# Exam 1 Study Guide. Differentiation and Anti-differentiation Rules from Calculus I

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1 Exm Stuy Guie Mth Clculus II, Winter 204 The following is list of importnt concepts from ech section tht will be teste on exm. This is not complete list of the mteril tht you shoul know for the course, but it is goo iniction of wht will be emphsize on exm. A thorough unerstning of ll of the following concepts will help you perform well on exm. Some plces to fin problems on these topics re the following: in the book, in the homework, on quizzes, n online (for exmple the COW webpge). Differentition n Anti-ifferentition Rules from Clculus I x (c) = 0 x (ex ) = e x x ln(x) = x x (xn ) = nx n x (x ) = x ln() x log (x) = xln() x sin(x) = cos(x) x csc(x) = csc(x)cot(x) x (sin (x)) = x 2 x (csc (x)) = x x 2 x cos(x) = sin(x) x sec(x) = sec(x)tn(x) x (cos (x)) = x 2 x (sec (x)) = x x 2 x tn(x) = sec2 (x) x cot(x) = csc2 (x) x (tn (x)) = +x 2 x (cot (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 cos(x)x = sin(x) +C sin(x)x = cos(x) +C sec 2 (x)x = tn(x) +C csc(x)cot(x)x = csc(x) +C sec(x) tn(x)x = sec(x) +C csc 2 (x) = cot(x) +C x 2 x = sin (x) +C x x 2 x = csc (x) +C x 2 x = cos (x) +C x x 2 x = sec (x) +C + x 2 = tn (x) +C + x 2 x = cot (x) +C.

2 Techniques of Integrtion Section 5.2 This section is on computing efinite integrls s limits of Riemnn sums n in terms of signe re. Interpret efinite integrl s signe re. Compute integrls using re formuls of of geometric figures. Section 5.3 This section is on the funmentl theorem of clculus. Use the funmentl theorem of clculus to compute efinite integrls, i.e. b where F(x) is n nti-erivtive of f (x) on [, b]. f (x)x = F(b) F(), Compute the inefinite integrls from the tble bove f (x)x = F(x) +C, where F(x) is n nti-erivtive of f (x). Section 5.5 This section is on u-substitution. u-substitution is the nti-ifferentition rule formulte from unoing the chin rule: Chin Rule x f (g(x)) = f (g(x))g (x) u-substitution f (g(x))g (x)x = f (g(x)) +C Use u-substitution to evlute inefinite integrls f (g(x))g (x)x = f (u)u = f (u) +C = f (g(x)) +C [ ] u = g(x) u-substitution u = g (x)x Use u-substitution 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-substitution u = g (x)x u(b) = g(b) Section 7. This section is on integrtion by prts.

3 Integrtion by prts is the nti-ifferentition rule formulte from unoing the prouct rule: Prouct 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) g(x) f (x)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 ] This formul sys tht uv = u v vu. When integrting by prts, we choose u n v must then be the reminer of the integrn. Then u is the ifferentil of u n v is chosen 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 techniques here re the sme s integrting by prts to evlute inefinite integrls. The ifference is tht we evlute the terms t the enpoints n b. Section 7.2 This section is on techniques for integrting trigonometric functions. Compute inefinite integrls of the form sin m (x)cos n (x)x when either m or n is o by splitting off erivtive term, using the Pythgoren theorem, n integrte using u-substitution. You shoul be ble to compute efinite integrls of this form s well. You re expecte to know the Pythgoren theorem, sin 2 (x) + cos 2 (x) =. Compute inefinite integrls of the form sin m (x)cos n (x)x when both m or n re even by using ouble ngle formuls. The ouble ngle formuls, cos 2 (x) = cos(2x) n sin2 (x) = 2 2 cos(2x) will be provie for you; you re not expecte to memorize them. You shoul be ble to compute efinite integrls of this form s well.

