PG. Topic Clculus Notes BC PG. Topic 6 CH. Limits / Continuity - Find Limits Grphiclly nd Numericlly - Evluting Limits Anlyticlly - Continuity nd One-sided Limits -5 Infinite Limits/ -5 Limits tht Approch Infinity 7 CH. Derivtives - Derivtive nd the Tngent Line Prolem - Bsic Derivtives nd Nottion - Product nd Quotient Rule - Chin Rule -5 IImplicit Differentition 5-6 Relted Rtes 6 CH. Applictions of Derivtives - Extrem on n Intervl - Rolle's Theorem nd Men Vlue Theorem - Incresing nd Decresing Functions nd st Derivtive Test - Concvity nd the nd Derivtive Test -6 A Summry of Curve Sketching -7 Optimiztion Prolems -8 Newton's Method -9 Differentils 5 CH. Integrtion - Antiderivtives nd Indefinite Integrtion - Riemnn Sum nd Definite Integrls - Fundmentl Theorem of Clculus -5 Integrtion y Sustitution -6 Numericl Integrtion Trp. Rule, Left, Right, nd Midpoint 9 CH. 5 Trnscendentl Functions 5- Nturl Logrithm Function nd Differentition 5- Nturl Logrithm Function nd Integrtion 5- Exponentil Functions: Differentition nd Integrtion 5-5 Bses other thn e nd Applictions 5-6 Inverse Trig. Functions: Differentition 5-7 Inverse Trig. Functions: Integrtion CH.6 Differentil Equtions 6- Slope Fields 6- Growth, Doule-Life, nd Hlf-Life formuls 6- Seprtion of Vriles 5 CH.7 Applictions of Integrtion 7- Are 7- Volume rotted out x-xis 7- Volume rotted out y-xis Volume rotted out other lines Volumes of known cross sections 7- Arclengths nd Surfce of Revolution 5 58 CH. 8 Integrtion Techniques 8- Bsic Integrtion Rules 8- Integrtion y Prts 8-5 Prtil Frctions 8-7 Indeterminte Forms nd L'Hopitl's Rule 8-8 Improper Integrls 59 75 CH. 9 Infinite Series 9- Sequences/ 9- Series nd Convergence 9- The Integrl Test nd p-series 9- Comprison of Series 9-5 Alternting Series 9-6 The Rtio nd Root Tests 9-7 Tylor Polynomils nd Approximtions 9-8 Power Series 9-9 Representtion of Functions y Power Series 9- Tylor nd Mclurin Series 76 8 CH. Prmetric nd Polr Coordintes - Plne Curves nd Prmetric Equtions - Prmetric Equtions nd Clculus - Polr Coordintes nd Polr Grphs -5 Are nd Arc Length in Polr Coordintes 8 Hooke's Lw 8 Finding Are Using Limits 8 Lgrnge Error Bound 8 89 AP Review 9 9 8- Trigonometric Integrls 9 9 8- Trigonometric Sustitution
Clcultor Procedures Clcultor Settings Your clcultor should e in Rdin mode when performing Clculus prolems. (Those of you in Physics need to e in Degree mode while in Physics clss. You will e switching modes dily). Push ZOOM for pre-set windows Grphing Windows ) ZDeciml is the est for trcing (.7 to.7). ( 6. to 6. for TI-89) ) ZStndrd shows most of the grph ( to ). ) ZoomFit will help you find the grph if the grph is not in your current window. Intersection TI - 8 / 8 TI - 89 Push CALC Intersection Push CALC Intersection ) Move cursor ner point of intersection. ) Move cursor ner point of intersection. ) Push enter for first eqution. ) Push enter for first eqution. ) Push enter for second eqution. ) Push enter for second eqution. ) Push enter for guess. ) Move cursor to left of intersection nd press enter. Push CALC Zero Zeroes ) Move cursor to the left of intercept (lower ound) nd push enter. ) Move cursor to the right of intercept (lower ound) nd push enter. ) Push enter for guess 5) Move cursor to right of intersection nd press enter. Derivtives Derivtive t ll points nd Derivtive Derivtive t point TI-8: NA NA nderiv( eqution, x, point) TI-8: NA NA d dx eqution) TI-89: d( eqution, x) d( eqution, x,) d( eqution, x) x = point Indefinite Integrl Integrls Definite Integrl TI-8: NA fnint eqution, x, lower ound, upper ound upper ound TI-8: NA eqution TI-89: eqution, x lower ound ( eqution, x, lower ound, upper ound) dx
CH. LIMITS ** When evluting limits, we re checking round the point tht we re pproching, NOT t the point. **Every time we find limit, we need to check from the left nd the right hnd side (Only if there is BREAK t tht point). - Finding Limits Grphiclly nd Numericlly No reking point Hole in the grph piece-wise function rdicls lim f x x = lim f x x = lim f x x = c = lim f x x + lim x f ( x) = lim f x x = = lim f x x + lim x f ( x) = Asymptotes Asymptotes Asymptotes lim f ( x) = x + lim x lim f x x f ( x) = = lim f ( x) = x + lim x lim f x x f ( x) = = lim f ( x) = x + lim x lim f x x f ( x) = = **If left nd right hnd limits DISAGREE, then the limit Does Not Exist (DNE) t tht point. **If left nd right hnd limits AGREE, then the limit exists t tht point s tht vlue. **Even if you cn plug in the vlue, the limit might not exist t tht point. It might not exist from the left or right side or the two sides will not gree. lim x + = right hnd limit lim f x x f ( x ) = left hnd limit
**Breking Points re points on the grph tht re undefined or where the grph is split into pieces. Breking Points : ) Holes (when the numertor nd denomintor equls ) ) Rdicls (when the rdicl equls ) ) Asymptotes (when the denomintor equls ) ) Piece-wise functions (the # where the grph is split) Note : In generl when doing limits, # = # = # = **In limits, if the two sides of grph don t gree, then the limit does not exist. - Anlyzing Limits Anlyticlly LIMITS AT NON - BREAKING POINTS ( Very esy. Just plug in the #) EX#: lim x x = EX# : lim x 5 x + = EX# : lim x x x + = HOLES IN THE GRAPH ( ) ( Fctor nd cncel or multiply y the conjugte nd cncel, then plug in #) EX#: lim x x + x 8 x = EX# : lim x x + 5 x + = EX# : x lim = EX# : lim x x x 5 x 6 6 x = TRIG. FUNCTIONS Trig. Identities to know : sinx = sin x cos x cosx = cos x sin x FACTS : lim x lim x sin x x sinx x = lim x = lim x cos x x cosx x = lim x = lim x tn x = x tnx = x EX#: EX# : EX# : lim x lim x lim x π sinx 5x tn5x 9x sin x x = EX# : lim x = EX#5 : lim x = EX#6 : lim x sin x tn x = x 8sin x cos x = x 7sin x tn x x =
RADICALS - Continuity