Review Topics for Midterm 1 in Calculus 1B

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1 Review Topics for Midterm 1 in Clculus 1B Instructor: Zvezdelin Stnkov 1. Definitions Be ble to write precise definitions for ny of the following concepts (where pproprite: both in words nd in symbols), to give emples of ech definition, to prove tht these definitions re stisfied in specific emples. Wherever pproprite, be ble to grph emples for ech definition. Wht is (1) du = u ()d? Is this formul? Wht does it men nd where is it used? () udv nd vdu? Are these formuls? Wht re they used for? (3) trigonometric substitution? When is it used? Geometric interprettion. Tble of trig. substitution. (4) trigonometric integrl? Emples. Which type of trigonometric integrls hve we lerned to solve? (5) rtionl function? Wht does it men to represent it in terms of prtil frctions? (6) long division? When do we need to use long division? (7) n irreducible qudrtic fctor? How do we complete the squre in such qudrtic polynomil? (8) rtionlizing substitution? When do we use it? (9) rtionlizing numertor or denomintor? Is this the sme s rtionlizing substitution? (10) Riemnn sum R n, L n or M n? An pproimtion T n or S n? How re these sums relted to one nother? Which re epected to be more precise thn the others? Wht is n error bound for sum nd how do we use it? Wht is the difference between n pproimtion, the error of the pproimtion, nd n estimte for the error of the pproimtion? (11) n improper integrl? How mny types of improper integrls re there? When is n improper integrl convergent nd when is it divergent? How cn we compre two improper integrls nd use knowledge of convergence/divergence for one of them to conclude convergence/divergence for the other? Why do we cre if n improper integrl converges or diverges? (1) n indetermincy? How mny different types of indetermincies re there? When do indetermincies rise? How re limits nd L Hospitl s rule relted to indetermincies? Where does this stuff pper in wht we hve lerned so fr?. Theorems Be ble to write wht ech of the following theorems (lws, propositions, corollries, etc.) sys. Be sure to understnd, distinguish nd stte the conditions (hypothesis) of ech theorem nd its conclusion. Be prepred to give emples for ech theorem, nd most importntly, to pply ech theorem ppropritely in problems. The ltter mens: decide which theorem to use, check (in writing!) tht ll conditions of your theorem re stisfied in the problem in question, nd then stte (in writing!) the conclusion of the theorem using the specifics of your problem. (1) The Fundmentl Theorem of Clculus: The definite integrl of f() equls ny ntiderivtive F () of f(), evluted t the ends of the intervl, i.e. b f()d = F (b) F () where F () = f(). () Tble for Direct Integrtion: must be committed to memory for efficient direct integrtion. Summrize your clssnotes nd the tetbook references in tble which includes the most common directly integrble functions. Do not miss formuls in 7.5: you do not wnt to rederive these formuls on the em, but rther quickly use them. (3) Substitution in Indefinite Integrls: f(g())g ()d = f(u)du (u = g(), du = u ()d.) (4) Substitution in Definite Integrls: b f(g())g ()d = u(b) u() f(u)du (u = g(), du = u ()d.) (5) Integrtion by Prts for indefinite integrls: udv = uv vdu, or: f()g ()d = f()g() g()f ()d.

