPdx % ( Rp@"Arial0U 0TiK0W0&z08 0 !0xXm50x(L(dv%    T KKKAJA LpConvert remainder    % ( F(GDICFGDICRp@"Arialz08=0x=Ծ=0=820 dv%    T`(KKKKAJA(LTsin % ( Rp@"Arial@| Q|Hm| 7=0x=Ծ=0z08=820 dv!PD|p|m|Z|m|dv%    TTLUKKKAJALLPn  % ( Rp@"Arial=@| Q|Hm| 8820 z08dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TT]lKKKAJA]LPx % ( Rp@"Arial0 @| Q|Hm| 9! PD|p|m|z08(dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    T`rKKKAJArLTor   % ( Rp@"Arial P@| Q|Hm| :! PD|p|m|z08dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    T`KKKAJALTcos % ( Rp@"Arial P@| Q|Hm| ;! PD|p|m|z08@dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TTKKKAJALPn  % ( Rp@"Arial P@| Q|Hm| <! PD|p|m|z08Xdv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial P@| Q|Hm| =! PD|p|m|z08䌥dv! PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TXKKKAJALP,  % ( Rp@"Arial0U 0 TiK0W0z08 0 !0xXm50x (t< dv%    TT(6KKKAJA(LPn % ( Rp@"ArialW0L|Q|Hm|U 0 !0z08m50x (t< dv!\P|p|m|Z|m|dv%    Tp>KKKAJA>LXis odd % ( F(GDIC1FGDICRp@"Arial0 R̋0ȧTiK0W0̋*ȧz08X 0 !0xXm50xД(<dv%    mRp@"Arial R̋0ȧTiK0W0̋*ȧz08X ww!4 0w! 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PD|p|m|Z|m|dv!PD|p|m|Z|m|dv%    TXKKKAJALP,  ( Rp@"Arial0U 0 TiK0W0"z080 0 !0xXm50x(ġxdv%    TT-;KKKAJA-LPn ( Rp@"ArialW0L|Q|Hm|U 0 !0z08m50x (ġxdv!\P|p|m|Z|m|dv%    TXC[KKKAJACLP&  ( Rp@"Arial L|Q|Hm|# m50x (ġz08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    TT\qKKKAJA\LPm ( Rp@"Arial L|Q|Hm|U ! \P|p|m|z08dv! \P|p|m|Z|m|dv!\P|p|m|Z|m|dv%    T|yKKKAJAyL\are even  ( F(GDICOFGDICRp@"Arial0s ^0 TiK0W0 z08 0 !0xXm50x (L( dv%    TdV@{KKKAJAuLTTrig  ( Rp@"Arial0U 0 TiK0W0z08 0 !0xXm50x (L( dv%    T|~wKKKAJAL\Identity   ( F(GDICOFGDICRp@"Arial0# 0  TiK0W0z08 0 !0xXm50x (L( dv%    Tnj.KKKAJAn LdSubstitution   ( F(GDICOFGDICRp@"Arial0s f0  TiK0W0z08 0 !0xXm50x (L( dv%    T,jKKKAJA, L`Expression  ( F(GDICNRFEMF+*@$??@<0H@A@, AàBǩsDàB@!b $$==_8882% % W$LT<% % $$AA( " FEMF+@ FGDICF(GDICFEMF+*@$??@4(DZ?A@, A#*CǩsD#*C@!b $$==_888% % W$ < % % $$AA( " FEMF+@ FGDICF(GDICFEMF+*@$??@4(DZ?A@, ACǩsDC@!b $$==_888% % W$<% % $$AA( " FEMF+@ FGDICF(GDIC13FEMF+*@$??@4(DZ?A@, Alj DǩsDlj D@!b $$==_888% % W$04#<#% % $$AA( " FEMF+@ FGDICF(GDICFEMF+*@$??@<0H@A@ , Alj%DǩsDlj%D@ !b $$==_8882% % W$[)<[)% % $$AA( " FEMF+@ FGDICF(GDICNFEMF+*@$??@ <0H@A@ , AàBAlj%D@  !b $$==_8882% % W$L[)% % $$AA( " FEMF+@ FGDICF(GDICOFEMF+*@$??@ 4(DZ?A@ , 0rCàB0rClj%D@  !b $$==_888% % W$N[)% % $$AA( " FEMF+@ FGDICF(GDICOFEMF+*@$??@4(DZ?A@, +DàB+Dlj%D@!b $$==_888% % W$N**[)% % $$AA( " FEMF+@ FGDICF(GDICNFEMF+*@$??@<0H@A@, ǩsDàBǩsDlj%D@!b $$==_8882% % W$L<<[)% % $$AA( " FEMF+@ FGDICF(GDICFEMF+*@$??@4(DZ?A@, A9.vCǩsD9.vC@!b $$==_888% % W$[<[% % $$AA( " FEMF+@ FGDICF(GDIC;=FEMF+*@$??@4(DZ?A@, ACǩsDC@!b $$==_888% % W$:><% % $$AA( " FEMF+@ FGDICF(GDICFEMF+*@$??@4(DZ?A@, ArCǩsDrC@!b $$==_888% % W$\<\% % $$AA( " FEMF+@ FGDICFGDICF4(EMF+*@$??!b Ld)??" 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S  0x ^ `  >*  0} ^   @*  0肥 ^ `  @*H  0޽h ? 3380___PPT10.qp~2 Default Design|   p|(  x  c $ȷ `    E <:c  3x  1 Example: -------------- dx x - x  6`F` <"z a BQn C___PPT9bZ p factor denominator rewrite fraction multiply through by common denominator solve one factor and substitute repeat with remaining factor(s substitute A and B and integrate ^       + @^ f 6   g B K  #  =  "  h B, 3 4+"$ i < P l8x - x  6 = (x  3) (x  2) j <Dq D  3x  1 A B ------------- = --------- + ---------- x  x  6 (x  3) (x  2)qFq k <  :3x  1 = A(x + 2) + B(x  3),  l <  <  Lwhen x = -2 then B = 7/5 m <h  l  Kwhen x = 3 then A = 8/5 n <  /  8 1 7 1 -- -------- dx + -- --------- dx 5 (x-3) 5 (x + 2) = (8/5) ln|x-3| + (7/5) ln|x+2| + C>Fn$& $$,1K o B  rp 8+"$ p Bx  sp 8+"$H  0޽h ? 