Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems? Review of Bsic Concepts. Double-ngle formuls. Your nswer should involve trig functions of θ, nd not of 2θ. cos(2θ) = cos 2 θ sin 2 θ sin(2θ) = 2 sin θ cos θ.. Hlf-ngle formuls. Your nswer should involve cos(2θ). cos 2 (θ) = 1 + cos (2θ) 2 sin 2 (θ) = 1 cos (2θ) 2. Since cos 2 θ + sin 2 θ = 1, we know tht the corresponding reltionship beween:. du u tngent (i.e., tn) nd secnt (i.e., sec) is 1 + tn 2 θ = sec 2 θ. cotngent (i.e., cot) nd cosecnt (i.e., csc) is 1 + cot 2 θ = csc 2 θ. Remember Your Clculus I Integrtion Bsics? u 0 = ln u + C. u n du n 1 = u n+1 n+1 +C. e u du = e u +C. u du 1 = = u ln. cos u du = sin u + C. sec 2 u du = tn u + C. sec u tn u du = sec u + C. sin u du = - cos u + C. csc 2 u du = - cot u + C. csc u cot u du = - csc u + C. tn u du =. cot u du =. sec u du = + C ln sec u or = - ln cos u + C - ln csc u or = ln sin u + C ln sec u + tn u or = - ln sec u tn u + C. csc u du = - ln csc u + cot u = or ln csc u cot u + C. 1 du >0 = sin ( ) -1 u + C 2 u 2. 1 du >0 = ( ) 1 2 +u 2 tn-1 u + C. ( ) 1 u du >0 1 = u u 2 2 sec-1 + C Prof. Girrdi Pge 1 of 11 Mth 142
Review of Bsic Concepts Integrtion from Clculus II. Integrtion by prts formul: u dv = uv vdu. To integrte f(x) g(x), where f nd g re polyonomils, 1st find its Prtil Frction Decomposition (PDF). If [degree of f] [degree of g], then one must first does long division. If [degree of f] < [degree of g] i.e., hve strictly bigger bottoms then first fctor y = g(x) into: liner fctors px + q nd irreducible qudrtic fctors x 2 + bx + c (to be sure it s irreducible, you need b 2 4c < 0 ). Next, collect up like terms nd follow the following rules. Rule 1: For ech fctor of the form (px+q) m where m 1, the decomposition of y = f(x) g(x) contins sum of m prtil frctions of the form, where ech A i is rel number, A 1 (px + q) 1 + A 2 (px + q) 2 +... + A m (px + q) m. Rule 2: For ech fctor of the form (x 2 + bx + c) n where n 1, the decomposition of y = f(x) g(x) contins sum of n prtil frctions of the form, where the A i s nd B i s re rel number, A 1 x + B 1 (x 2 + bx + c) 1 + A 2 x + B 2 (x 2 + bx + c) 2 +... + d Anx+Bn (x 2 +bx+c) n.. Trig. Substitution. (Recll tht the integrnd is the function you re integrting.) Here, is constnt nd > 0. if the integrnd involves 2 u 2, then one mkes the substitution u = sin θ. if the integrnd involves 2 +u 2, then one mkes the substitution u = tn θ. if the integrnd involves u 2 2, then one mkes the substitution u = sec θ. Improper Integrls 0. Fill-in-the boxes. Below,, b, c R with < c < b.. If f : [0, ) R is continuous, then we define the improper integrl 0 0 t t 0 f (x) dx.. If f : (, 0] R is continuous, then we define the improper integrl 0 0 0 t t f (x) dx. Prof. Girrdi Pge 2 of 11 Mth 142
. If f : (, ) R is continuous, then we define the improper integrl [ t 0 t ] f (x) dx + [ s s 0 Review of Bsic Concepts ] f (x) dx.. If f : (, b] R is continuous, then we define the improper integrl t + t f (x) dx.. If f : [, b) R is continuous, then we define the improper integrl t b t f (x) dx.. If f : [, c) (c, b] R is continuous, then we define the improper integrl [ t c t ] f (x) dx. An improper integrl s bove converges precisely when + [ s c + s ] f (x) dx. ech of the its involves converges to finite number.. An improper integrl s bove diverges precisely when the improper integrl does not converge.. An improper integrl s bove diverges to precisely when t lest one of the involved its diverges to AND ech of the other involved its either diverges to or converges to finite number.. An improper integrl s bove diverges to precisely when t lest one of the involved its diverges to AND ech of the other involved its either diverges to or converges to finite number. Prof. Girrdi Pge 3 of 11 Mth 142
