Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8


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1 Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections , 5.0. Applictions of Integrtion Chpter 6 including sections nd section Infinite Sequences nd Series Chpter 8 including sections ; sections limited to lecture mteril. Chpter 5: Integrls Chpter 5 is n introduction to integrl clculus. In this chpter, we studied the definition of the integrl in terms of Riemnn sums nd we developed the notions of definite nd indefinite integrls. Lter, we studied techniques for evluting both of these kinds integrls. Sigm Nottion A convenient shorthnd for writing long (nd even infinite) sums is the soclled sigm nottion. The expression n is bbrevited n i= i i.e., n i = n, i= but this representtion of the sum is not unique. There re importnt formuls to know: i= = n, i= i = n(n+), i= i = n(n+)(n+) i= i3 = nd rules ( n(n+) 6, ), i= kf(i) = k n i= f(i), i= [f(i) ± g(i)] = n i= f(i) ± n i= g(i) Riemnn Sums nd Definite Integrls Our introduction to definite integrls begins with Riemnn sums. First, we divide the intervl into [, b] into n subintervls, ech of width x = b n. Next, we determined the endpoints of these subintervls to be x i = + i x.
2 Mth 3 Fll 0 for j = 0,,,..., n. In prticulr, x 0 = nd x n = b. Then Riemnn sum for function f on the intervl [, b] is defined to be n f(x i ) x = f(x ) x + f(x ) x f(x n) x i= where x i is n rbitrry smple point in the ith subintervl [x i, x i ] for i =,,..., n. We defined the definite integrl on [, b] to be f(x) dx = lim n n f(x i ) x We we compute definite integrls using the limit definition bove, we often tke x i to be the righthnd endpoint, x i, of the ith subintervl. (Signed) re: The definite integrl f(t) dt is the signed re of the region in the plne bounded by the grph of y = f(x) nd the lines y = 0, x =, nd x = b. Remember tht re bove the xxis is positive, but re below the xxis is negtive. Properties of Definite Integrls: [f(t) + g(t)] dt = f(t) dt + g(t) dt nd kf(t) dt = k f(t) dt = 0; i= b f(t) dt = b f(t) dt; nd f(t) dt = c If m f(x) M on [, b], then m(b ) f(x) dx M(b ) Evlution Theorem: Suppose f is continuous on [, b]. Then f(t) dt = F (b) F (), f(t) dt f(t) dt + c f(t) dt where F is ny ntiderivtive of f, (i.e., F (x) = f(x)). The Fundmentl Theorem of Clculus: If f is continuous on the intervl [, b], then the function F defined by F (x) = x f(t) dt is differentible on [, b] nd F (x) = d [ x ] f(t) dt = f(x). dx Indefinite Integrls Integrl Formuls: Know the integrls:. x n dx = un+ + C, if n n +. x dx = ln x + C, 3. e x dx = e x + C, 4. sin x dx = cos x + C, 5. cos x dx = sin x + C, 6. sec x dx = tn x + C,
3 Mth 3 Fll 0 7. sec x tn x dx = sec x + C, 8. + x dx = tn x + C, 9. x dx = sin x + C. Techniques of Integrtion: This is wht sections 5.5, 5.6 nd 5.7 were ll bout. There re severl techniques, so we need wy to decide which to use. Here is strtegy to use when the integrl isn t in one of the bsic forms bove:. Simplify: multiply, cncel, use trigonometric identities, etc.. Substitution: is there n obvious substitution? If the integrl hs the form f(g(x))g (x) dx, then use f(g(x))g (x) dx = f(u) du, where u = g(x). 3. Integrtion by Prts: use u dv = uv v du to replce difficult integrl with something esier. 4. Clssify: is the integrl of one of the types we sw in Section 5.7? trig functions only? try using trig identities sum/difference of squres? rdicls? consider trig substitution rtionl function? try seprting the frction into the sum of rtionl functions whose denomintors re the fctors of the originl 5. Be cretive: nother substitution, prts more thn once, mnipulte the integrnd (rtionlize the denomintor, identities, multiply by just the right form of ), or combine severl methods. Additionl Techniques of Integrtion Trigonometric Integrls For integrls of the form sin m (x) cos n (x) dx, consider the following strtegies: If m is odd, sve one sine fctor nd use sin x = cos x to express the remining fctors in terms of cosine. If n is odd, sve one cosine fctor nd use cos x = sin x to express the remining fctors in terms of sine. If both m nd n re even, use the hlfngle identities sin x = cos x cos x = + cos x to try to simplify. It is often helpful to recll sin x = sin x cos x. 3
