EXAMPLE3Finds (5r + 1)-5 d.THEOREM 5the Substitution Rulefag(r) isadifferemtiable fumctionwhcse range is an inlerval I,and Jis contimuous on /,thnWesubstiute=5t+ andd= 5dhThenSolutionNgt)g'(x)dr-[(n) dusee (5r + 1)-5 dt =seeudiLau-5el-3a tanr +CnvscSupposethat Fis theantiderivative off,then tan(5t + 1) +CShimtoSr + ISorn量Fg()=F(g(0)g/()Chain RalrThe Substitution Rale provides the following substitution method io evaluate the imtegnl= f(g(x)-g(x)P-1ng(t)g'(d dIf we make the substitutiom a = g(x), then[ng()g(),一/层目g(0),when f and g" are continuous functions:Substitute= gtx)and dr = (du/dt) dx= g(cx) drto obeain the integral=Fig(a) +Cteriad TheammHn=F(a)+C[o) dn.[Fr(a)dFuslarsttad Thoam2Inegrate withresp to-JrmaF-/3Replacbyt)inthreut4-31Some TipsEXAMPLE4Find cos (78 + 3) a8.We let u = 7e + 3 so that dv 7 do. The constamt factoe 7 is nSolutissmetr+ 1. Sometimes we observe that a power ofx appears in the integrandthe do term in themtegral.We for it by multiplying and dividing by 7uing thesmc procedure ain Exmple 2Ththat is onceless thanthe powerofx appearinginthe argument ofafunction we want to integrate.This observationimmediately suggestscos (70 + 3) d6 = +,eos (70 + 3)-7 d6we try a substitution for the higher power ofx.Wecanverifythissolutionby differentiating and[2ea=[e+xa-e.idaaduchecking that we obtainthe(/3)-originalfunctioncos(70+3)-fiw+cedar+Integrate wiahtespect tot= +sin(76 + 3) +:CThere is another apgrooch to ihis proelemWitha = 78 + 3 and dr = 7 as bce+cReplace wu by sfore, we solve forde to obtain dy = (1/7) dlk. Then the integral becomesThe successofthe substitutionoos(78 + 3) 0 -osa-famethoddepends onfinding a+sina + Csubstitutionthatwill changeanintegral we cannot evaluate号 sin(7 + 3) + Cdirectlyintothe onethatwe canSlide 4- 34Evaluating Indefinite IntegralsExercisesEXAMPLETV2x+idxEvaluateEvaluatc theindefinite integralsinExercises116by using the givensubstitutions to reduce the integrals to standard form.Method1 Let u-2x+1, then x=(1/2) (u-1), and dx=1/2 du, So0s(2x)+C8.rsin(2)de,w=2r2[rV2r +1d =1 / (u DVrdr 1 / (u 1)a/Par Sbme--/an-wy duMultiply term-cos+c10 (-c)sind,=1-c(sn-号wn)+c专bategrete9raw=16M-+11.V-r(2r+1)-(2x+p+CRoplaceuby2r +112.12(04+ 42+ 1PG + 20), = y*+4p2+1Method 2 Let u = 2x+1, then x= ,(u -I, and dx=u-du. So*+4y+y+CJx2x+dx=(a-1)-uud=,(-)d2-Vesin(rn = 1)d ==113,4111.s(2x+1)"_))+C:"+C:(2x+1)*"+C25101066()-sin(2 2)+CSlide 4- 35Slide 4-38
2016/11/15 6 Slide 4 - 31 Suppose that F is the antiderivative of f, then Slide 4 - 32 Slide 4 - 33 We can verify this solution by differentiating and checking that we obtain the original function cos(7θ+3) The success of the substitution method depends on finding a substitution that will change an integral we cannot evaluate directly into the one that we can. Slide 4 - 34 1. Sometimes we observe that a power of x appears in the integrand that is once less than the power of x appearing in the argument of a function we want to integrate. This observation immediately suggests we try a substitution for the higher power of x. Slide 4 - 35 Method 1 Let u=2x+1, then x=(1/2) (u-1), and dx=1/2 du. So Method 2 Let , then , and dx=u·du. So u x 2 1 1 2 ( 1) 2 x u 1 1 2 4 2 2 1 ( 1) ( ) 2 2 x x dx u u u du u u du 1 1 1 1 1 1 1 5 3 5 3 5/2 3/2 ( ) (2 1) (2 1) 2 5 3 10 6 10 6 u u C u u C x x C Slide 4 - 36 Exercises 1 2 cos(2 ) 4 x C 2 3 (1 cos ) 3 2 t C 3 6 1 r C 4 2 3 ( 4 1) y y C 1 1 3/2 3/2 ( 1) sin(2 2) 3 6 x x C
ExercisesSome TipsE1siar+c14.t，=-cos 2. An integrand may fequire some algebraic manipulafion before2x4the sutbstitution method can be applied.edx+cot(20)+C15,.ese 28 cot28 d8(a)Mulnigly by te'/e) -+ea2+1≥. Using u = cot 20b. Using cse 2Letw-en-e-c0r(20)+C,-一+1山-edcot(20)+Ccse*(20)+C,(a)-(6)-= tan'μ + Ctntegrale withrespectto= tan"i(e") + Ceplacea by e16.V5r+8h. Using i = V5r + 8seext=(sx)d=Se1-S0C1± Using r = 5r + 8omotb)(a)=/5x+8+c(6)=95x+8+ct stnsecx+tan x-/赠-(= In[a] + C = In[see x + tan | + C..Slide 4-37Some TipsSome Tips+3. We can use the subsitution method ofintegrafion as an 3. We can use the substitution method aof integrafion as anexplorationtool: Substitutefor the mosttroublesome part of theexploration tool:Substituteforthemosttroublesome partof theintegrand and seehow things work out.integrand and seehowthingswork out.2d2:dEXAMPLE8EvaluateEXAMPLE8EvaluateV+1W+ISolution 1: Substitute ur = +* + 1Solution 2: Substitute w = V + linstead2=dduLetu-+1-[-[W41-tV-2 +/u-l/ duloae fim /n-d=3/ud+℃-tategrate=3.