2 (6) Integrtion by Prts for definite integrls: b udv = uv b b vdu, or b f()g ()d = f()g() b b g()f ()d. (7) Trigonometric Identities. () Pythgorus Theorem: sin + cos = 1; replcement: cos = 1 sin ; sin = 1 cos. (b) Hlf-ngle formuls (reduction of degree): sin = 1 (1 cos ) nd cos = 1 (1 + cos ). (c) Double-ngle formuls: sin = sin cos, cos = cos sin. (d) Turning products into sums: sin cos y = 1 (sin( y) + sin( + y)); sin sin y = 1 (cos( y) cos( + y)); cos cos y = 1 (cos( y) + cos( + y)). (8) Prtil Frction Representtion Theorem. Any rtionl function f()/g() cn be written uniquely s sum of simpler frctions. Specificlly, ny liner fctor of g() contributes s follows: f() ( ) n = A 1 + A ( ) + A 3 ( ) A n 1 ( ) n 1 + A n ( ) n And ny irredicible qudrtic fctor contributes s follows: f() ( + b + c) n = B 1 + C 1 + b + c + B + C ( + b + c) + + B n 1 + C n 1 ( + b + c) n 1 + B n + C n ( + b + c) n. However, if degf() deg g(), then long division is in order to peel off polynomil before ny of the bove prtil frction representtions re used. To ctully integrte one of the prtil frctions with n irreducible qudrtic in denomintor, one must complete the squre in the denomintor, mke mild liner substitution, nd then split the integrl ppropritely long the numertor before finlly integrting. (9) Trpezoidl nd Simpson s Approimtions. We cn pproimte the re under continuous function f() on [, b] by trpezoids nd by prbols: () T n = [f( 0 ) + f( 1 ) + f( ) + + f( n 1 ) + f( n )] (b) S n = [f( 0 ) + 4f( 1 ) + f( ) + 4f( 3 ) + f( 4 ) + + f( n ) + 4f( n 1 ) + f( n )] 3 Note tht Simpson s pproimtion works only with even number of subintervls of [, b], hence the even nottion n for the inde of S. (10) Error Bounds for Integrl Approimtions. Let f() be function on [, b] whose second derivtive cn be bounded by number K: f () K for ll [, b]. Then the errors E Tn nd E Mn of the trpezoidl nd midpoint pproimtions of b f()d re bounded s follows: E Tn K(b )3 K(b )3 1n, nd E Mn 4n If the fourth derivtive of f() cn be bounded by number N: f (4) () N for ll [, b], then the error E Sn of the Simpson s pproimtion of b f()d is bounded s follows: N(b )5 E Sn 180n 4 (here n cn be only even.) Note tht the bove formuls re not pproimtions of the integrl, neither re they the ect errors of T n, M n or S n, respectively. These formuls give us n upper bound for the errors in these pproimtions, i.e. they tell us the mimum (worse) error tht ech of T n, M n or S n cn hve. The ctul errors could be much smller, but usully there is no wy to find the ctul errors; hence, we contend ourselves with knowing bound for the errors. (11) Comprison Theorem for improper integrls. If f() nd g() re continuous functions on [, ), (, b] or (, ), nd f() g() 0 for ll in the domin intervl, nd () If f() d converges, then g() d lso converges, nd g() d < f() d. (b) If g() d diverges, then f() d lso diverges. Note tht the bove is written only for the intervl [, ), but nlogous sttements work for ny infinite invervl.

3 3. Problem Solving Techniques (1) How do we guess u() nd u () in the substitution rule? There isn t ny bullet-proof method for guessing (tht s why it s clled guessing fterll), but here re severl tips tht will suffice in ll problems of Clculus II. By pplying the Substitution Rule we re mking n ttempt to reverse the effect of differentition by the Chin Rule. Thus, we wnt to represent our function s if it were the result of Chin Rule: f(g()) g (), where u = g() = blh; in other words, we re trying to write our function s product of u-prt nd u ()d-prt. The key to correct guessing is to remember tht, whtever u() turns out to be, the derivtive u () must pper multiplied by d! () Try to locte u (): ll cndidtes for u () re mong the things with which d is multiplied. (b) For emple, 3 cos( 3 )d yields two possibilities for u (): u () = 3 nd u () = cos( 3 ). Now let s recll tht we need to determine lso u(), i.e. we hve to do direct integrtion on our chosen u (). Obviously, the choice u () = cos( 3 ) is bd becuse we don t know how to directly integrte cos( 3 ). However, the choice u () = 3 works very well, since we cn integrte immeditely: u() = 3 + C, nd notice tht 3 ppers in the rest of the epression. Thus, we set u() = 3 nd du = ( 3 ) d = 3 d, nd the substitution rule yields: 3 cos( 3 )d = cos( 3 )(3 d) = cos(u)du = sin(u) + C = sin( 3 ) + C. Don t forget to check the nswer by differentition!! (c) Recll tht denomintors my lso yield something relevnt for u (). For emple, in rctn 1+ d it will be unsuitble to set u () = rctn() (why?), but it will be very suitble to set u () = 1 1+ : then u() = 1 1+ d = rctn() nd du = 1 1+ d so tht ( ) rctn d = rctn 1 + d = udu = u + C = rctn () + C. Don t forget to check the nswer by differentition!! (d) Sometimes