Review of Bsic Concepts Sequences. Let { n } n=1 be sequence of rel numbers. Complete the below sentences. The it of { n } n=1 is the rel number L provided for ech ɛ > 0 there exists nturl number N so tht if the nturl number n stisfies n > N then L n < ɛ. If the it of { n } n=1 is L R, then we denote this by n = L. { n } n=1 converges provided there exists rel number L so tht n = L. { n } n=1 diverges provided { n } n=1 does not converge.. Prctice tking bsic its. (Importnt, e.g., for Rtio nd Root Tests.) 5n 17 + 6n 2 + 1 7n 18 + 9n 3 + 5 = 0 36n 17 6n 2 1 4n 17 + 9n 3 + 5 = 36 or 9 4-5n 18 + 6n 2 + 1 36n = DNE or - 7n 17 + 9n 3 + 5 17 6n 2 1 4n 17 + 9n 3 + 5 = 36 or 3 4 Cn you do similr problems?. Commonly Occurring Limits Thoms Book 10.1, Theorem 5 pge 578 ln n (1) n = 0 (2) n n = 1 (3) c 1/n = 1 (c > 0) (4) c n = 0 ( c < 1) (5) ( 1 + c n) n = e c (c R) x n (6) n! = 0 (c R). Let < r <. (Needed for Geometric Series. Wrning, don t confuse sequences with series.) If r < 1, then r n = 0. If r = 1, then r n = 1. If r > 1, then r n = DNE (tends to ). If r = -1, then r n = DNE (oscilltes between 1 nd 1). If r < -1, then r n = DNE (r 2n while r 2n+1 ). Prof. Girrdi Pge 4 of 11 Mth 142
Series. In this section, ll series re understood to be n=1, unless otherwise indicted. Review of Bsic Concepts. For forml series n=1 n, where ech n R, consider the corresponding sequence {s N } N=1 of prtil sums, so s N = N n=1 n. Then the forml series n : converges if nd only if the N s N exists in R converges to L R if nd only if the N s N exists in R nd equls L R diverges if nd only if the N s N does not exist in R. Now ssume, furthermore, tht n 0 for ech n. Then the sequence {s N } N=1 of prtil sums either is bounded bove (by some finite number), in which cse the series n converges or is not bounded bove (by some finite number), in which cse the series n diverges to +.. Stte the n th -term test for n rbitrry series n. If n 0 (which includes the cse tht n does not exist), then n diverges.. Fix r R. For N 17, let s N = N n=17 rn (Note the sum strts t 17). Then, for N > 17, s N = r 17 + r 18 +... + r N (your nswer cn hve... s but not sign) r s N = r 18 +... + r N + r N+1 (your nswer cn hve... s but not sign) (1 r) s N = r 17 r N+1 (your nswer should hve neither... s nor sign) r nd if r 1, then s N = 17 r N+1 1 r (your nswer should hve neither... s nor sign). Geometric Series where < r <. The series r n (hint: look t the previous questions): converges if nd only if r < 1 diverges if nd only if r 1.. p-series where 0 < p <. The series 1 n p converges if nd only if p > 1. diverges if nd only if p 1. This cn be shown by using the integrl test here, nme the test one uses nd compring 1 1 (the hrd to compute series) to n p (the esy to compute improper integrl) dx. x p x=1 Prof. Girrdi Pge 5 of 11 Mth 142
