Math 113 Exam 1Review


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1 Mth 113 Exm 1Review September 26, 2016 Exm 1 covers in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between Curves This section covers how to compute the re between two curves. The min result is the next theorem. Theorem 0.1 The re A of the region bounded by the curves y = f(x), y = g(x), nd the lines x =, x = b, where f nd g re continuous nd f(x) g(x) for ll x b, is A = [f(x) g(x)]dx. To find the re between two curves you need to do the following. 1. Find the set of points such tht f(x) = g(x). 2. Let [x 0, x 1 ] be n intervl such tht f(x 0 ) = g(x 0 ), f(x 1 ) = g(x 1 ), nd for ll x 0 < x < x 1 we know tht f(x) g(x). Then fix point x (x 0, x 1 ) nd see if f(x) or g(x) is greter. Whichever is greter will be greter throughout the entire intervl (by the Intermedite Vlue Theorem). 3. Integrte over the intervls using the bove informtion from (2). 1
2 The other typicl problems involve finding the re of two curves over given intervl in the xxis. 6.2 Volumes In this section we introduce computing volume by integrting the cross sectionl re. Theorem 0.2 Let S be solid tht lies between x = nd x = b. If the crosssectionl re of S in the plne P x, through x nd perpendiculr to the xxis, is A(x), where A is continuous function, then the volume of S is A(x)dx. You should be ble to use this when the crosssectionl re is simple geometric object such s squre or equilterl tringle. So you need to know the formuls for the re of squre, circle, nd tringle (especilly equilterl tringle). However, the min ppliction is when we revolve grph round n xis or line. Theorem 0.3 Let f nd g be continuous functions over n intervl [, b] where 0 g(x) f(x) for ll x b. If we revolve the grph of the functions bout the xxis nd find the volume of the solid S between the surfces, then π[f(x)] 2 π[g(x)] 2 dx = π [f(x)] 2 [g(x)] 2 dx. You should lso be ble to compute when we revolve round the yxis (so we integrte dy), nd you should be ble to modify the formul if we revolve round line such s y = c. If we know tht c f(x) g(x) for ll x b, then we hve π[f(x) c] 2 π[g(x) c] 2 dx = π [f(x) c] 2 [g(x) c] 2 dx. 2
3 6.3 Volumes by Cylindricl Shells Theorem 0.4 Let f(x) be continuous function for x b, where 0 < b, nd ssume tht 0 f(x) for ll x b. Then the volume of the solid obtined by rotting bout the yxis the region under the curve y = f(x) from to b, is 2πxf(x)dx. Notice for cylindricl shells tht if we revolve bout the yxis or line x = c we integrte dx, wheres, in the wsher method we would integrte dy. Also, if we revolve bout the xxis or line y = c for cylindricl shells tht we integrte dy, wheres, in the wsher method we would integrte dx. If we know tht 0 g(x) f(x) for x b nd both f nd g re continuous for the intervl [, b], then we hve 2πx[f(x) g(x)]dx. Also, if we revolve round the line x = c nd we know tht c, 0 f(x) for ll x b, nd f is continuous on [, b], then we hve 2π(x c)f(x)dx. In deciding which method to use sometimes it will not mtter, but often it does. You should consider the following ides. 1. Is it esy to solve for the necessry vribles? 2. Are the limits of integrtion esier to find with one method? 3. Does one form integrte esier for the given expression? 4. Is the region more esily described using top nd bottom boundries (y = f(x)) or left nd right boundries (x = g(y))? 3
4 6.4 Work From physics we know tht Work = Force x Distnce. If we hve n object tht we re moving from point to b nd t point x we need to use force of f(x) to move it, then the eqution becomes W = f(x)dx. The bove expression will be defined so long s the function f is continuous over [, b]. There re three min pplictions of this tht you should be prepred to nswer. 1. For spring we know from Hooke s lw tht the force is proportionl to x. So if we strt with x = 0 being the position t rest for the spring, then f(x) = kx nd k is positive constnt clled the spring constnt. These problems consist of two prts. In the first prt you will use some informtion given by the problem to find the constnt k. Then you will use this vlue to nswer the question. 2. Moving cble, rope, etc tht hngs verticlly. In these cses there will be weighted object whose weight chnges in terms of x. This is the form of chin or rope tht wys certin mount per length, or bucket tht is leking s it is lifted. In ech of these there re two importnt spects. First, fix coordinte system for the problem. The set up will depend on the coordintes, but not the nswer. Second, find the formul for the force t given point x given your coordinte system. Now tht this is done the eqution should be simple integrl s expressed bove. 3. Moving wter out some sort of geometric pool. The first step is to tke the geometric object nd reduce to one vrible. Now find the volume of thin slice of the wter A(x) x where A(x) is the cross sectionl re depending on the point x nd your given coordinte system. Then force= g density volume where g is the grvittionl constnt so we usully hve n expression of the form g densitya(x)dx. 4