we hve redjust u () nd u() by constnts. For emple, in cos( 3 7)d we cn initilly guess u () = so tht u() = C. But this u() does not conveniently pper in the rest of the function! Insted, we would hve liked tht u() = 3 7: this is hrmless little wish, which cn be stisfied on the spot by setting du = ( 3 7) d = 3 d. We cn then solve du = 3 d: d = 1 3du, nd substitute in the integrl: 1 cos( 3 7)d = 3 cos(u)du = 1 3 sin(u) + C = 1 3 sin(3 7) + C. Don t forget to check the nswer by differentition!! (e) We lern from the bove tht fter hving mde choice for u (), nd hence finding u(), we cn redjust u() by replcing it by ny convenient epression of the form C 1 u() + C where C 1 nd C re some suitble constnts. This forces slight chnge in u (): multipliction by the constnt C 1. To summrize: lwys keep in mind tht certin degrees of freedom in our choices for u() nd u () eist: for u() there re degrees of freedom, nd for u () there is only 1 degree of freedom. Be wre tht mong ll possible choices for u() nd u () there is one most convenient choice: it is dictted by wht we would like u() to be in our originl function. (f) Sometimes it my pper s if there re no cndidtes for the role of u (). For emple, in e 6 d, d is multiplied only by e 6, so wht cn u () be? It cn be constnt: it cn be ny constnt tht we like. So, we look for possible u(): it looks like u = 6 would be nice; hence u () = 6d, d = 1 6du, nd e 6 d = 1 6 eu du = 1 6 eu + C = 1 6 e6 + C. 3

4 Don t forget to check the nswer by differentition!! (g) Finlly, some problems re trickier thn we would hve liked. The trouble there, surprisingly, is not in the guessing of u() nd u (), but rther in trying to get rid of ll s nd replcing them by u s. For emple, in d, it is cler tht we definitely wnt to get rid of + 1 under the rdicl, so we substitute u() = + 1, du = d, but then how do we get rid of the remining 4 in our function? We solve for 4 from our substitution eqution: u = + 1 = u 1 4 = (u 1), so tht our integrl will look like: +1 5 d= +1 4 d= u(u 1) du = 1 u(u u+1)du= 1 (u 5/ u 3/ +u 1/ )du Finish this emple s we did in clss, nd check with differentition. (h) Continuing with the men strek of substitution rule problems, check out some of the review problem in Chpter 5. Don t forget lso the trick in #41 ( 5.5) of splitting the integrl s sum of two integrls; lso, don t forget tht functions mybe even/odd, thus simplifying integrtion on symmetric intervls [, ]. () How do we locte u nd v in Integrtion by Prts? The ide of Integrtion by Prts is to represent our function s product of two functions f() nd g () such tht it is esier to integrte f ()g()d thn the originl integrl f()g ()d. It is usully obvious how to represent our originl function s product of two functions. Wht is not obvious is to decide which of these functions will ply the role of u = f() nd which will ply the role of v = g(). As rule of thumb, the function which hs simpler derivtive plys the role of u = f(), nd the function which hs simple enough integrl plys the role of v = g(). () Polynomils re ecellent cndidtes for u = f() becuse their derivtives re simpler functions: they re polynomils of lower degrees! (b) The clssicl inverse functions such s inverse trig functions (rcsin(), rccos(), rctn(), rccot()) nd logrithmic functions (ln()) re lso ecellent cndidtes for u = f() becuse we know nothing bout their integrls, but we do know lot bout their derivtives. (c) Eponentil functions (e.g. e ) nd trigonometric functions like sin nd cos cn be used both s u = f() nd v = g(), depending on the needs of the other involved function. The reson for this neutrl behvior of e, sin nd cos is tht both their derivtives nd integrls re s complicted s the originl functions themselves. (d) Emple of product of polynomil nd trig. function: 3 cos d. We set u = 3 becuse the derivtive u = 3 is n esier function. Hence v = cos, v = sin. 3 cos d = uv vdu = 3 sin 3 sin d. We cn now gin pply Integrtion by Prts on the lst integrl by setting u = 3, v = sin. This will further reduce the degree of the polynomil: we ll end up hving to find 6 cos d. One lst integrtion by prts with u = 6 nd v = cos will reduce everything to finding 6 sin d. Mke sure you complete this whole clcultion crefully nd substitute bck ll your nswers into the originl integrl. (e) Emple of product of polynomil nd n eponentil function: e d. By nlogy with (d), we set u = nd v = e, so tht u = 1 nd v = e : e d = e e d = e e + C. The ide here ws gin to reduce the power of the polynomil. (f) Wht if there is no product? For emple, rcsin d? We recll one of the gretest tricks in mthemtics: multiplying by 1! (The other trick is dding 0. :)) Thus, we cn set u = rcsin, v = 1, so tht u = 1/ 1 nd v = : rcsin d = rcsin 4 1 d.