Review of Bsic Concepts Tests for Positive-Termed Series (so for n where n 0) 0.1. Stte the Integrl Test with Reminder Estimte for positive-termed series n. Let f : [1, ) R be so tht Then (1) n = f (n) for ech n N (2) f is positive function (3) f is continuous function (4) f is decresing (nonincresing is lso ok) function. n converges if nd only if nd if n converges, then ( ) ( N ) 0 k k k=1 k=1 x= x=1 f(x) dx converges. x= x=n f(x) dx. 0.2. Stte the Direct Comprison Test for positive-termed series n. 0 If n c n when n 17 nd c n converges, then n converges. (only n c n is lso ok b/c given n 0) If 0 d n n (need 0 d n prt here) when n 17 nd dn diverges, then n diverges. Hint: sing the song to yourself. 0.3. Stte the Limit Comprison Test for positive-termed series n. Let b n > 0 nd L = n bn. If 0 < L <, then [ b n converges n converges ] If L = 0, then [ b n converges = n converges ]. If L =, then [ b n diverges = n diverges ]. Gol: cleverly pick positive b n s so tht you know wht b n does (converges or diverges) nd the sequence { n bn }n converges. 0.4. Helpful Intuition Fill in the 3 boxes using: e x, ln x, x q. Use ech once, nd only once. Consider positive power q > 0. There is (some big number) N q > 0 so tht if x N q then ln x x q e x. Prof. Girrdi Pge 6 of 11 Mth 142
Review of Bsic Concepts Tests for Arbitrry-Termed Series (so for n where < n < ) 0.5. By definition, for n rbitrry series n, (fill in these 3 boxes with convergent or divergent). n is bsolutely convergent if nd only if n is convergent. n is conditionlly convergent if nd only if n is convergent nd n is divergent. n is divergent if nd only if n is divergent. 0.6. Stte the Rtio nd Root Tests for rbitrry-termed series n with < n <. Let ρ = n+1 n or ρ = n 1 n. If ρ < 1 then n converges bsolutely. If ρ > 1 then n diverges. If ρ = 1 then the test is inconclusive. 0.7. Stte the Alternting Series Test (AST) & Alternting Series Estimtion Theorem. Let Then (1) u n 0 for ech n N (2) u n = 0 (3) u n > (lso ok ) u n+1 for ech n N. the series ( 1) n u n converges. (lso ok: ( 1) n+1 u n converges or ( 1) n 1 u n converges) nd we cn estimte (i.e., pproximte) the infinite sum n=1 ( 1)n u n by the finite sum N k=1 ( 1)k u k nd the error (i.e. reminder) stisfies N ( 1) k u k ( 1) k u k u N+1. k=1 k=1 Prof. Girrdi Pge 7 of 11 Mth 142
Review of Bsic Concepts Condsider (forml) power series with rdius of convergence R [0, ]. h(x) = Power Series (Here x 0 R is fixed nd { n } is fixed sequence of rel numbers.) n (x x 0 ) n, (1.1) Without ny other further informtion on { n }, nswer the following questions.. The choices for the next 4 boxes re: AC, CC, DIVG, nything. Here, AC stnds for: lwys bsolutely convergent CC stnds for: lwys conditionlly convergent DIVG stnds for: is lwys divergent nything stnds for: cn do nything, i.e., there re exmples showing tht it cn be AC, CC, or DIVG. (1) At the center x = x 0, the power series in (1.1) AC. (2) For x R such tht x x 0 < R, the power series in (1.1) AC. (3) For x R such tht x x 0 > R, the power series in (1.1) DIVG. (4) If R > 0, then for the endpoints x = x 0 ±R, the power series in (1.1) nything.. For this prt, fill in the 7 boxes. Let R > 0 nd consider the function y = h(x) defined by the power series in (1.1). (1) The function y = h(x) is lwys differentible on the intervl (x 0 R, x 0 + R) (mke this intervl s lrge s it cn be, but still keeping the sttement true). Furthermore, if x is in this intervl, then h (x) = n= 1 n n (x x 0 ) n 1. (1.2) Wht cn you sy bout the rdius of convergence of the power series in (1.2)? The power series in (1.2) hs the sme rduis of convergence s the power series in (1.1). (2) The function y = h(x) lwys hs n ntiderivtive on the intervl (x 0 R, x 0 + R) (mke this intervl s lrge s it cn be, but still keeping the sttement true). Futhermore, if α nd β re in this intervl, then x=β x=α h(x) dx = n= 0 n n + 1 (x x 0) n+1 x=β x=α. Prof. Girrdi Pge 8 of 11 Mth 142