号+CIndirgralc号u+C号(+1/+CReplaewbyt-+1s=号(2*+ 1)/ +CRepleeu liy 2+1Slide 4- 39Slide 4- 40ExercisesExercisesin(+cos(++1)+C21.+o29,Va+vsin(2#+1)31.L22,/Eos (3 + 4± =sin(3 +4)+Ccos(2+1)2cos(21+1)s(l--sin--n+cc(3x+2d=an(3+2)+C33.23.25-sine1+c34,cos(Vi+3d=2sin(F+3)+C35.26+cCosVo36.+(2. [----+c8sinyoVosiVaalgebraic-(1-)+(38.manpulationSlide 4- 41slide 4- 42
2016/11/15 7 Slide 4 - 37 Exercises 1 1 2 sin( ) 2 4 C x x 2 1 1 ( ) cot (2 ) 4 a C 2 2 1 ( ) csc (2 ) 4 b C 1 2 ( ) 5 8 5 a x C 2 2 ( ) 5 8 5 b x C 2 2 2 2 1 (1 cot (2 )) 4 1 1 cot (2 ) ( ) 4 4 C C Slide 4 - 38 2. An integrand may require some algebraic manipulation before the substitution method can be applied. Slide 4 - 39 3. We can use the substitution method of integration as an exploration tool: Substitute for the most troublesome part of the integrand and see how things work out. Slide 4 - 40 3. We can use the substitution method of integration as an exploration tool: Substitute for the most troublesome part of the integrand and see how things work out. Slide 4 - 41 Exercises 2 1 C x 1 sin(3 4) 3 z C 1 tan(3 2) 3 x C 1 6 sin 2 3 x C 1 8 tan 4 2 x C 3 6 ( 1) 18 r C Slide 4 - 42 Exercises 2 3/2 cos( 1) 3 x C 1 1 2 cos(2 1) C t 1 sin( 1) C t 2sin( 3) t C 1 2 cos 4 C 2 sin C 2 1 3/2 (1 ) 3 C x algebraic manipulation
ExercisesProperties of Indefinite Integrals18tan67.Table of fundamental integralb12+tnJee'xit--cotx+CJkdr =kr+C (K is a constant)tan x,followed byw u,then by w2+uaHh.M= tan'x.followedby y =2+6fr'dreex+C+C (a+1)see.xtan+Ce.μ=2+tanxa+12+tan'x[些=m|x]+Csee.x+Creotxa68V+ sin* (x 1)sin (x 1)cos (x = 1) dJa'dr-,_+C (a>0, a±1)aresinx+Ca. u = + 1, followed by ur = sinu, then by w = I + vVi-b. u= sin (x - 1),followed byv =1 +Jed=e+cT+1+c. u= 1 + sin (x - 1)Jcosxdk=sinx+Csinh xdx =coshx+C-=(1+sin'(x-1)32+CJsin xde--cosx+csinhx+CnshJsee' xdr =tanr+Clide 4- 431004-45.3The area under the graph ofa positive function, the distance traveled by a moving object thatdoesn't change direction, and the average value of a nongative function over an intervalcan all be approximated by finite sums. First we subdivide the interval into subintervals,treating the appropriate fiunction f as if it were constant over each particular subinterval.Estimating withFinite SumsThen we multiply the width of each subinterval by the valuc of f at some point within it,and add these products togethet. Ir the interval [a, b] is subdivided ino a subinterals arequal widths Ar (b a)/n, and if f(ca) is the value of f at the chosen point e in theth subinteral, this process givesafinitesumofefomf(ei) Ar + f(cg) dr + f(cg) Ar +*-*+ f(c,) Ar.Thechoicesforthec could maximize orminimize the value of finthe kth subinterval,orgive some value in between. The true value lies somewhere between the approximationsgiven by upper sums and lower suns. The finite sum approximations we looked at im-proved as we took more subintervals of thinner width.slide 4- 45Slide 4- 48Suppose we want to find the area of the shaded region R that lies above the x-axis, belowthe graph of y = 1 - x, and between the vertical lines x = 0 and = 1 (Figure 5.1).-1-r1.TheArea Under theGraph of a0.5Positive Function0.5FIGURE 5.1 The area of the regioR cannot be found by a simpleformula,determiming the eaictarea ofceptforapproxmatingSlide 4- 47
2016/11/15 8 Slide 4 - 43 Exercises 3 6 2 tan C x 1 2 3/2 (1 sin ( 1)) 3 x C •44Slide 4 - 44 Properties of Indefinite Integrals Table of fundamental integrals kdx kx C (k is a constant) 2 csc cot xdx x C 1 1 x x dx C ( 1) sec tan sec x xdx x C ln | | dx x C x csc cot sec x xdx x C ln x x a a dx C a (a 0, a 1) 2 arcsin 1 dx x C x x x e dx e C 2 arctan 1 dx x C x cos sin xdx x C sinh cosh xdx x C sin cos xdx x C cosh sinh xdx x C 2 sec tan xdx x C Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 4 - 45 5.3 Estimating with Finite Sums Slide 4 - 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 47 1. The Area Under the Graph of a Positive Function Slide 4 - 48 No method for determining the exact area of R except for approximating