5 Now we use the substitution rule: set u = 1 so tht du = d, nd d = 1 1 u du = 1 u 1/ du = u 1/ + C = 1 + C. Finlly, rcsin d = rcsin C. Check by differentition!! (g) An emple of product of n eponentil nd trigonometric function: e cos d. Since neither e nor cos hs simpler derivtive (or integrl), it doesn t mtter how we pply integrtion by prts here. So, sy, u = e, v = cos, u = e, v = sin : e cos d = e sin e sin d. OK, we didn t get nything simpler, but wht if we venture once gin nd do similr integrtion by prts: u = e, v = sin, so tht u = e, v = cos : e sin d = e cos + e cos d. Substituting bove, e cos d = e sin ( e cos + e cos d) = e (sin + cos ) e cos d e cos d = 1 e (sin + cos ). (1) (h) For etr chllenge, review E. 6 in 7.1, nd do #4-44 in 7.1. (3) When do we decide to use trigonometric substitution? When the integrnd hs rdicl of the form: +, or. Ech of these rdicls requires different trig. substitution. Check your clssnotes for detils on the individul substitutions. It is lwys useful to drw right tringle nd lbel it ppropritely ccording to the Pythgorus theorem, so tht the unwnted rdicl is epressed s one of the sides of the tringle. The gol of the trigonometric substitution is to end with trig. integrl, e.g. for some integer n nd m: sin n cos m d. A word of cution: sometimes, rdicls cn be delt with by direct u-substitution, even if they contin rdicls of the bove form. So, before you embrk on trig. substitution, mke sure you re not missing simpler substitution nd fster wy to solve the problem. (4) How do we compute trigonometric integrls? By trig. integrls, we understnd integrls of form in (1). We hve studied two methods for solving trig. integrls: () Peel off, replce nd do u-substitution: this works if one (or both) trig. functions pper in odd powers. For emple, if the sine function ppers in odd power (e.g. sin 7 ), then peel off one sine s sin d, replce the remining sines (e.g. sin 6 = (sin ) 3 = (1 cos ) 3 ), nd substitute the cosine function (e.g. u = cos, du = sin d): sin 7 cos 4 d = sin 6 cos 4 sin d = (1 cos ) 3 cos 4 sin d = (1 u ) 3 u 4 du. Multiply through nd integrte the resulting polynomil directly. Don t forget to substitute bck cos. As generl rule, whtever function you peel off, you ll end up substituting the other function. 5

6 If you hve both sine nd cosine ppering in odd powers, usully it is better to peel off the function in the smller power in order to hve simpler multipliction of polynomils t the end: sin 7 cos 5 d = sin 7 cos 4 cos d (finish it off! don t be lzy!) (b) Reduce powers, multiply through, then see if peel off will work or gin reduction of power, nd so on. Reduction of powers works only if both trig. functions pper in even powers, e.g. ( ) 3 ( ) 1 cos() 1 + cos() sin 6 cos 4 d = (sin ) 3 (cos ) d = d = = 1 5 (1 3 cos() + 3 cos () cos 3 ())(1 + cos() + cos () ) d = = 1 5 (1 cos() cos () + 5 cos 3 () cos 4 () cos 5 () ) d = = 1 5 ( cos() + 5 cos 3 () cos 5 () ) d + (1 cos () cos 4 () ) d The integrls in the first btch will be delt with by peel off, while the integrls in the second btch will fll under reduction of power. (Why don t you check the multiplictions bove nd let me know if there is typo?... Checking who s reding this... :)) Note tht if the integrnd hs only one trig. function, it still flls within one of the two methods described bove, e.g. sin 5 d will peel off, while sin 6 d will require reduction of the power 6. A further finessing of the bove method cn be done if the integrnd is frction of trig. functions. If n odd power is in the numertor, the peel off method will turn the integrnd into rtionl function, nd then splitting of the integrl will finish the job, e.g. cos 3 cos sin 4 d = 1 sin 1 u sin 4 cos d = sin 4 cos d = u 4 du = u 4 du u du... If n odd power is in the denomintor, then we cn force peel off in the numertor by multiplying ppropritely top nd bottom: cos 4 sin 4 sin d = cos 4 sin 3 d = cos 4 (1 cos