4 Mth 3 Fll 0 Trigonometric Substitutions Chnging the vrible of n integrnd from x to θ using trigonometric substitution llows us to simplify the integrtion of forms with rdicls tht would otherwise not be menble to ordinry substitutions. The trig substitutions with which you should be fmilir re: For x use x = sin θ For + x use x = tn θ For x use x = sec θ. Prtil Frctions We integrte rtionl functions (rtios of polynomils) by expressing them s sums of simpler frctions, clled prtil frctions tht we lredy know how to integrte. We hve limited our considertion to prtil frctions in which the denomintor is qudrtic nd the numertor is liner form such s the following (from Exmple 4 of Section 5.7): 5x 4 x + x dx. Hence, you should be ble to reexpress this problem in the form A x + B x + dx identifying A nd B nd completing the integrtion. Improper Integrls We hve considered two types of improper integrls: integrls on infinite intervls nd integrls with discontinuous integrnds. Type Improper Integrls Integrls on infinite intervls hve the forms f(x) dx = lim t t f(x) dx nd f(x) dx = t lim t f(x) dx An improper integrl is convergent if the limit exists nd divergent otherwise. If both f(x) dx nd f(x) dx re convergent, then we define f(x) dx = Of prticulr interest is the integrl p. x p f(x) dx + f(x) dx dx which is convergent if p > nd divergent if 4
5 Mth 3 Fll 0 Type Improper Integrls We considered three cses involving discontinuities. If f is continuous on [, b) nd is discontinuous t b, then f(x) dx = lim f(x) dx, t b if this limit exists (s finite number). Similrly, f(x) dx = lim f(x) dx. t + If f hs discontinuity t c, where < c < b, nd both c f(x) dx nd c f(x) dx re convergent, then we define f(x) dx = c f(x) dx + c f(x) dx. Hence, n integrl over n intervl where f hs discontinuity will diverge if the integrl over either hlf diverges. Chpter 6: Applictions of Integrtion Are Between Curves Given two functions f(x) nd g(x) such tht g(x) f(x) on the intervl x b, the re between the grphs of y = f(x) nd y = g(x) between x = nd x = b is Volumes A = [f(x) g(x)] dx. Alwys be sure to crefully consider the solid of interest to determine whether it will be esier to integrte with respect to x or y. In the descriptions below, it is ssumed tht the integrtion will be with respect to x. When integrting with respect to y, simply interchnge the roles of x nd y. Disk method Given function y = f(x), if we spin it bout the xxis we obtin solid of volume V = π[f(x)] dx. Wsher method When there re holes in the solid of revolution, we use the wsher method V = π ( [f(x)] [g(x)] ) dx where f(x) is function determining the outside rdius nd g(x) the inside rdius. When the revolution is crried out round line other thn the x or y xis, cre must lwys be tken to determine the pproprite rdius function(s). 5
6 Mth 3 Fll 0 Method of Cylindricl Shells For volume problems for which neither the method of disks or wshers pplies, consider the method of cylindricl shells. The volume of solid obtined by rotting bout the yxis the region under the curve y = f(x) from to b is V = πxf(x) dx When the revolution is crried out round line other thn the x or y xis, cre must lwys be tken to determine the pproprite rdius function(s). In more complicted situtions (e.g., torus), cre must lso be tken to ensure tht the height function is correctly determined. Arc Length We hve the formuls for rc length of curve ( ) dy L = + dx, dx ( ) dx L = + dy, dy (dx ) ( ) dy L = + dt, dt dt where we denote the expression to the right of the integrl sign by ds, the differentil of rc length. Here re some guidelines for deciding which version of ds to use: ( If the problem is written in terms of prmetric equtions, then use ds = dx ) ( ) dt + dy dt dt. If the problem is given in terms of rectngulr coordintes x nd y, then:. if y = f(x) nd you cn t solve for x (e.g., y = x x ), then use ds = +. if x = g(y) nd you cn t solve for y (e.g., x = ln sec y ), then use ds = ( dy dx) dx + ( dx dy ) dy; 3. if the eqution my be expressed giving y s function of x, y = f(x), nd x s function of y, x = g(y) (e.g., y = 3x /3 ), then try both formuls nd see which is simpler. In ll these problems, your first priority should be to simplify the expression under the squre root, since this is where the most difficulties rise. You should expect to hve to use techniques developed in chpter 5 to evlute some of these integrls (e.g., trig substitution). Averge Vlue of Function The verge vlue of function f(x) over the closed intervl [, b] is fve = b f(x) dx. 6
