Exam 1 Study Guide. Differentiation and Anti-differentiation Rules from Calculus I
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1 Exm Stuy Guie Mth 26 - Clulus II, Fll 205 The following is list of importnt onepts from eh setion tht will be teste on exm. This is not omplete list of the mteril tht you shoul know for the ourse, but it is goo inition of wht will be emphsize on exm. A thorough unerstning of ll of the following onepts will help you perform well on exm. Some ples to fin problems on these topis re the following: in the book, in the homework, on quizzes, n online (for exmple the COW webpge). Differentition n Anti-ifferentition Rules from Clulus I x () = 0 x (ex ) = e x x ln(x) = x x (xn ) = nx n x (x ) = x ln() x log (x) = xln() x sin(x) = os(x) x s(x) = s(x)ot(x) x (sin (x)) = x 2 x (s (x)) = x x 2 x os(x) = sin(x) x se(x) = se(x)tn(x) x (os (x)) = x 2 x (se (x)) = x x 2 x tn(x) = se2 (x) x ot(x) = s2 (x) x (tn (x)) = +x 2 x (ot (x)) = +x 2. k x = k x +C e x x = e x +C x = ln( x ) +C x x n x = n + nxn+ +C; n x x = x ln() +C os(x)x = sin(x) +C sin(x)x = os(x) +C se 2 (x)x = tn(x) +C s(x)ot(x)x = s(x) +C se(x) tn(x)x = se(x) +C s 2 (x) = ot(x) +C x 2 x = sin (x) +C x x 2 x = s (x) +C x 2 x = os (x) +C x x 2 x = se (x) +C + x 2 = tn (x) +C + x 2 x = ot (x) +C.
2 Tehniques of Integrtion Setion 5.5 This setion is on. is the nti-ifferentition rule formulte from unoing the hin rule: Chin Rule x f (g(x)) = f (g(x))g (x) f (g(x))g (x)x = f (g(x)) +C Use to evlute inefinite integrls f (g(x))g (x)x = f (u)u = f (u) +C = f (g(x)) +C u = g(x) u = g (x)x Use to evlute efinite integrls b g(b) f (g(x))g (x)x = f (u)u = f (g()) f (g(b)) g() u = g(x) u() = g() u = g (x)x u(b) = g(b) Setion 5.6 This setion is on integrtion by prts. Integrtion by prts is the nti-ifferentition rule formulte from unoing the prout rule: Prout Rule x ( f (x) g(x)) = g(x) f (x) + f (x)g (x) Integrtion by prts f (x)g (x)x = f (x)g(x) Use integrtion by prts to evlute inefinite integrls f (x)g (x)x = f (x)g(x) g(x) f (x)x u = f (x) v = g(x) Integrtion by prts u = f (x)x v = g (x)x g(x) f (x)x This formul sys tht uv = u v vu. When integrting by prts, we hoose u n v must then be the reminer of the integrn. Then u is the ifferentil of u n v is hosen by nti-ifferentiting g (x) so the v is the ifferentil of v. Use integrtion by prts to evlute efinite integrls b f (x)g b b (x)x = f (x)g(x) g(x) f (x)x u = f (x) v = g(x) Integrtion by prts u = f (x)x v = g (x)x The tehniques here re the sme s integrting by prts to evlute inefinite integrls. The ifferene is tht we evlute the terms t the enpoints n b.