nd One-Sided Limits ( You must first check tht the limit exists on the side(s) you re checking) If # mkes rdicl negtive, the limit will not exist t tht #. When we check t the reking point (the # tht mkes the rdicl zero) there re two possile nswers: ) if the limit works from the side tht you re checking. ) DNE if the limit does not work from the side tht you re checking. EX #: lim x 5 + x 5 = EX # : lim x 5 = EX # : lim x 5 x 5 x 5 = EX # : lim 6 x = EX #5 : lim 6 x = EX #6 : lim 6 x = x x 5 x f ( x) = x x < x + x < 9 x PIECE - WISE FUNCTIONS The reking points re nd. EX #: EX # : EX #7 : lim x + lim x lim x 7 + = EX # : lim f x f x x + = EX #5 : lim f x x = EX #8 : lim x 5 = EX # : lim f x x f ( x) = f ( x) = EX #6 : lim f ( x) = x f ( x) = EX #9 : lim f ( x) = x CONTINUITY Continuous functions hve no reks in them. Discontinuous functions hve reks in them (Asymptotes or Holes / Open Circles). ** To check for continuity t, you must check left hnd limits lim x s well s the vlue of the function t tht point f nd right hnd limits lim f x x + f ( x). If ll three re equl then the function is continuous t. If f ( ) = lim x f ( x) x + f ( x ) then the function is continuous t. If f ( ) is not equl to either one - sided limit, then the function is not continuous (discontinuous) t. Continuous t 5 Discontinuous t
-5 Infinite Limits (Since the point DNE we hve to check point tht is close on the side we re pproching) ASYMPTOTES # There re three possile nswers when checking ner the reking point (the # tht mkes ottom = zero) ) If we get positive nswer the limit pproches ) If we get negtive nswer the limit pproches ) DNE If we get positive nswer on one side nd negtive nswer on the other side, then the limit DNE EX #: EX # : lim = EX # : lim = EX # : lim x 7 + x 7 x 7 x 7 x 7 lim x = EX #5 : lim cos x ( x ) x x x 7 = = EX #6 : lim tn x = x π + Check the powers of the numertor nd denomintor. -5 Limits t Infinity LIMITS THAT APPROACH INFINITY ) If the denomintor (ottom) is igger power the limit =. ) If the numertor (top) is igger power the limit = or -. ) If powers re the sme the limit = EX#:lim x x + 5x 7 = EX# : lim x coefficient of the highest power of numertor coefficient of the highest power of denomintor 6x x + = EX# : lim x x x = EX# : lim x 8 x x 8 = EX#5 : lim x 8x + 5 = EX#6 : lim x x 7x + 8x 7 = EX#7 : lim x 6x x = EX#8 : lim = x FINDING VERTICAL ASYMPTOTES AND HOLES A verticl symptote is the # tht mkes only the denomintor =. A hole occurs t the points tht mke the numertor nd denomintor = t the sme time. EX#: ) f ( x) = x x + ) f ( x) = 8x x + 9 c) f x = ( x + 5) ( x 7) ( x 7) d) f x = ( x +) ( x + ) ( x + ) ( x 6) vert.sym. hole vert.sym. hole vert.sym. hole vert.sym. holes 6
CH. DIFFERENTIATION - The Derivtive y Definition nd the Tngent Line Prolem Derivtive t the point (, f ( ) ) Derivtive t ll points f ( x) = lim h f(x+h) f ( x) f x + h h l f ( ) = lim h f ( ) f + h h f(x) f(x+h) f(x) l f(x+h)=f(x) l x x+h x x+h x Line l is secnt line f ( x) slope of secnt line l = f x + h x + h x h Line l is secnt line h f ( x) f x + h lim mens tht the distnce h is pproching nd h h the points get closer to ech other nd the two points ecome the sme point nd line l is now tngent line. Line l is tngent line The derivtive of function finds the slope of the tngent line! EX #: f ( x) = = x x + Find f ( x) nd f x f ( ). Use f x = lim h f ( x) f x + h h EX # : f ( ) = = x x + Find f ( ) Use f ( ) = lim f x h f ( ) f + h h 7
tngent line to curve The derivtive finds the slope of the tngent line. The norml line is perpendiculr to the tngent line. norml line to curve EX # : f ( x) = x Find eqution of the tngent line nd norml line t x =. Eqution of the tngent line : Eqution of the norml line : 8
- Bsic Differentition Rules, Nottion nd Rtes of Chnge Properties of Derivtives Derivtive is rte of chnge; it finds the chnge in y over the chnge in x, dy, which is slope. dx st derivtive mx. nd min., incresing nd decresing, slope of the tngent line to the curve, nd velocity. nd derivtive inflection points, concvity, nd ccelertion. *Power Rule * Constnt Rule f (x) = x n f (x) = c f ( x) = nx n f x = EX : y = x y = 8 y = x y = EX#: y = 5x Find dy dx nd dy dx x=. Newton Nottion Derivtive nottion Leiniz Nottion y = x y = x y = x dy dx = x y = 6x d y dx = 6x y = 6 y ( ) = d y dx = 6 d y dx = EX# : Find f ( x) for ech. ) f (x) = 5 ) f (x) = 8 x c) f (x) = x d) f (x) = x 7 EX : Slope of the tngent line to the curve Given f ( x) = x x Find eqution of the tngent line nd norml line t x =. = x x f ( x) = 6x = 8 f ( ) = f x f Eqution of Line (point - slope form): y y = m x x Eqution of the tngent line : y 8 = x Eqution of the norml line : y 8 = x 9
EX # : Find the slope, write the eqution of the tngent line nd the eqution of the norml line t x =. = 5x 7 Eqution of the tngent line Eqution of the norml line f x *Trig. Functions STEPS : Derivtive of the trig. function Derivtive of the ngle Function Derivtive Function Derivtive sin x cos x tn x sec x cos x sin x cot x csc x Function sec x csc x Derivtive sec x tn x csc x cot x EX # : Find ech derivtive ) y = sin( 8x) ) y = cos x c) y = tn( 7x ) d) y = sec( 6x + 5) e) y = csc( x 5 ) f ) y = cot( 9x ) g) y = cos( sin x) h) y = sec( tn x) Use s( t) = 6t + v t + s. s( t) = ending height s = initil height v = initil velocity t = time EX #5 : To estimte the height of uilding, weight is dropped from the top of the uilding into pool t ground level. How high is the uilding if the splsh is seen seconds fter the weight is dropped? EX #6 : A red ll is thrown upwrd from uilding feet ove the ground with n initil velocity of ft/s. At the sme time, lue ll is thrown downwrd from height of 5 feet with n initil velocity of ft/s. Which ll hits the ground first? How much fster?