7 Mth 3 Fll 0 Probbility The probbility density function f models the probbility tht X lies between nd b by P ( X b) = f(x) dx. The probbility density function of rndom vrible X must stify the conditions f(x) 0 nd f(x) dx =, since probbilities re mesured on scle from 0 to. The men of rndom vrible X with probbility density function f is given by µ = while the medin is the number m such tht m f(x) dx = xf(x) dx, m f(x) dx =. In more generlity, we defined the pth quntile for rndom vrible s qp f(x) dx = p. The probbility density function most commonly used to model witing time X is the exponentil density { 0 x < 0 f(x) = β e x/β x 0. The prmeter β of the exponentil density is lso its men. The most importnt distribution of ll is the norml distribution with probbility density function f(x) = σ π e (x µ) /(σ ) The density is defined for < X < nd hs prmeters men µ nd stndrd devition σ. Chpter 8: Infinite Sequences nd Series Sequences A sequence is function f : {,, 3,...} R, which we often think of s n ordered list of rel numbers,, 3,..., n,..., where n = f(n). We sy tht sequence converges to the limit L if the terms n get rbitrrily close to L s n. Sometimes the terms of sequence re determined by n explicit formul, but sometimes they re defined recursively by specifying the vlue of nd then indicting how to obtin n+ from n. In this cse, if lim n = L nd n+ = f( n ), then we must hve tht L = f(l), which enbles us to find the vlue of the limit if we first know tht the limit exists. To nswer this question, we often use the Monotonic Sequence Theorem, which sttes tht bounded, monotonic sequence converges. 7
8 Mth 3 Fll 0 Numericl Series A series is obtined from sequence { n } by dding the terms: n = n + is the series ssocited with the sequence { n }. While we cnnot compute the infinite sum directly, we cn compute the prtil sums s n = n nd check for convergence of this new sequence {s n } s. If it converges to the limit s, we sy tht the series converges with sum s, nd write n = s. If the sequence of prtil sums fils to converge, we sy tht the series n diverges. Below is strtegy for determining if series converges:. Is the series one of the bsic types? Geometric series: n= rn or, equivlently, n=0 rn, converges to r when r < nd diverges when r. pseries: n= n p converges if p > nd diverges when p.. Test for Divergence: If lim n n 0, then n diverges. 3. Clssify: does n involve fctorils, products, nd/or quotients? try the Rtio Test (RT) is it lternting? use the Alternting Series Test (AST) does it look like geometric or pseries? try the Comprison Test (CT) or Limit Comprison Test (LCT); f(n) converges if nd only if f(x) dx converges (Integrl Test). Power Series Next we introduced power series, i.e., series of the form n=0 = c nx n. One importnt exmple of power series tht we studied is the geometric power series: x n = + x + x + x x n + n=0 For x within the rdius of convergence R, rules of polynomils lso pply to power series, e.g., ddition, multipliction, nd termbyterm differentition nd integrtion. The most importnt clss of exmples of power series re the Tylor series for functions. The Tylor series of the function f bout is f (n) () f(x) = (x ) n. n! For the specil cse = 0, we obtin the Mclurin series for f t : f (n) (0) f(x) = x n. n! n=0 n=0 You re responsible for knowing the following Mclurin series, both of which re convergent for ll rel numbers x: e x = + x + x! + x3 3! + nd sin x = x x3 3! + x5 5! x7 7! + 8
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