Tylor/Mclurin Polynomils nd Series Review of Bsic Concepts Let y = f(x) be function with derivtives of ll orders in n intervl I contining x 0. Let y = P N (x) be the N th -order Tylor polynomil of y = f(x) bout x 0. Let y = R N (x) be the N th -order Tylor reminder of y = f(x) bout x 0. Let y = P (x) be the Tylor series of y = f(x) bout x 0. Let c n be the n th Tylor coefficient of y = f(x) bout x 0.. The formul for c n is c n = f (n) (x 0 ) n! b. In open form (i.e., with... nd without -sign) P N (x) = f(x 0 ) + f (x 0 )(x x 0 ) + f (2) (x 0 ) 2! (x x 0 ) 2 + f (3) (x 0 ) 3! c. In closed form (i.e., with -sign nd without... ) (x x 0 ) 3 + + f (N) (x 0 ) (x x 0 ) N N! P N (x) = N f (n) (x 0 ) n! (x x 0 ) n d. In open form (i.e., with... nd without -sign) P (x) = f(x 0 ) + f (x 0 )(x x 0 ) + f (2) (x 0 ) 2! e. In closed form (i.e., with -sign nd without... ) (x x 0 ) 2 + + f (n) (x 0 ) (x x 0 ) n +... n! P (x) = f (n) (x 0 ) n! (x x 0 ) n f. We know tht f(x) = P N (x) + R N (x). Tylor s BIG Theorem tells us tht, for ech x I, R N (x) = f (N+1) (c) (N + 1)! (x x 0 ) (N+1) for some c between x nd x 0. g. A Mclurin series is Tylor series with the center specificlly specified s x 0 = 0. Prof. Girrdi Pge 9 of 11 Mth 142
Review of Bsic Concepts Commonly Used Tylor Series. Here, expnsion refers to the power series expnsion tht is the Mclurin series.. An expnsion for y = e x is x n n!, which is vlid precisely when x (, ).. An expnsion for y = cos x is ( 1) n x2n (2n)!, which is vlid precisely when x (, ).. An expnsion for y = sin x is ( 1) n x 2n+1 (2n + 1)!, which is vlid precisely when x (, ).. An expnsion for y = 1 1 x is x n, which is vlid precisely when x ( 1, 1).. An expnsion for y = ln(1+x) is ( 1) n=1 n+1 xn n, which is vlid precisely when x ( 1, 1].. An expnsion for y = rctn x is ( 1) n x2n+1 2n + 1, which is vlid precisely when x [ 1, 1]. Prof. Girrdi Pge 10 of 11 Mth 142
Review of Bsic Concepts Prmetric Curves In this prt, fill in the 4 boxes. Consider the curve C prmeterized by x = x (t) for t b. y = y (t) 1) Express dy dx in terms of derivtives with respect to t. Answer: dy dx = 2) The tngent line to C when t = t 0 is y = mx + b where m is 3) Express d2 y dx 2 using derivtives with respect to t. Answer: d 2 y dx 2 = 4) The rc length of C, expressed s on integrl with respect to t, is t=b (dx ) 2 ( ) 2 Arc Length = dy + dt dt dt t= dy dx dy dt dx dt evluted t t = t 0. ( ) d dy dt dx dx dt Polr Coordintes. Here, CC stnds for Crtresin coordintes ( while PC) stnds for polr ( coordintes. ). A point with PC (r, θ) lso hs PC r, θ + 2π s well s r, θ + π.. A point P R 2 with CC (x, y) nd PC (r, θ) stisfies the following. x = r cos θ & y = r sin θ & r 2 = x 2 + y 2 & tn θ = { y x if x 0 DNE if x = 0.. The period of f(θ) = cos(kθ) nd of f(θ) = sin(kθ) is 2π k. To sketch these grphs, we divide the period by 4 nd mke the chrt, in order to detect the mx/min/zero s of the function r = f(θ).. Now consider sufficiently nice function r = f(θ) which determines curve in the plne. The the re bounded by polr curves r = f(θ) nd the rys θ = α nd θ = β is Are = θ=β θ=α The rc length of the polr curves r = f(θ) is Arc Length = θ=β θ=α 1 2 [f(θ)]2 dθ. r 2 + ( ) 2 dr dθ. dθ Prof. Girrdi Pge 11 of 11 Mth 142