sin d = ) u 4 (1 u ) du = u 4 (1 u) (1 + u) du. The method of prtil frctions will finish the job here. (5) How do we pply prtil frctions to integrtion? The method of prtil frctions pplies f() when the integrnd is rtionl functions, i.e. rtio of two polynomils s in g() d. f() r() () If deg f() deg g(), then pply long division to peel off polynomil: = h() + g() g(). Here h() is the result of the division of f() by g(), nd r() is the reminder of this division. Thus, the integrl becomes: ( f() g() d = h() + r() ) r() d = h()d + g() g() d. We know how to integrte h(), while deg r() < deg g(), so the second integrl is solved using prtil frctions s described below. (b) If deg f() < deg g(), fctor g() s product of liner nd irreducible qudrtic polynomils. Then use the Prtil Frction Representtion Theorem to write the integrnd f()/g() s sum of severl simple frctions. Net, get rid of ll denomintors, equte ll coefficients nd mke system of liner equtions. Solve this system to find the unknown constnts A, B, C,... Finlly, plug these constnts into your prtil frction representtion of f() nd integrte ech such frction seprtely (see (c) below). 6

7 Wrning: sy, you re integrting ( 1) (+) d. A JEOPARDY question: The degree of the top is bigger thn degree of the bottom, so long division is in order; wht should we do first: the long division nd then prtil frctions, or first prtil frctions nd then long division? Answer: neither of the bove becuse, even though we hve to do long division before prtil frctions, we cnnot long divide by fctored polynomil. So, multiply through the denomintor, then do long division, nd then the new frction will hve the sme denomintor, so you cn fctor it bck s it ws before: ( 1) ( + ) d = d = + 5 ( + 1)d d = = ( 1) ( + ) d. Only now the lst frction is in form suitble for prtil frctions. Is there mistke bove? Finish the integrtion. (c) Recll the direct nd u-substitution integrtion formuls: 1 ( ) n+1 1 d = + C when n 1; d = ln( ) + C; ( ) n n + 1 ( ) 1 ( + ) d = 1 ( ) rctn + C; ( + ) n d = 1 du u n =... Wht ws substituted in the lst integrl? Finish the integrtion there nd mke it into formul to use for prtil frctions in cse of dire need. :) (d) Sometimes irreducible qudrtic fctors require completion of the squre, s in d: = 3( 4 + 8) = 3( ) = 3(( ) + 4). So, our integrl becomes: + 3(( ) + 4) d = 1 u u + 4 du = 1 3 u u + 4 du u + 4 du. where u =, du = d, = u +, nd we used splitting of the frction for the lst equlity. Now, for the first integrl we will use substitution: t = u + 4, while for the second integrl we will use direct formul of integrtion (involving rctn). Finish the integrtion nd don t forget to substitute bck. The generl formul for completion of squre is s follows: ( ) ( b b + b + c = + b + ) + c = ( + b ) + c ( ) b. In other words, wht you complete the squre with, ( b ), is obtined by tking the coefficient b of, dividing it by, nd then squring the frction b/. Wrning: the bove formul works only if the coefficient of is 1! If hs some other coefficient, e.g. 3 s in the emple bove, then you must first fctor out 3 from the whole epression, nd then pply the formul for completion of the squre. (e) The creful reder will hve noticed tht one type of frction is missing from the formuls in (c) (this is just for bonus): 1 ( + ) n d = A A ( + ) + A 3 ( + ) A n 1 ( + ) n 1 + A nrctn( + ) + C. This is of no use unless we find out the constnts A 1, A,..., A n. To this end, differentite RHS nd set this derivtive to equl the integrnd 1/( + ). Then follow the well-known pth: get rid of denomintors, equte coefficients to obtin system of equtions, solve this system for A 1, A,..., A n, nd voil - plugging these numbers bove yields the integrl we were looking for. 7