4 Compute inefinite integrls of the form tn m (x)sec 2k (x)x n tn 2k+ (x)sec n (x)x by splitting off erivtive term, using the Pythgoren theorem, n integrte using u-substitution. You shoul be ble to compute efinite integrls of this form s well. You re expecte to know nother version of the Pythgoren theorem, obtine in the following wy: strt with the Pythgoren theorem, sin 2 (x) + cos 2 (x) =, n ivie both sie by cos 2 (x) to obtin sin 2 (x) cos 2 (x) + cos2 (x) cos 2 (x) = cos 2 (x) tn 2 (x) + = sec 2 (x). Section 7.4 This section is on integrting rtionl functions using prtil frction ecomposition. Use prtil frction ecomposition to rewrite rtionl functions into sum of functions tht you cn integrte irectly. This my involve polynomil long ivision (or synthetic ivision if you prefer). You will be expecte to be ble to ecompose rtionl functions whose enomintors hve liner terms, repete liner terms, irreucible qurtic terms, n combintions thereof. You will not be expecte to hnle repete irreucible qurtic terms. In prticulr, the enomintors of the functions you will be expecte to integrte will contin terms of the following forms: Liner term Repete liner term Irreucible qurtic term (x ) (x ) k (x ) For exmple, you re expecte to be ble to integrte x 4 (x )(x 3) 3 (x 2 + ). This prticulr problem woul be too long for n exm question, but you shoul know ll of the techniques involve. You shoul be ble to compute both efinite n inefinite integrls of this form. Applictions of Integrtion Section 6. This section is on computing re between curves. Given two functions f (x) n g(x) on n intervl [,b], you shoul be ble to compute the re between the two curves on [,b] b A = f (x) g(x) x. In prctice, we brek up the intervl [,b] in to sections where f (x) is bove g(x) n vice vers, then integrte the top function minus the bottom function on ech intervl.

5 Given two functions f (y) n g(y) on n intervl [c,], you shoul be ble to compute the re between the two curves on [c,] A = f (y) g(y) y. c This is essentilly the sme problem s the lst bullet item turne on its sie. We o everything in the y irection inste of the x irection. You shoul be ble to ientify regions efine by their bouning curves. Tht is, you shoul be ble to ientify region given the curve 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. Section 6.2 This section is on computing volumes of solis. You shoul be ble to use the formul b V = A(x)x where V is the volume of the soli with cross-sectionl re A(x) for x in [,b]. You shoul be ble to use this formul when soli is escribe by its bse n cross-sectionl region s well s using revolutions of solis. Use the bove formul (using the isk or wsher metho) to compute the volume of solis of revolution obtine by revolving given region bout horizontl or verticl xis. There re two formultions of this formul to compute volumes. Revolving bout horizontl xis Revolving bout verticl xis -Axis of revolution is of the form y = -Axis of revolution is of the form x = (incluing the x-xis, which is y = 0) (incluing the y-xis, which is x = 0) -Verticl cross-sectionl cuts -Horizontl cross-sectionl cuts (prllel to the y-xis) (prllel to the x-xis) -Fix x n compute cross-sectionl re A(x) -Fix y n compute cross-sectionl 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 c -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 c Section 6.4 This section is on computing work. You shoul be fmilir with the formuls W = F (work = force istnce) n F = m (force = mss )

6 You shoul be ble to set up n compute integrls to fin the mount of work one in vrious pplictions problems like the ones in section 6.4 in the text. This inclues using Hooke s lw to compute the force function for spring, computing work given force function, work one to pump wter out of tnk, work one to lift chin/rope, etc. Section 6.5 This section is on computing verge vlues of functions. Given function f (x) on [,b], the verge vlue of f (x) on [,b] is f [,b] = b f (x)x. b You shoul know this eqution n be ble to compute integrl verges using the integrtion techniques mentione bove. Know the interprettion of the verge vlue of function in terms of the re uner curve. In prticulr, f [,b] is the height of the rectngle of with b necessry to enclose the sme re unerneth f (x) on [,b]. Tht is, b f [,b] (b ) = f (x)x ( ) ( ) Are of rectngle with height = Are uner f (x) on [,b]. f [,b] n with b.

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