*Product Rule - Product nd Quotient Rule STEPS : Derivtive of First eqution Second eqution + Derivtive of Second eqution First eqution y = f (x) g(x) *Quotient Rule 5 STEPS : y = f (x) g(x) dy dx = f (x)g(x) + g (x) f (x) Derivtive of Top eqution Bottom eqution Derivtive of Bottom eqution Top eqution ( Bottom eqution) dy dx = f (x)g(x) g (x) f (x) g(x) - Chin Rule STEPS : ) Power in front ) Lower power y ) Multiply y derivtive of inside OR Derivtive of outside function Derivtive of inside function y = ( f (x)) n OR y = f g( x) y = f ( g( x) ) g x y = n( f (x)) n f (x) OR y = f x OR y = f x y = f ( x) y = f ( x ) x EX #: f ( x) = x tn x EX # : y = ( x 5) 6 EX # : f ( x) = sin x cos x EX # : y = x x 5
-5 Implicit Differentition The differentile functions we hve encountered so fr cn e descried y equtions in which "y" is expressed in terms of "x". We cn lso find the derivtive of the eqution expressed in terms of x nd y. *Implicit Differentition: function in terms of x s nd y s must write dy dx everytime you tke deriv. of y EX : x xy + y = 7 x y() + dy dx x + 6y dy dx = x y x dy dy + 6y dx dx = x dy dy + 6y dx dx = x + y dy ( x + 6y) = x + y dx dy dx = x + y x + 6y EX#: x + y = 6 (, 7) Find the eqution of the tngent line t the given point. EX # : Find first two derivtives of x y = 5 EX# : Find dy dx. 5x xy + y = x EX# : Find dy dx. ( Use shortcut) 5x xy + y = x
Questions from CH. TEST Derivtive of picture = f ( ) = f ( ) = f = f ( 7) = f ( 8) = f 6 5 6 7 8 9 5 - Eqution of the tngent line t x = Eqution of the tngent line t x = Derivtive of chrt x 6 8 6 f ( x) 58 6 7 6 6 69 6 7 67 EX : f ( ) = 6 6 6 = = EX : f ( ) = 7 6 = EX#: f = EX# : f ( ) = How to red Derivtive y Definition Prolems This prolem mens tke derivtive of cos x. This prolem mens tke derivtive of x t x =. EX : lim h cos( x + h) cos x = sin x EX : lim h h ( + h) 8 = h Solve ech EX#: lim h 5( x + h) 5x h = EX# : lim h ( + h) 8 = h EX# : lim h tn( x + h) tn x = EX# : lim h h sin( π 6 + h ) = h
.6 Relted Rtes We tke derivtives with respect to t which llows us to find velocity. Here is how you tke derivtive with respect to t: derivtive of x is dx dt, derivtive of y is y dy dt, derivtive of r dr is r dt, derivtive of t is t dt dt = t dv V mens volume ; mens rte of chnge of volume (how fst the volume is chnging) dt dr r mens rdius ; mens rte of chnge of rdius (how fst the rdius is chnging) dt dx is how fst x is chnging; dy is how fst y is chnging dt dt Volume of sphere Surfce Are of sphere Are of circle Circumference of circle V = πr A = πr A = πr C = πr dv dt = πr dr dt da dt = 8πr dr dt da dt = πr dr dt dc dt = π dr dt Volume of cylinder Volume of cone V = πr h V = πr h use r h = r is not vrile in cylinder ecuse its' vlue is lwys the sme Due to similr tringles, the rtio of the rdius to the height is lwys the sme. Replce r or h depending on wht you re looking for. EX #: Suppose sphericl lloon is inflted t the rte of in / min. How fst is the rdius of the lloon chnging when the rdius is 7 inches? EX # : Wter is poured into cylinder with rdius 5 t the rte of in / s. How fst is the height of the wter chnging when the height is inches? 5
EX # : Wter is leking out of cone with dimeter inches nd height 9 inches t the rte of 5 in / s. How fst is the rdius of the wter chnging when the rdius is in? 9 EX # : A 7 foot ldder is lening ginst the wll of house. The se of the ldder is pulled wy t ft. per second. ) How fst is the ldder sliding down the wll when the se of the ldder is 5 ft. from the wll? ) How fst is the re of the tringle formed chnging t this time? 7 y θ x c) How fst is the ngle etween the ottom of the ldder nd the floor chnging t this time? EX #5 : A person 6 ft. tll wlks directly wy from streetlight tht is feet ove the ground. The person is wlking wy from the light t constnt rte of feet per second. ) At wht rte, in feet per second, is the length of the shdow chnging? 6 x y ) At wht rte, in feet per second, is the tip of the shdow chnging? 5
CH. APPLICATIONS OF DIFFERENTIATION - Extrem on n Intervl Let f e continuous on closed ounded intervl [,]. Then f hs n solute mximum nd solute minimum on the intervl [,]. Procedure for finding solute mx. nd solute min. : Compute the vlues of f t ll criticl points (when f ( x) = ) in, The lrgest of these vlues is the solute mximum vlue of f on [,]. The smllest of these vlues is the solute minimum vlue of f on [,]. nd t the endpoints nd. Find the extreme vlues (mx. nd min.) of f on given intervl nd determine t which #'s in they occur. EX#: Let f x = x x ; [,] There re criticl points tht show up s undefined in the derivtive, ut work in the originl eqution. I cll these specil points Hrdpoints. The textooks don't hve nme for these points so I gve them one. Find the extreme vlues (mx. nd min.) of f on given intervl nd determine t which #'s in they occur. EX# : f ( x) = x ; [ 8, 7] EX# : f ( x) = x ; [, ] 6
*Men-Vlue Theorem - Men Vlue Theorem nd Rolle's Theorem (Only pplies if the function is continuous nd differentile) f() Slope of tngent line = slope of line etween two points f () f () f (c) = According to the Men Vlue Theorem, there must e numer c etween nd tht the slope of the tngent line t c is the sme s the slope etween points (, f ()) nd (, f ()). The slope of secnt line from nd is the sme s slope of tngent line through c. f() f() 5 f() c 5 Use the Men Vlue Theorem to find ll vlues of c in the open intervl (, ) such tht f (c) = EX #: f ( x) = sin x [, π ] EX # : f ( x) = x + 7 [, ] f () f () EX # : f ( x) = x [, 5] 7
- Incresing nd Decresing Functions nd the First Derivtive Test Properties of First Derivtive incresing : slopes of tngent lines re positive (derivtive is positive.) f decresing : slopes of tngent lines re negtive (derivtive is negtive). f ( x) > ( x) < ( = ) f ( x) = mximum point : Slopes switch from positive to negtive t mximum point. found y setting f x minimum point : Slopes switch from negtive to positive t minimum point. found y setting ***st Derivtive Test ) If f chnges from positive to negtive t the criticl point c, then f hs reltive mximum vlue t c. ) If f chnges from negtive to positive t the criticl point c, then f hs reltive minimum vlue t c. - Concvity nd the Second Derivtive Test Properties of Second Derivtive concve up : slopes of tngent lines re incresing (nd Derivtive is positive). f concve down : slopes of tngent lines re decresing (nd Derivtive is negtive). f inflection points : points where the grph switches concvity. found y setting f x ( x) > ( x) < ( = ) slopes of tngent line switch from incresing to decresing or vice vers. ***nd Derivtive Test (Alternte method for finding rel. mx. nd rel. min.) Set f ( c) = to find your criticl points c. Plug your criticl point(s) into f ( x). ) If f ( c) <, then c is rel. mx. ) If f ( c) >, then c is rel. min. c) If f ( c) =, then from this test lone we cn't drw ny conclusions out reltive extreme vlue t c incresing/concve down slopes re positive nd decresing incresing/concve up slopes re positive nd incresing decresing/concve up slopes re negtive nd incresing decresing/concve down slopes re negtive nd decresing M M M is Mximum; slopes switch from positive to negtive m m is Minimum; slopes switch from negtive to positive I is n inflection point; slopes switch from decresing to incresing I I m, I m M is Mximum; m is minimum; I is n inflection point 8