8 (6) How do we decide which method to use in solving integrls? This is by fr the hrdest, but most rewrding, prt of solving integrls. You re given problem, but not given strtegy or hint how to do it. It is up to you to mke educted choices nd try them out until one works out. Think hed of the gme : wht kind of functions cn you integrte for sure, so try to reduce the given integrnds to functions tht you like. () Simplify the integrnd if possible. This my involve one or more of the following: cncelling, multiplyling through, replcing trig. functions only by the bsic trig. functions sin nd cos, replcing epressions using formuls (e.g. double ngle formuls), etc. (b) Look for obvious substitutions: this mens tht you see some function in the numertor, which you identify s the derivtive of your substitution, e.g. sin cos + 7 d = 1 u du. Wht did I substitute? Finish the problem. (c) Describe the type of function you re integrting nd try to find method tht corresponds to this type of functions. (This is only if the first two steps didn t yield solution.) (i) Trigonometric functions: sin n cos m for integers n nd m. Use the methods described erlier: peel off or reduction of power. (ii) Rtionl functions: f()/g() for two polynomils f() nd g(). Use prtil frctions (with long division, if necessry). (iii) Product of functions: f() g(). Try integrtion by prts: you need to decide which will be u nd which will be v. As generl rule, u should be the function whose derivtive is simpler function, while v should be function whose ntiderivtive is not hrd to find even though it my be more complicted thn v itself. Look t your clssnotes for emples of different types in which we decided on u nd v. Even if there is only one function, f(), we my sometimes find it dvntgeous to think of it s product of 1 f(), nd set u = f() nd v = 1. This is especilly helpful with horrible functions like f() = ln, rctn, rcsin, etc. We would relly like to get rid of these functions nd replce them by their derivtives, nd integrtion by prts will do ectly tht. (iv) Rdicls. These cn be delt with in different wys. (A) +, or usully cll for trig. substitution. (B) n f() usully clls for full rtionlizing substitution: i.e. u = n f(), u n = f(), n u n 1 du = f ()d. Note how we delt with the differentil opertors du nd d: this is shortcut, which proves to be very useful in such problems nd sves lot of trouble. (C) n f() sometimes clls for n under-the-rdicl substitution: i.e. u = f(), du = f ()d, etc. (D) Rtionlizing numertor or denomintor lso counts s wy to del with rdicls. Recll how we rtionlize epressions of the form... ±... (d) If everything else fils, (i) Try unobvious substitution: this mens tht you don t yet see u () in the numertor, but you still would like to substitute some function u = f(). Try it out nd del with the consequences lter. It my or my not work out. A prime emple of this is E. in 7.5. (ii) Do something unepected: mnipulte (leglly!) the integrnd, e.g. force rtionlizing substitution of (1 + sin ) in the numertor by multiplying top nd bottom by (1 sin ); use trig. identities, split the integrl into two or more integrls using splitting of frctions, etc. (iii) Recll n erlier problem to which you cn reduce the present integrl. 8

9 (iv) Don t give up. Try ll over gin from the beginning. You must hve missed something since the em problem cn t be tht hrd! (7) How do we find the bounds K nd N for f () nd f (4) () on [, b] in order to pply the error bound formuls for T n, M n nd S n? There is no single recipe for finding K nd N, just like there is no single wy for proving inequlities. Here re some ides, which by no mens summrize ll possibilities. () If f() is frction f() = g()/h(), try to find n upper bound for the numertor nd lower bound for the denomintor, i.e. g() A nd h() B for ll [, b]. Then conclude tht f() A/B. In other words, incresing the numertor nd decresing the denomintor will increse the frction. Wrning: we hve used the bsolutely vlue signs everywhere on the involved functions in order to mke sure tht ll involved numbers re positive nd hence there is no chnge of inequlity signs on the