3 Setion 5.7 This setion is on tehniques for integrting trigonometri funtions. Compute inefinite integrls of the form sin m (x)os n (x)x when either m or n is o by splitting off erivtive term, using the Pythgoren theorem, n integrte using. You shoul be ble to ompute efinite integrls of this form s well. You re expete to know the Pythgoren theorem, sin 2 (x) + os 2 (x) =. Compute inefinite integrls of the form sin m (x)os n (x)x when both m or n re even by using ouble ngle formuls. The ouble ngle formuls, os 2 (x) = os(2x) n sin2 (x) = 2 2 os(2x) will be provie for you; you re not expete to memorize them. You shoul be ble to ompute efinite integrls of this form s well. Compute inefinite integrls of the form tn m (x)se 2k (x)x n tn 2k+ (x)se n (x)x by splitting off erivtive term, using the Pythgoren theorem, n integrte using. You shoul be ble to ompute efinite integrls of this form s well. You re expete to know nother version of the Pythgoren theorem, obtine in the following wy: strt with the Pythgoren theorem, sin 2 (x) + os 2 (x) =, n ivie both sie by os 2 (x) to obtin sin 2 (x) os 2 (x) + os2 (x) os 2 (x) = os 2 (x) tn 2 (x) + = se 2 (x). Setion 5.7b This setion is on integrting using trigonometri substitution. Use trig substitution to integrte funtions tht hve terms in involving 2 x 2, 2 + x 2, or x 2 2. Assoite to eh of these forms is n pproprite substitution n version of the Pythgoren theorem tht will be useful in simplifying the integrn. They re esribe in the tble below. Form Substitution Differentil Pythgoren ientity 2 x 2 x = sin(θ) x = os(θ)θ 2 x 2 = 2 ( sin 2 (θ)) = 2 os 2 (θ) 2 + x 2 x = tn(θ) x = se 2 (θ)θ 2 + x 2 = 2 ( + tn 2 (θ)) = 2 se 2 (θ) x 2 2 x = se(θ) x = se(θ)tn(θ)θ x 2 2 = 2 (se 2 (θ) ) = 2 tn 2 (θ)
4 Setion 5.7 n Appenix G These setions re on integrting rtionl funtions using prtil frtion eomposition. Use prtil frtion eomposition to rewrite rtionl funtions into sum of funtions tht you n integrte iretly. This my involve polynomil long ivision (or syntheti ivision if you prefer). You will be expete to be ble to eompose rtionl funtions whose enomintors hve liner terms, repete liner terms, irreuible qurti terms, n ombintions thereof. You will not be expete to hnle repete irreuible qurti terms. In prtiulr, the enomintors of the funtions you will be expete to integrte will ontin terms of the following forms: Liner term Repete liner term Irreuible qurti term (x ) (x ) k (x ) For exmple, you re expete to be ble to integrte x 4 (x )(x 3) 3 (x 2 + ). This prtiulr problem woul be too long for n exm question, but you shoul know ll of the tehniques involve. You shoul be ble to ompute both efinite n inefinite integrls of this form. Applitions of Integrtion Setion 6. This setion is on omputing re between urves. Given two funtions f (x) n g(x) on n intervl [,b], you shoul be ble to ompute the re between the two urves on [,b] b A = f (x) g(x) x. In prtie, we brek up the intervl [,b] in to setions where f (x) is bove g(x) n vie vers, then integrte the top funtion minus the bottom funtion on eh intervl. Given two funtions f (y) n g(y) on n intervl [,], you shoul be ble to ompute the re between the two urves on [,] A = f (y) g(y) y. This is essentilly the sme problem s the lst bullet item turne on its sie. We o everything in the y iretion inste of the x iretion. You shoul be ble to ientify regions efine by their bouning urves. Tht is, you shoul be ble to ientify region given the urve tht form its bounry. For exmple, you shoul be ble to ientify the region boune by y = sin(x), y =, x = π/2, n x = 3π/2. You will not be teste on fining res of regions efine by prmetri urves.
5 Setion 6.2 n 6.3 This setion is on omputing volumes of solis using the isk/wsher or shell metho. You shoul be ble to use the formul b V = A(x)x where V is the volume of the soli with ross-setionl re A(x) for x in [,b]. You will not be testing on fining the re of regions where the bse n the typil ross-setion re esribe in wors (e.g. you will not be ske questions like 40, 4, n 42 from this setion). Use the bove formul (using the isk or wsher metho) to ompute the volume of solis of revolution obtine by revolving given region bout horizontl or vertil xis. There re two formultions of this formul to ompute volumes. Disk/Wsher Metho Revolving bout horizontl xis Revolving bout vertil xis -Axis of revolution is of the form y = -Axis of revolution is of the form x = (inluing the x-xis, whih is y = 0) (inluing the y-xis, whih is x = 0) -Vertil ross-setionl uts -Horizontl ross-setionl uts (perpeniulr to the xis of revolution) (perpeniulr to the xis of revolution) -Fix x n ompute ross-setionl re A(x) -Fix y n ompute ross-setionl re A(y) -Integrte in x -Integrte in y -Integrting with isks: b -Integrting with isks: V = πr(x) 2 x V = πr(y) 2 y -Integrting with wshers: -Integrting with wshers: b V = π(r(x) 2 r(x) 2 )x V = π(r(y) 2 r(y) 2 )y Cylinril Shell Metho Revolving bout horizontl xis Revolving bout vertil xis -Axis of revolution is of the form y = -Axis of revolution is of the form x = (inluing the x-xis, whih is y = 0) (inluing the y-xis, whih is x = 0) -Horizontl ross-setionl uts -Vertil ross-setionl uts (prllel to the xis of revolution) (prllel to the prllel to the xis of revolution) -Fix y n ompute ross-setionl re A(y) -Fix x n ompute ross-setionl re A(x) -Integrte in y -Integrte in x -Integrting with shells: -Integrting with isks: b V = 2πR(y)H(y)y V = 2πR(x)H(x)x
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