EX #: From [,7] tell me out the function. (Use grph ove) List the x - coordintes for ech : Find ech : On which intervl(s) is the grph: Inflection points As. mx. vlue incresing/concve up Reltive mximum As. min. vlue incresing/concve down Reltive minimum As. mx. vlue occurs t decresing/concve up Hrd points As. min. vlue occurs t decresing/concve down EX # : f ( x) = x x is incresing nd decresing. Find reltive extreme vlues nd determine the intervls on which f x Find ll inflection points nd the intervls on which f ( x)is concve up nd concve down. 5 6 7 EX # : f ( x) = ( x ) + 7x 7 is incresing nd decresing. Find reltive extreme vlues nd determine the intervls on which f x Find ll inflection points nd the intervls on which f ( x)is concve up nd concve down. 9
.6 A Summry of Curve Sketching GRAPHING TRIG. REVIEW x - intercepts : where grph crosses the x-xis. The # tht mkes y =. To find the x-intercept, set the numertor = (plug in zero for y). y - intercepts : where grph crosses the y-xis. The # found when x =. To find the y-intercept, plug in zero for x. Holes : the # tht mkes oth the numertor nd denomintor =. To find the hole in the grph, you plug the # into the remining function fter cnceling out the like fctors. Find the hole first efore finding the verticl symptote. verticl symptotes : An undefined point on the grph. A grph will never cross the verticl symptote. To find the verticl symptote, set the denomintor =. The verticl symptote is x = the # tht mkes only the denomintor =. horizontl symptotes : The grph will pproch the horizontl symptote s x pproches nd. To find the horizontl symptote, check the highest powers of the numertor nd the denomintor. ) If the denomintor (ottom) is igger power the horizontl symptote is y =. ) If the numertor (top) is igger power there is no horizontl symptote (there is different kind of symptote). ) If powers re the sme the horizontl symptote is y = slnt symptotes : qudrtic symptotes : coefficient of the highest power of numertor coefficient of the highest power of denomintor Occur when the numertor is one power higher thn the denomintor. To find the slnt symptote, you must use long division to divide the denomintor into the numertor. The quotient is your slnt symptote. Occur when the numertor is two powers higher thn the denomintor. To find the qudrtic symptote, you must use long division to divide the denomintor into the numertor. The quotient is your qudrtic symptote. Sketch the grph ( Lel the mximum, minimum nd inflection points) EX#: y = x x y = x 6x y = 6x 6 x int y int v.sym. h.sym. rel.mx. rel.min. inc. dec. inf.pts. conc.up conc.down (, ) (, ) NONE NONE (, ) (, ) (, ) (, ) (, ) (, ) (,) (, ) (, ) x x = x 6x = 6x 6 = = x( x ) = 6( x ) = x x x =, x =, x = + + f '(x) - f "(x) - + - M - - - I - - -5 m
Note ll relevnt properties of f nd sketch the grph ( Lel the mximum, minimum nd inflection points) EX# : y = x( x ) x int y int v.sym. h.sym. rel.mx. rel.min. inc. dec. inf.pts. conc.up conc.down 5-5 - - - - - 5 - - - -5 Note ll relevnt properties of f nd sketch the grph ( Lel the mximum, minimum nd inflection points) EX# : y = x ( x +) x int y int v.sym. h.sym. rel.mx. rel.min. inc. dec. inf.pts. conc.up conc.down 5-5 - - - - - 5 - - - -5
-7 Optimiztion Prolems ) Drw nd lel picture. ) Write eqution sed on fct given nd write eqution for wht you need to mximize or minimize. ) Plug in fct eqution into the eqution you wnt to mximize or minimize. ) Tke derivtive nd set equl to zero. 5) Find remining informtion. EX#: An open ox of mximum volume is to e mde from squre piece of mteril, inches on side, y cutting equl squres from the corners nd turning up the sides. How much should you cut off from the corners? Wht is the mximum volume of your ox? X X X X X X X X -X X X X X X X -X X X EX# : A frmer plns to fence rectngulr psture djcent to river. The frmer hs 56 feet of fence in which to enclose the psture. Wht dimensions should e used so tht the enclosed re will e mximum? Wht is the mximum re? RIVER Y Y X
EX# : A crte, open t the top, hs verticl sides, squre ottom nd volume of 86 ft. Wht dimensions give us minimum surfce re? Wht is the surfce re? Y X X EX# : A rectngle is ounded y the x-xis nd the semicircle y = 6 x. Wht length nd width should the rectngle hve so tht its re is mximum? Y= 6-X Y X -5 5 X
-8 Newton's Method Newton's method is used to pproximte zero of function. *Newton s Method c f ( c) f ( c) where c is the pproximtion for the zero. EX# : If Newton s method is used to pproximte the rel root of x + x =, then first pproximtion x = would led to third pproximtion of x = f (x) = x + x x = EX# : Given f (x) = x nd first pproximtion x = 5. Find third pproximtion x =. -9 Differentils Differentils re tngent lines. Tngent lines hug closely long grph ner the point of tngency. Sometimes it's esier to use the differentil to pproximte vlue on grph s opposed to using the grph itself. = ln x EX# : xy + y x = 7 EX# : f x ) Find the eqution of the tngent line t ( e, ). ) Find the eqution of the tngent line t (, ). ) Use the differentil to pprox. f ( ). ) Use the differentil to pprox. f (.). c) f ( ) = c) f (.) =
CH. INTEGRATION - Antiderivtives nd Indefinite Integrtion ** Integrls : They find the ntiderivtives, sum/totl, Are nd Volume. Integrtion Formuls ( f ( x) + g( x) ) dx = f ( x) dx + g( x) dx c f ( x) dx = c f ( x) dx where c is constnt. Constnts cn move in nd out of n integrl. f x dx = F x + C (nottion) *Integrl of constnt dx = x + C EX : 5 dx = 5x + C EX : π dx = π x + C *Polynomils x dx = x + + + C EX : x dx = x + C EX : x 6 dx = x7 7 + C *Frctions Bring up denomintor, then tke integrl dx x dx = x x + + + C EX : x dx x dx = x + C = x + C *Trig Functions ( Alwys divide y derivtive of the ngle) sin x dx = cos x + C EX : sin 7x dx = cos x dx = sin x + C EX : cos x dx = cos 7x + C 7 sin x + C 5