wy to obtining the bounds A, B nd A/B. (b) When sin nd cos re involved, the well-known bounds sin 1 nd cos 1 for ll re sometimes useful. (c) When f() is incresing on [, b] (e.g. check f () 0 for ll [, b]), then f() f() f(b), i.e. minimum occurs in the beginning nd mimum occurs t the end of the intervl. This observtion pplies eqully well to prts of f() which re obviously incresing. (Anlogous ides pply if the involved functions re decresing.: the mimum is obtined t the beginning of the intervl.) For instnce, f() = ( + 5)/( ) on [ 4, 3] cn be bounded s follows: since + 5 is positive nd decresing on [ 4, 3] (why?) nd is negtive nd incresing on [ 4, 3] (why?), then + 5 ttins its mimum t the beginning = 4 of the intervl, nd being positive nd decresing, ttins its minimum t the end = 3 of the intervl, i.e. + 5 = + 5 ( 4) + 5 = 1 nd 3 = 5. Using gin the ide of incresing the numertor nd decresing the denomintor, we rrived t f() = for ll [ 4, 3]. (d) When f() or prt of it involves sums/differences of other functions, then the tringle inequlity is useful: ±b + b nd ±b b. For instnce, if f() = (sin 7)/( 3 4+cos ) on [10, 0], f() = sin 7 sin cos cos = Note tht in the denomintor we used tht 3 4 is incresing on [10, 0] (why? check derivtive), hence its minimum vlue is obtined t = 10, while the minimum of cos is 1; so putting this together, we rrived t cos ( ) 1 = 959. (8) How do we use the error bounds for T n, M n nd S n? If problem sks to clculte b f()d ccurte to within 0.01 using, sy, Simpson s pproimtions, we must go through severl steps: () Clculte f (4) (); (b) Find n upper bound N for f (4) () on [, b], i.e. f (4) () N for ll [, b]; (c) Set the error bound inequlities N(b )5 E Sn 180n nd solve for n: 4 N(b ) n. Round up the number on the LHS (left-hnd side) to the nerest integer, e.g. if n, we round up to n = 893. Yet, for the Simpson s pproimtion we need n to be even, so n = 894. (d) Clculte S n for your nswer for n from (c), e.g. S 894. Conclude tht S 894 is within 0.01 of the ect vlue of b f()d. 9

10 Devise similr lgorithms when using midpoint M n or trpezoidl T n pproimtions. (9) How do we compute improper integrls? How do we decide when they converge or diverge? () First, decide if indeed your integrl is improper. There could be one or more of the following resons. (i) The intervl of integrtion is infinite, e.g. [, ), (, b], or (, + ). In ech cse, we turn the integrl into limit ordinry (proper) integrls by replcing the ± by new vrible t, nd letting t ± : f()d = f()d = lim c t t f()d + f()d; c b f()d = f()d = c lim t t lim t b t f()d + lim f()d; t t c f()d. Note tht the splitting point c in the third cse cn be chosen s it is convenient for our clcultions, so it is our responsibility to choose suitble c. If the function f() is odd or even, it is resonble to choose c = 0. (ii) The function f() hs n infinite discontinuity in the intervl of integrtion [, b]. So, find where f() hs this discontinuity, sy, t = c, split the integrl into two prts, replce c in ech prt by t, nd let t c + nd t c, s pproprite: b f()d = c f()d + b c f()d = lim t c t f()d + lim t c + b t f()d. Wht one hs to remember here is gin tht we represent the originl improper integrl s limit of ordinry proper integrls. (b) After you hve determined why the integrl is improper nd you hve replced it by limits of proper integrls, it is obvious how to proceed: clculte ech (proper) integrl using stndrd integrtion techniques, nd then tke the corresponding limits to figure out the finl vlue of the originl improper integrl. (c) In cse there re two or more limits of proper integrls involved, do not forget bout the iron curtin between the vrious integrls tht come up in the clcultion: you hve to perform ll clcultions seprtely on both sides of the curtin. If you get even one infinity (plus or minus) somewhere, then forget the whole thing - the originl integrl diverges nd tht s it! The only cse when the iron