3380___PPT10.qp~2r0 e> sign Template Slide Titles FMicrosoft Equation 3.0 DS Equation Equation.39qXII x+7x 2 "x"6dx  +"_1104565243FV9V9Ole CompObjfObjInfoEquation Native t_11045652851FV9V9Ole CompObjf !$'*-0149<?BEHKNPQRSUVWXYZ[^abcdgjklmpstuvy|}~ FMicrosoft Equation 3.0 DS Equation Equation.39qڌfObjInfo@Equation Native AdxII dx 16"1+x() 2 +" FMicrosoft Equation 3.0 DS Equation Equation.39qTII dx "_1104565972FHY9HY9Ole CCompObjDfObjInfoFEquation Native Gp_1264797293dO)HY9]9Ole IEPRINTP4D4x"x 2 +" dO)"Microsoft PowerPoint PresentationMSPresentationPowerPoint.Show.89q %_ ~Chris HeadleeChris HeadleeOh+'0  Xl1'% EMF4D\KhCF, EMF+@F\PEMF+"@ @ $@ 0@?!@ @ F(GDICFXLEMF+*@$?? @$yD;D!b $$==% % V0.x>.x>% % $$AA" FEMF+@ FGDICF(GDIC1FGDICF(GDICUFGDICF(GDIC1BFGDICRp@Arial0 fg0 ɀ\{TiK0W0fg ɀ0PU0z08g2~01~\{!0l|pЀm50lq Ѐ(\{ ( Ѐdv%    T 7KKKAJA/L|Strategy for Integration       % ( F(GDIC1lFGDICRp@Arial>0#0-!{~*Lih0-0T000PU0z08h -!{~\{Ѐdvdvdvdv@Arialw i ip i|0 idv%    TT<sGKKKAJA<LP"  % ( Rp@Arial|-w(!`|-w 0w j(!|6w;w 0w G-nwNnw j80PU0z08g j0 D\{@Arial\{Ѐm5000h8<hр Ѐdv%    T,bsKKKAJAb%LKnow basic integration table (pg 506)a            % ( Rp@Arial>0#0-!{~*Li ~h0-0T000PU0z08h -!{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TT<GKKKAJA<LP"  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w PG-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8<hр Ѐdv%    TbmKKKAJAb"LSimplify the Integrand if Possible         % ( Rp@Arial>0#0-!{~*Lih0-0T000PU0z08h -!{~\{Ѐdvdvdvdv@Arialw y yp y|0 ydv%    TT<GKKKAJA<LP"  % ( Rp@Arial|-w(!`|-w 0w z(!|6w;w 0w G-nwNnw z80PU0z08g z0 D\{@Arial\{Ѐm5000h8<hр Ѐdv%    T boKKKAJAb LLook for an Obvious Substitution        % ( Rp@Arial>0#0-!{~*Lih0-0T000PU0z08h -!{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TT<G(KKKAJA<"LP"  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w PG-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8<"hр Ѐdv%    TTb(KKKAJAb",LClassify the Integrand According to its Form              % ( Rp@Arial>0#0-{~*Lih0-0T000PU0z08h -{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TTn2}QKKKAJAnLLP  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w G-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8nLhрЀdv%    T2RKKKAJALL|Trigonometric Functionsg    % ( Rp@Arial>0#0-{~*Lih0-0T000PU0z08h -{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TTnZ}yKKKAJAntLP  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w PG-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8nthрЀdv%    TZzKKKAJAtLpRational Functions   % ( Rp@Arial>0#0-{~*Lih0-0T000PU0z08h -{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TTn}KKKAJAnLP  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w G-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8nhр Ѐdv%    TKKKAJALtIntegration by Parts    % ( Rp@Arial>0#0-{~*Lih0-0T000PU0z08h -{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TTn}KKKAJAnLP  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w PG-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8nhрЀdv%    T|KKKAJAL\Radicals % ( Rp@Arial>0#0-!{~*Lih0-0T000PU0z08h -!{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TT<GKKKAJA<LP"  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w G-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8<hр Ѐdv%    TbKKKAJAb L`Try Again    % ( Rp@Arial>0#0-{~*Lih0-0T000PU0z08h -{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TTn} KKKAJAnLP  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w PG-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8nhрЀdv%    TD!KKKAJA)LBasically Only Two Methods of Integrationt     % ( Rp@Arial>0#0-{~*Li~h0-0T000PU0z08h -{~\{Ѐdvdvdvdv@Arialw  p |0 dv%    TT(CKKKAJA?LP"  % ( Rp@Arial|-w(!`|-w 0w (!|6w;w 0w G-nwNnw 80PU0z08g 0 D\{@Arial\{Ѐm5000h8?hр Ѐdv%    T'IDKKKAJA? 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