EX#: ) x 8 dx = ) x dx = c) 9 dx = d) 7 dx = x EX# : ) ( x 7x + ) dx = ) x + 8x dx = x EX# : ) cosx dx = ) sin 7x dx = Symols used position : x( t), y( t), s t velocity : v t ccelertion : ( t) Derivtive Rectiliner Motion position velocity ccelertion Integrl EX# : = t 6t 6 x( ) = t > v t ) Find position t ny time t. ) Find ccelertion t ny time t. c) Find velocity t t = 7. d) Find ccelertion t t = 7. e) Is the speed inc. or dec. t t = 7? f ) When is the prticle t rest? 6
st Fundmentl Theorem of Clculus - Riemnn Sums nd Definite Integrls After you tke the integrl, just plug in the top # minus the ottom #. dx f x = F x Are = F( ) F( ) ( top eqution ottom eqution) dx EX #: Find Are of ech shded region ) ) c) y=x EX # : Evlute x x + 5 dx = EX # : Evlute 6 dx = x EX # : Find Are of the region etween y = sin x nd the x-xis from [,π ] y=sinx - - π π 5 π π 7
- The Fundmentl Theorem of Clculus Averge Vlue (use this when you re sked to find the verge of nything) If f is integrle on the closed intervl [,], then the verge vlue of f on the intervl is Averge vlue = f (x) dx EX#: Find the verge vlue of f (x) on the closed intervl. EX # : Find verge ccelertion from, ) f ( x) = x, [ ] ) f x = x, [ ] v(t) = t + t 5 [ ] nd Fundmentl Theorem of Clculus (When tking the derivtive of n integrl) Plug in the vrile on top times its derivtive minus plug in the vrile on ottom times its derivtive. EX # : ) d dx x d dx x dt f t = f x Evlute ech. d dx t dt = ) x f ( t) dt = f ( x) f ( ) = f x d dx t dt = x c) F x x dt = t 7 + = d) F x F ( x) = F x sin x = + t dt = = x - - 5 6 EX# : ) f x dx = ) f x dx = c) f x dx = 6 6 d) f x dx = e) f x dx = f ) f x dx = - 6 8 6 ( + )
*Sustitution -5 Integrtion y Sustitution When integrting we usully let u = the prt in the prenthesis, the prt under the rdicl, the denomintor, the exponent, or the ngle of the trig. function. f g( x) g ( x)dx = f ( u) du = F u Let u = g( x) du = g ( x)dx + C + C = F g( x) EX#: x ( x + 5) sin x dx = EX# : cos x dx = EX# : x ( x 5 ) 7 dx = EX# : xsin x dx = EX#5 : x x + dx = EX#6 : x 6 x 7 + 8 dx = 5 EX#7 : x x + 9 dx = EX#8 : x x 6 dx = 9
-6 Numericl Integrtion (Approximting Are) We pproximte Are using rectngles (left, right, nd midpoint) nd trpezoids. *Riemnn Sums 5 ) Left edge Rectngles f (x) = x + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = n left height of ech rectngle Totl Are= 8.75 A = + 5 + + ) Right edge Rectngles f (x) = x + from [, ] using sudivisions (Find re of ech rectngle nd dd together) - 5 A = n right height of ech rectngle A = Totl Are = 6 8 5.75 c) Midpoint Rectngles f (x) = x + from [, ] using sudivisions A = (Find re of ech rectngle nd dd together) n midpt. height of ech rectngle A = Totl Are = 8.65 5 + + + 5 7 6 + 5 6 + 6 + 65 6-5 *Trpezoidl Rule (used to pproximte re under curve, using trpezoids). Are f (x ) + f (x ) + f (x ) + f (x )... f (x n ) + f (x n ) n where n is the numer of sudivisions. [ ] - d) f (x) = x + Approximte the re under the curve from [, ] A = f () + f using the trpezoidl rule with sudivisions. + f () + f + f () = + 5 + () + + 5 = 76 = 76 6 = =.75 All you re doing is finding the re of the trpezoids nd dding them together! 5 - dx e) Actul Are = x + = x + x 8 = + + = =.667
*Approximting Are when given dt only ( no eqution given) To estimte the re of plot of lnd, surveyor tkes severl mesurements. The mesurements re tken every 5 feet for the ft. long plot of lnd, where y represents the distnce cross the lnd t ech 5 ft. increment. x 5 5 6 75 9 5 y 58 6 7 6 6 69 6 7 67 ) Estimte using Trpezoidl Rule ) Estimte using Midpoint sudivisions A 8 f ( ) + f ( 5) +...+ f ( 5) + f ( ) A [ ] A [ 6+ 6 + 69 + 7 ] A 5 58 +6 + + + +8 + +8 + 67 A 785.5 A 798 f ( 5) + f ( 5) + f ( 75) + f ( 5) c) Estimte Avg. vlue using Trpezoidl Rule d) Wht re you finding in prt c? 65.75 The verge distnce cross the lnd. Avg.Vlue 785.5 e) Estimte using Left Endpoint f) Estimte using Right Endpoint A f ( ) + f ( 5) + f ( ) +...+ f ( 5) 8 A 8 A 5 58 + 6+ 7 + 6 + 6 + 69 + 6+ 7 f ( 5) + f ( ) + f ( 5)...+ f ( ) [ ] A 5[ 6+ 7 + 6 + 6 + 69 + 6+ 7 + 67] A 7785 A 79 *Approximting Are when given dt only ( no eqution given) Unequl sudivisions: You must find ech Are seprtely. x 5 y 5 ) Estimte using Trpezoids A = ( + )h A! ( + ) + ( +) + ( +5 ) 5! ) Estimte using Left Endpoint( A = width left height) A! ( )+ ( )+5( )!! Trpezoids shown c) Estimte using Right Endpoint A = width right height A! ( )+ ( )+5 5
EX #: To estimte the surfce re of his pool, mn tkes severl mesurements. The mesurements re tken every 5 feet for the 5 ft. long pool, where y represents the distnce cross the pool t ech 5 ft. increment. x y 5 7 5 9 5 7 5 5 5 5 8 5 5 5 5 Estimte ech Are using sudivisions 5 5 5 5 5 5 ) Use Trpezoidl Rule ) Estimte Avg. vlue using Trpezoidl Rule c) Use Right Endpoint d) Use Left Endpoint e) Use 5 Midpoint sudivisions 8 5 6 5 x y 5 8 5 5 5 EX# : Estimte ech Are using unequl sudivisions 5 ) Use Trpezoids ) Estimte Avg. vlue using Trpezoids c) Use Right Endpoint d) Use Left Endpoint
CH. 5 Logrithmic,Exponentil, nd other Trnscendentl Functions 5- The Nturl Logrithm Function nd Differentition Properties of Logrithms lim + n n = e e = nturl numer n logrithmic form exponentil form Log Lws : y = ln x y y = yln x ln x + ln y = ln xy y = ln x e y = x ln 8 = ln = ln ln + ln5 = ln ln x ln y = ln x y ln 7 ln = ln 7 Chnge of Bse Lw : y = log x y = ln x ln or log x log (,) - y=e x (,e) (,) (e,) y=lnx Memorize these grphs. They re inverses of ech other so they re symmetric out the line y=x. Memorize : lne = ln = 5 Fct : You cn t tke ln/log of negtive # or zero. We use logrithms to solve ny prolem tht hs vrile in the exponent. EX : e 5 x = lne 5 x = ln 5x lne = ln x = ln 5 EX : ln x = e ln x = e x = e *Derivtive of Nturl Log ( STEPS: divided y function i Derivtive of function) y = ln( f (x)) y = f (x) f (x) Find derivtive of ech EX#: d dx ln(x) = EX# : d dx ln(x ) = EX# : d dx ln(x5 ) = EX# : d log x = EX#5 : f x dx = ln x x + x =
5- The Nturl Logrithmic Function: Integrtion f x dx = ln f x f x dx = ln x + C x Integrls of other four trig. functions tn x dx = + C top is the derivtive of the ottom sin x dx = sin x cos x dx = ln cos x + C cos x cos x cot x dx = dx = ln sin x + C sin x sec x dx = sec x(sec x + tn x) dx = sec x + tn x sec x + sec x tn x dx = ln sec x + tn x + C sec x + tn x csc x(csc x + cot x) csc csc x dx = x + csc x cot x dx = csc x + cot x dx = (csc x + csc x cot x) csc x + cot x dx csc x + cot x = ln csc x + cot x + C EX#: x dx = EX# : x 5 x dx = EX# : x 5 dx = 9x + EX# : sec x dx = EX#5 : tn x dx = EX#6 : x 6 x x + dx = 7 EX#7 : x + x + dx = x + EX : tn5x dx = ln cos5x + C EX : sec9 5 x dx = ln sec9x + tn9x + C 9 EX : csc8x dx = ln csc8x + cot 8x + C EX : cotx dx = 8 ln sinx + C