curtin will go down, is when everywhere you get finite numbers: only then you re llowed to put them together to come up with the finl vlue for the originl improper integrl (which will be convergent in this cse.) (10) How do we use the Comprison Theorem for improper integrls? Sy, you re trying to find whether b f()d converges or diverges, but you cnnot (or do not wnt to) compute the ntiderivtive of f(). The only solution is if you know of nother improper integrl which cn be compred to the originl integrl. Thus, look crefully t your function f(): cn you simplify it to some similr function g() whose ntiderivtive (nd hence integrl) you know how to find esily? Emples of such twin functions re s follows: f() = e nd g() = e, f() = 4.78 nd g() = 5 or g() = 4, etc. In ny cse, clculte b g()d, nd then compre both g() to f(), nd b g()d to b f()d. Be creful when the Comprison Theorem pplies nd when it doesn t: e.g. cn we conclude nything for sure bout f() d if g() d diverges nd 0 f() g()? if g() d converges nd 0 g() f()? 10

11 4. Useful Formuls nd Miscellneous Fcts (1) The formul describing circle of rdius r s function: y = ± r or = ± r y. () Qudrtic formul for solving qudrtic equtions. (3) Formuls for re nd circumference of circle, the volume nd surfce re of sphere of rdius r: πr, πr, 4 3 πr3, nd 4πr, respectively. How bout the corresponding formuls for ellipse / + y /b = 1 nd the body resulting from revolving this ellipse bout the -is? As n etr chllenge, how bout clculting the volume nd surfce re of torus (#61 in 6.)? (4) Formuls for res between curves nd for volumes of solids of revolution. (5) Frction mnipultions, eponentil nd logrithmic mnipultions nd formuls. Mnipultions with the sigm nottion: going bck nd forth between sigm nottion nd epnded nottion. (6) Eponentil nd logrithmic mnipultions nd formuls, e.g. n b = ( ) n, n b ln + ln b = ln b, ln ln b = ln b, c ln = ln(c ). (7) Algebric Formuls, e.g. (A ± B) = A ± AB + B, (A + B) 3 = A 3 + 3A B + 3AB + B 3, (A B) 3 = A 3 3A B + 3AB B 3. (8) Differentition Lws. 5. Problems for Review The em will be bsed on Howework nd clss problems. Review ll homework problems, nd ll your clssnotes nd discussion notes. Such thorough review should be enough to do well on the em. If you wnt to give yourself mock-em, select 6 representtive hrd problems from vrious HW ssignments, give yourself 1 hour nd 15 minutes, nd then compre your solutions to the HW solutions. If you didn t mnge to do some problems, nlyze for yourself wht went wrong, which res, concepts nd theorems you should study in more depth, nd if you rn out of time, think bout how to mnge your time better during the upcoming em. 6. No Clcultors during the Em. Chet Sheet nd Studying for the Em No Clcultors will be llowed during the em. Anyone cught using clcultor will be disqulified from the em. If you find ny typos in this review hndout: don t pnic! You know tht there re lwys few typos plnted on purpose to check if students re reding crefully nd redoing the stuff themselves. :) For the em, you re llowed to hve chet sheet - one side of regulr sheet. You cn write whtever you wish there, under the following conditions: The whole chet sheet must be hndwritten by your own hnd! No eroing, no copying, (nd for tht mtter, no tering pges from the tetbook nd psting them onto your chet sheet.) Any violtion of these rules will disqulify your chet sheet nd my end in your own disqulifiction from the em. I my decide to rndomly check your chet sheets, so let s ply it fir nd squre. :) Don t be freksurus! Strt studying for the em severl dys in dvnce, nd prepre your chet sheet t lest dys in dvnce. This will give you enough time to become fmilir with your chet sheet nd be ble to use it more efficiently on the em. Do NOT overstudy on the dy of the em!! No sleeping the night before the em due to crmming, or more thn 3 hours of mth study on the dy of the em is counterproductive! No kidding! 11