The upper sum is obtained by taking the heightThe lower sum is obtained by taking the heightofeach rectangle as the maximum value of f(x)ofeach rectangle as the minimum value off(x)y=l-y= 1 32for a point x in each base interval of thefor a point x in each base interval of therectangle.rectangle.The length of each.The sum of the areas of the n rectangles issubinterval is.xf(Gla...4The sum of the areas of the n rectangles is-Sorecf(c)s..24ZoIn each of our computed sums, the interval [a, b] over which the function f is definedThe lkength ofeachwas subdivided into n subintervals of equal width (also ealled length) Js (b - a)/n,and / was evaluatod at a point in cach subinterval: , in the first subinterval, cz in the see-subnterval isAxond subinterval, and so on. The finite sums then all take the form(ei)x + f(e) + f(e) x +*++ f(e) r.By taking more and more rectangles, with each rectangle thinner than before, it appearsthat these finite sums give better and better approximations to the true area of the region Rlide 4- 49Slide 4-50When we don' know an antiderivative for the velocity function (c), we can apply thesame principle of approximating the distance traveled with finitesums ina way similartour estimates forarea discussed before, Wesubdivide the intenal[a, ] into short time intervals on each of which the velocity is considered to be fairly constant. Then we approxi-mate the distance traveled on each time subinterval with the usual distance formuladistance = velocity × timeand add the results across [a, b].2.The Distance Traveled byaSuppose the subdivided intenal looks likeMoving Object Without Changing-ar--4r--r-Direction+ (sec).with the subintervals all of equal length r, Pick a number tj in the first interval. If r isso small that the velocity barely changes over a short time interval of duration ,then thdistance traveled in the first time interval is about u(r) t, If ta is a mumber in the secondinterval, the distance traveled in the second time interval is about u(t) At. The sum of thedistances traveled over all the time intervals isD (h) Ar + o(r) A +*+ () r,where x is the total mumber of subintervals.Slide 4- 51EXAMPLE2The velocity function of a projectile fired straight into the air is(b)Thieebinasflengh,ihuluteghtndpointsgvingowerf(r) =160 -9.&r m/sec.Use the summation technique just described to estimate howfar the projectile rises during the first 3 sec, How close-do the sums come to the exact.value of 435.9 m?-ar-jSolutionWeexploretheresults for differentnumbers of intervals anddifferent choicesofvaluationoints.Notice that isdceasing, sochoosing tenointsgives anWith f evalunted at t = 1, 2, and 3, we haveper sum estimate; choosing right endpoints gives a lower sum estimate.D Jn) r + F(n) r + f(a) r(a)Thee subintenuls of length 1,with fevauatedar lefi endpointsghing anupper sam:= [160 9.8(1)](1) + [160 9.8(2)](1) + [160 9.8(3)](1)格= 421.2.X02[-ar-With evaluated at =0, 1,and 2, we hveD () + () + f) = [160 9.8(0)](1) + [160 9.8(1)](1) + [160 9.8(2)](1) 450.6.Slide 4- 53Slide 4- 54
2016/11/15 9 Slide 4 - 49 1 2 ( ) ( ) ( ) n f c x f c x f c x The sum of the areas of the n rectangles is C1 C2C3 Cn The length of each subinterval is ⊿x The lower sum is obtained by taking the height of each rectangle as the minimum value of f(x) for a point x in each base interval of the rectangle. Slide 4 - 50 1 2 ( ) ( ) ( ) n f c x f c x f c x The sum of the areas of the n rectangles is C1 C2 C3 Cn The length of each subinterval is ⊿x The upper sum is obtained by taking the height of each rectangle as the maximum value of f(x) for a point x in each base interval of the rectangle. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 4 - 51 2. The Distance Traveled by a Moving Object Without Changing Direction Slide 4 - 52 Slide 4 - 53 Slide 4 - 54