5- Inverse Functions INVERSES : To find n inverse, f ( x), you switch the x's nd y's nd solve for y. EX : f ( x) = x + Find f ( x). EX : f ( x) = e x Find f ( x). Inverse : x = y + Inverse : x = e y lnx = lne y x = y so f ( x) = x Fcts out inverses ) When you plug one inverse into the other you lwys get the nswer x nd vice vers. ( f f ) x EX : = x nd ( f f ) x = f f ( x) f x ( f f ) x = x = f f ( x) = e x f ( x) = ln x = e ln x = x ( f f )( x) = lne x = x ln x = y so f ( x) = ln x y=e x ) Grphs tht re inverses re symmetric out the line y = x. dy finds the slope of the tngent line. dx dx dy = dy dx (,) - (,e) (,) (e,) y=lnx 5 dx dy finds derivtive ( the slope of the tngent line) of the inverse. Formul for finding derivtive of the inverse t point : g ( y) = f x (where g( y) is the inverse of f ( x)) EX #: Let y = x + x. If h is the inverse function of f, then h ( ) = A) B) C) D) E) EX # : Let f e differentile function such tht f The function g is differentile nd g x A) B) 5 E) The vlue of g 7 =, f ( 8) =, f ( ) = 7, nd f ( 8) = 5. = f ( x) for ll x. Wht is the vlue of g ( )? C) 8 D) cnnot e determined from the informtion given. 5
Trig. Review 5- Exponentil Functions:Differentition nd Integrtion e ln x = e ln = e ln 5 = Exponentil Differentition Constnt Vrile d dx ex = e x lne = e x ( steps : itself, multiplied y deriv. of exponent, multiplied y ln of se) d dx ex = d dx e5x = Exponentil Integrtion Constnt Vrile e x dx = ex lne + C = ex + C e x dx = ( steps : itself, divided y deriv. of exponent, divided y ln of se) e5x dx = EX#: Find derivtive of ech ) y = e 7x ) y = e x sin x EX# : Evlute ech integrl ) e x e x dx = ) e x dx = + ln EX# : x e x dx = 6
5-5 Bses other thn e nd Applictions *Derivtive of Constnt Vrile y = f (x) y = f (x) f (x) ln ( steps : itself, multiplied y derivtive of exponent, multiplied y ln of se) EX#: d dx 7x = EX# : d dx x = EX# : f ( t) = t 5 t f ( t) = *Integrl of Constnt Vrile x dx = x ln + C ( steps : itself, divided y deriv. of exponent, divided y ln of se) EX# : 9 x dx = EX#5 : 5 x dx = EX#6 : 8 x dx = EX#7 : x 7 x dx = *Derivtive of Vrile Vrile y = f (x) g(x) ( tke ln of oth sides then tke derivtive of oth sides) dy ln y = g( x)ln f ( x) y dx = g ( x)ln f ( x) + f ( x) f ( x) g( x) dy dx = f x ( ( x) )g g EX#8 : y = x sin x EX#9 : y = ( x) 5 x ( x)ln f x + f ( x) f ( x) g( x) *Derivtive of Vrile Vrile y = f (x) g(x) ( lternte wy) Need to chnge y = f ( x) g( x) to y = e ln f (x)g(x) y = e g(x)ln f (x) then tke derivtive. EX : y = x sin x sin x ln x y = e y = e sin xln x cos x lnx + x sin x = xsin x cos x lnx + sin x x *Derivtive of Vrile Vrile (Shortcut) ( steps : itself, multiplied y product rule of exponent nd ln of se) 7
sinθ = sinθ = 5-6 Inverse Trigonometric Functions: Differentition mens give me ll the nswers. θ = π 6 + nπ = 5π + nπ ; n Z 6 ; [, π ] mens give me only the nswers from, π [ ]. θ = π 6, 5π 6 rcsin mens give me the principl (first) nswer only etween π x π. θ = π 6 EX #: Trig. Review EX # : Trig. Review EX # : Trig. Review ) rcsin = ) rcsin = ) tn rccos = ) sin rccos x = c) rcsin = d) rcsin = e) rctn = f ) rctn g) rcsin = h) rcsin ( ) = i) rctn = j) rctn = *Inverse Trig. Functions y = rcsin f ( x) y = y = rctn f ( x) y = y = rcsec f ( x) y = f ( x) + f x f x = ) cos rccsc 7 8 f ( x) y = rccos f x = ) sec [ rctn x ] = ( ) f ( x) y = rccot f ( x) f ( x) f ( x) y = rccsc f x y = f ( x) y = + f x y = f x f ( x) ( ) f ( x) f ( x) f ( x) Tke derivtive of ech EX#: y = rcsin x EX# : y = rctn x EX# : y = rcsec6x EX# : y = rcsine x EX#5 : y = rctn( ln x) EX#6 : y = rcsec( sin x) 8
5-7 Inverse Trigonometric Functions: Integrtion *Inverse Trig Functions dx = rcsin x x + C dx = + x rctn x + C x x dx = rcsec x + C OR OR OR = rccos x + C = rccot x + C = rccsc x + C Find vrile v nd constnt. The top MUST e the derivtive of the vrile v. EX#: 5 x dx = EX# : dx = EX# : x + 9 x x 6 dx = EX# : dx = EX#5 : 5 x 7 dx = 9x + 6 EX#6 : dx = EX#7 : x 6x 9 dx = x 8x EX#8 : x dx = x + 9
Drw the slope field for ech EX #: 6- Slope Fields nd Euler's Method dy dx = x y EX # : dy dx = x y - - - - - - Here re the slope fields for the given differentil equtions. Sketch the solution for the given point. EX # : dy dx = y ( 8 6 y) EX # : dy dx = x + y Euler's Method New y = Old y + dx dy dx dx = chnge in x (the step), dy = derivtive(slope) t the point dx dy EX #5 : y = x y pssing thru (, ), step of h =. EX #6 : dx = xy Find f (.) = Find f () = pssing thru (, ), step of h =. 5
6- Differentil Equtions: Growth nd Decy *Growth Formul ( Cn e used t ny time) y = C e kt ( Comes from y = ky ) *Compound Continuous Formul (Money eqution) A = P e rt *Doule - Life Formul (Use only when douling is mentioned) y = C () t d *Hlf - Life Formul (Use only when hlf - life is mentioned) A, y = ending mount C, P = initil mount t = time d = doule-life time h = hlf-life time k = growth constnt y = C t h EX #: A certin kind of lge doules every 6 dys. If the eginning popultion of the lge is, wht will the popultion e fter weeks? EX # : If I invest $, compounded continuously for 5 yers nd it grows to $5,. At wht rte ws the money invested? EX # : SHHS popultion in 98 ws people. SHHS popultion in ws 7 people. Wht will the popultion e in 5 t the sme growth rte? (Round nswer to nerest whole #) EX # : Cron hs hlf-life of 57 yers. We mesure the mount of cron in tree nd it hs % less cron thn when it ws plnted. How old is the tree? (Round nswer to nerest whole #)
6- Seprtion of Vriles nd the Logistic Eqution (used when you re given the DIFFERENTIAL EQUATIONS Seprting Vriles derivtive nd you need to find the originl eqution. We seprte the x's nd y's nd tke the integrl). EX #: Find the generl solution given dy dx = x + y EX # : Find the prticulr solution of ech: ) xy dy ln x = y() = 8 ) dx y x y = y() = dy EX # : AP Test BC#6 dx = x y ) Find the prticulr solution y = f x given f ( ) = ) Drw slope field t points indicted. - - - - If the rte of growth of something is proportionl to itself ( y = ky), then it is the growth formul ( y = C e kt ). Proof : y = ky dy dy = ky dt y = k dt e ln y = e kt+c y = e kt e C y = C e kt dy = k dt y ln y = kt + C
Logisticl Growth dy dt = ky y L L y = ; = L Y + e kt Y dp dt = kp P L P t = L + e kt ; = L P P k = constnt of proportionlity L = crrying cpcity Y = initil mount k = constnt of proportionlity L = crrying cpcity P = initil popultion EX #: ) If P dp dt = P 5 P =, wht is the lim P( t)? t If P =, wht is the lim t P( t)? ) If P( ) =, for wht vlue of P( t) is the popultion growing the fstest? c) Find originl eqution if P( ) =. dp P EX # : = P dt 5 ) lim P( t)? t P( ) = ) Find P( t). c) At wht time is the popultion growing the fstest?