5 Tht is now the force. The distnce we move depends on the coordintes you hve chosen but re some expression of the form x, x c, x, or x c depending on how you set up the problem. Remrk 0.5 Keep in mind the grders do not wnt to grde problem bsed on the lrge number of wys student will set up the eqution. This will very likely be multiple choice problem, or they will give you the coordintes so they don t hve to think of ech wy the students will set up the problem. 6.5 Averge Vlue of Function Theorem 0.6 Let f be continuous on the intervl [, b]. Then the verge vlue of f over [, b] is 1 b f(x)dx. b The problems involving this expression re usully quite stright forwrd. Mke sure not to forget to divide by the b term, especilly in the multiple choice prt of the exm. Theorem 0.7 (Men Vlue Theorem for Integrls) Let f be continuous on [, b], then there exists some c [, b] such tht f(c) = 1 b f(x)dx. Agin most problems re quite stright forwrd for this types of equtions. Mke sure tht you re creful the wording so tht you find the expression sked for. For instnce, either the verge or vlue c tht chieves the verge. 7.1 Integrtion by Prts Using the product rule nd the Fundmentl Theorem of Clculus we fid the formul for integrtion by prts is u dv = uv v du 5
6 or it cn lso be written s f(x)g (x)dx = f(x)g(x) g(x)f (x)dx. When the integrl is definite integrl we hve f(x)g (x)dx = f(x)g(x) b g(x)f (x)dx. As rule of thumb you should let u be chosen bsed on the following priority (roughly): 1. logrithms, 2. polynomils, 3. trigonometric functions, 4. nd exponentil functions. Also, remember the following ides discussed in clss. If you hve to perform integrtion by prts twice, then don t chnge the order of your choice of u nd dv or you will undue ll your work. You my need to combine this method with other methods such s substitution. There re lso times when you strt with n integrl of the form f(x)g (x)dx nd fter performing integrtion by prts twice you rrive t n integrl on the right side of the eqution of the form c f(x)g (x)dx where c is positive constnt. In this cse you should dd this quntity bck to the originl eqution to solve. 7.2 Trigonometric Integrls You need to know bsic trigonometric identities for this section such s sin 2 x + cos 2 x = 1, sec 2 x = 1 + tn 2 x, 6
7 csc 2 x = 1 + cot 2 x, 2 sin x cos x = sin 2x, sin 2 x = cos 2 x = 1 cos 2x 2, nd 1+cos 2x 2. Additionlly, you need to know the derivtives nd ntiderivtives of the 6 bsic trigonometric functions. For equtions of the form sin n x cos m xdx then you do the following: 1. if one is odd, then sve 1 fctor nd chnge the rest to tht form, 2. if both re even, then use the hlf ngle formuls bove. For equtions of the form tn m x sec n xdx then you do the following: 1. if n is even, then rewrite the other sec n 2 x in terms of tn nd solve with usubstitution with u = tn x, 2. if m is odd, then pull out tn x sec x, rewrite the remining tn m 1 x in terms of sec nd solve with usubstitution with u = sec x. Similr sttements hold for csc nd cot. Other fct to remember re tn xdx = ln sec x + C nd sec xdx = ln sec x + tn x + C. 7.3 Trigonometric Substitution These integrls re integrls with one of the following forms tht cnnot be solved by simple substitution x 2 use the substitution x = sin θ (or x = cos θ) nd the identity 1 sin 2 θ = cos 2 θ (likewise the identity 1 cos 2 θ = sin 2 θ). 7
8 x 2 use the substitution x = tn θ nd the identity 1 + tn 2 θ = sec 2 θ. 3. x 2 2 use the substitution x = sec θ nd the identity sec 2 θ 1 = tn 2 θ. The other component you need to sometimes need to compute is the limits of integrtion. To do this drw the tringle ssocited with the substitution to compute the inverse trigonometric vlues to use with the new limits of integrtion. 8
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