8 6 - *Are CH.7 Appliction of Integrtion 7- Are of Region Between Two Curves A = [top eqution-ottom eqution] dx f(x) g(x) f(x) g(x) h(x) c c Are from to c Are from to c c A = [ f (x) g(x)] dx + [g(x) f (x)] dx A = [ f (x) h(x)] dx + [g(x) h(x)] dx Steps to find Are : ) Find out where the equtions meet. ) Find out which eqution is on top. c EX #: f ( x) = x + g( x) = x Find Are of region etween grphs from [,] 5-5 - - - - - -5 5 EX # : - - - -5 = x + g( x) = e x Find the Are of the enclosed region. f x 8 7 6 5 5 - - - - - - EX # : Find the Are of the region ounded y the grphs of x = y nd x = y +. 5-5 - - - - - -5 5 - - - -5 5-6
7- Volume: The Disc method Volume V = A( x ) dx where A( x ) is the Are of the cross section. ( This is the formul for ll volume prolems. ) Volume if cross section rotted is circle( A = π r ) V = π ( top function) ottom function dx EX #: rdius f ( x) = x [, 9] Find the volume of the enclosed region etween f x nd the x-xis rotted out the x-xis. - - 5 6 7 8 9 EX # : = 5x x g( x) = x f x Find the volume of the enclosed region rotted out the x-xis 7 6 5 - - 5 6 7-5
7- Volume: The Shell method Volume if cross section rotted is cylinder A = πrh V = π [ ] x top function ottom function dx rdius height EX #: f ( x) = x [, 5] Find the volume of the enclosed region etween f x nd the x-xis rotted out the y-xis -5 - - - 5 - - -5 5 - - - -5-6 5 EX # : f ( x) = x f ( x) = x [,] Find the volume of the enclosed region revolved out the y-xis 5-5 - - - - - - - - -5 5-5 5 6
Volumes: Rottions out other lines *Volume rotted out : ( Verticl cross section) the x - xis ( g(x) ) the y - xis [ ] π f (x) dx π x f (x) g(x) dx the line y = k ( g(x) + k) the line x = c π f (x) + k dx π x + c the line y = m [ f (x) g(x) ] the line x = d dx y=m x=-c y=-k f(x) g(x) x=d ( m f (x)) π m g(x) dx π d x [ f (x) g(x) ] dx EX # : f ( x) = x x g( x) = x [, ] Find the volume of the solid formed when rotting the enclosed region out the given lines. the x - xis the line y = the line y = 8 the y - xis the line x = the line x = EX # : f ( x) = x [, ] Find the volume of the solid formed when rotting the enclosed region out the given lines. the x - xis the line y = 5 the line y = 6 the y - xis the line x = the line x = 7
Volumes of known cross sections *Volume (Region is not rotted) V = A( x) is the Are of the cross section. dx where A x - Sometimes we will find the volume of regions tht hve different cross sections (not circle or cylinder). - These regions re not rotted ut come out t us. - We must first find the Are of the cross section, then tke it's integrl. A = multiplier ( top eqution ottom eqution) dx EX# : Let R e the region in the first qudrnt elow f x nd ove g( x) from x = to x =. Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the x-xis re : Squres A = s f ( x) g( x) dx f ( x) g( x) V = f ( x) g( x) Equilterl Δ's A = s f(x) g(x) f ( x) g( x) Semicircle A = πr V = ( f ( x) g( x) ) dx V = ( f ( x) g( x) ) dx f ( x) g( x) = dimeter f ( x) g x V = dx π f ( x) g x V = π ( f ( x) g( x) ) 8 dx = rdius Rectngle with h = 5 ( A = 5h) f ( x) g( x) ( g( x) ) V = 5 f ( x) g( x) 5 f x Rectngle with h = x ( A = h = ( x) ) f ( x) g( x) x V = f x f x ( g( x) )dx V = 5 f x dx ( g( x) ) x dx 8 ( g( x) )
Isosceles Right Tringle A = h f(x)-g(x) f(x)-g(x) V = V = ( f ( x) g( x) ) ( f ( x) g( x) )dx V = ( f ( x) g( x) ) dx Leg is se g( x) g( x) f x f x dx V = ( f ( x) g( x) ) dx Hyp. is se Multipliers for other figures : -6-9( SL): Regulr Hexgon : Regulr Polygon : -6-9( LL): -6-9( HYP) : 8 # of sides tn( hlf of the interior ngle) EX#: Let R e the region in the first qudrnt under the grph of y = x for x 7. Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the x-xis (verticl cross sections) re : f( x) = x R 5 7 ) Squres ) Equilterl Δ's c) Semicircle d) Rectngle with h = 5 e) Rectngle with h = 6 x f ) Isos.Rt. ( Leg is se) Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the y-xis (horizontl cross sections) re : g) Squres h) Isos.Rt. (Hypotenuse is se) 9
7- Arclengths nd Surfce of Revolution This formul cn e used to find the length of n rc or the distnce prticle trvels long n rc. Let the function y = f ( x) represent the smooth curve on the intervl [, ]. The rclength of f etween nd is: ( ) s = + f x dx d c Also, for smooth curve given x = g( y), the rclength etween c nd d is: d ( ) s = + g y c dy EX #: Find the rclength of the grph of y = ln cos x from x = to x = π. EX # : Set up nd find the rclength of the grph of ( y ) = x from [, 8]. 5 5 8 EX # : Set up nd find perimeter of the shded region. f( x) = x 5 5
Preview for CH.8 Procedures for fitting integrnds to Bsic Rules Technique Expnd (numertor) Seprte numertor Exmple ( + e x ) dx = + e x + e x dx = x + e x + e x + C + x dx = x + x + + x x + dx = rctn x + ln x + + C Complete the squre dx = x + 8x + dx = x x x + dx = 5 rctn x + 5 + C + 5 dx = rcsin x ( x ) + C Divide improper rtionl function x dx = x + x + dx = x rctn x + C Add nd sutrct terms in the numertor x dx = x + x + x + x + x + x + x + dx = x + x + x + dx = ln x + x + + x + x + + C Use trigonometric Identities tn x = sec x sin x = cos x cot x = csc x Multiply nd Divide y Pythgoren conjugte dx = + sin x = sin x cos x dx = sin x + sin x sin x dx = dx cos x sin x cos x dx = sec x sec x tn x = tn x sec x + C sin x dx sin x Derivtives nd Integrls of Trig. Functions Integrl: -cosx sinx -ln cosx -ln cscx+cotx ln secx+tnx ln sinx Trig. Function: sinx cosx tnx cscx secx cotx Derivtive: cosx -sinx sec x -cscx cotx secx tnx -csc x 5
8- Bsic Integrtion Rules *Integrl ( top is higher or sme power thn ottom) *Two prolems in One ( Must divide ottom eqution into top eqution) ( Split top inomil into two integrls) EX#: x + x + dx = EX# : x + + x dx = x + *Complete the squre *Add nd Sutrct to numertor ( Turn prolem into rctngent) Turn prolem into nturl logrithm EX# : dx = EX# : x 6x + x dx = x 6x + 9 5
8- Integrtion y Prts *Integrtion y Prts (used when tking n integrl of product nd the products hve nothing to do with ech other) Alwys pick the function whose derivtive goes wy to e u. There re two specil cses. Cse : When ln x, rcsin x or rctn x is in the prolem they must e u. Cse : When neither eqution goes wy, either eqution cn e u (the eqution we pick s u must e u oth times) nd we perform int. y prts twice nd dd to other side. f ( x) g ( x)dx = f ( x)g x g( x) f ( x) dx more simply u dv = uv v du EX : xe x dx = x e x e x dx *Specil cse : EX : x ln x dx = = xe x e x + C = x u = x dv = e x dx u = ln x dv = x dx x ln x x dx ln x x 9 + C du = dx v = e x du = x v = x *Tulr method EX : x cos x dx = x sin x + x cos x sin x + C EX : x dx = x x ln x ( ln ) Deriv. Integrl Deriv. Integrl x + cos x x + x x sin x x + cos x ln sin x x ( ln ) *Specil cse ( neither function's derivtive goes wy so we use integrtion y prts twice nd dd integrl to the other side) st time EX : e x sin x dx = e x + e x cos x dx u = e x dv = sin x = e x cos x + e x sin x du = e x dx v = cos x e x sin x dx nd time e x sin x dx = e x cos x + e x sin x u = e x dv = cos x e x sin x dx = ex cos x + e x sin x + C du = e x dx v = sin x 5
EX#: x cos x dx = EX# : ln x dx = EX# : rctn x dx = EX# : x sin x dx = EX#5 : e x sin x dx = 5
8-5 Prtil Frctions Prtil frctions re used when you cn fctor the denomintor. Simple Prtil frctions Eqution Liner Fctor dx = ( Liner Fctor ) A + Liner Fctor B ( Liner Fctor ) dx EX : x + dx = x ( x +) = A + x 5 + ( x ) B dx x + = A x + ( x +) + B( x ) 5 x + dx Let x = = 5B B = 5 = 5 ln(x ) 5 ln(x +) + C Let x = = 5A A = 5 Specil Cses : (Not on AP Test) Eqution Liner Fctor dx = A + Liner Fctor B ( Liner Fctor) dx EX : x 7 x + 5 dx = = A B + ( x + 5) ( x + 5) + C x + 5 + dx Let x = 5 = C x + 5 x + 5 x + 5 + ( x + 5) dx dx x 7 = A x + 5 + B x + 5 + C = x + 5 Let x = 7 = 5A + 5B + C 5 = 5A + 5B = x + 5 + 6 + C x + 5 Let x = 6 = 6A + 6B + C 6 = 6A + 6B By su./elim. A = B = Eqution dx = x + # Liner Fctor Ax + B + x + # C ( Liner Fctor) dx EX : x + x dx = x + 9 x = Ax + B + x + 9 9 x + 5 + x + 9 = 9 6 ln ( x + 9) + 7 rctn x + 7 C ( x ) dx x + x = ( Ax + B) ( x ) + C( x + 9) 7 7 dx Let x = 7 = C C = ( x ) + C Let x = = B + 9C B = 5 ln x Let x = = A B +C A = 9 55
EX#: dx x 5x + 6 EX# : x + dx x + x + EX# : 5x + x + 6 x + x + x dx EX# : x x 8 x + x x dx 56
8-7 Indeterminte Forms nd L'Hôpitl's Rule Norml Forms Indeterminte Forms nd,,,, ***If limit is in indeterminte form, we convert it to norml form then use L'Hôpitl's Rule. L'Hôpitl's Rule: If lim x = or f x g x, then lim x = lim x f x g x f g ( x) x EX#: x + x lim x x sin x = EX# : lim = x x EX# : e x lim x x = EX# : lim x x e x = EX#5 : e x x lim = x x EX#6 : lim x ln x = x + EX#7 : lim x x = x EX#8 : 8 lim x + x 9 x x = 57
f ( x) dx = c lim f x = dx dx = f ( x) dx + f ( x) lim F( x) 8-8 Improper Integrls = lim F( ) F( ) f x dx where c is undefined f x c dx = lim c c dx + lim c + If f x dx = finite # then the function converges. If f x dx = then the function diverges. f x c f x dx EX#: dx = x EX# : dx = x EX# : e x dx = EX# : dx = x + EX#5 : dx = x 58
Sequence : CH.9 INFINITE SERIES 9- Sequences A sequence is list of numers, clled terms, in definite order. Sequences of ojects re most commonly denoted using rces. If the limit of sequence exists, then we sy the sequence converges. Otherwise the sequence diverges. nth term of n rithmetic sequence : n = + (n )d Recursively defined sequence : Ech susequent term depends on the previous term. EX : k+ = ( k ) ; = EX#: k+ = 5 k + ; = = 6 = = 8 5 = = = = 5 = Most fmous recursively defined sequence is the Fioncci Sequence k+ = k + k+ ; =, =,,,, 5, 8,,,,,,... Fioncci Petls petls lily, iris 5 petls uttercup, wild rose, lrkspur, columine 8 petls delphiniums petls rgwort, corn mrigold, cinerri petls ster, lck-eyed susn, chicory petls plntin, pytethrum 55, 89 petls michelms disies, the stercee fmily Humns exhiit Fioncci chrcteristics. The Golden Rtio is seen in the proportions in the sections of finger. EX# : Find lim n. n ) n = n + 5 n 7 ) n = 7n 8n c) n = 6n + 8 9n + d) n = 5 n EX# : Simplify ech )! 9! ) 7!! c) ( n +)! n! d) ( n +)! ( n + )! e) ( n + )! n! f ) 6 n 6 g) x + n+ n+ h) ( x + ) n x n+ 5 n x n 5 n+ 59
Infinite series: n = + + + +...+ n +... n= 9- Series nd Convergence Definition of Convergent nd Divergent series : For the infinite series n, the nth prtil sum is given y S n = + + + +...+ n. If the sequence of prtil sums {S n } converges to S, then the series n converges. The limit S is clled the sum of the series. S = + + + +...+ n +... If {S n } diverges, then the series diverges. Geometric series: r n = + r + r + r +...+ r n +... n= ***Geometric series Test : A geometric series r n converges iff r <. A geometric series r n diverges iff r. n=m If geometric series converges, it converges to the Sum : S = r n=m nth term test (Used to show immedite divergence) If lim n n = then it my converge. (ottom power is greter nd we must proceed nd use different test, ecuse this test cnnot prove convergence). If lim n n then it diverges (top power is the sme or greter thn the ottom power). EX #: Test for Convergence ) ) n n= n c) n= n= n n + EX # : A ll is dropped from height of 6 feet nd egins ouncing. The height of ech ounce is the height of the previous ounce. Find the totl verticl distnce trvelled y the ll. 6
***Telescoping Series re convergent Find the sum of the following telescoping series. EX # : n n + n= ( Limit = nd the terms get smller s we pproch ) EX # : n= n n + EX #5 : n= n(n +) Repeting decimls..........8... 6
9- Integrl Test / p-series ***Integrl Test : If f is positive, continuous nd decresing for x nd n = f ( n), then n nd n= f x dx either oth converge or oth diverge. EX #: Use Integrl Test to test for convergence ) n ) n + n= n= n + ***p-series: n= n p ) Converges if p > ) Diverges if < p If p =, it is clled the hrmonic series. is the divergent hrmonic series n n= EX # : Use p-series to test for convergence ) ) n n= c) n n= 7 n n= EX # : Review (Test ech series for convergence) ) 7 9 n= n ) n= n 7 n + 5 c) 6 6 n n + n= d) n= 7 n 6