Calculus AB Bible. (2nd most important book in the world) (Written and compiled by Doug Graham)


 Madlyn Carson
 4 years ago
 Views:
Transcription
1 PG. Clculus AB Bile (nd most importnt ook in the world) (Written nd compiled y Doug Grhm) Topic Limits Continuity 6 Derivtive y Definition 7 8 Derivtive Formuls Relted Rtes Properties of Derivtives Applictions of Derivtives Optimiztion Prolems 7 Integrls/Sustitution 7 Properties of Logrithms 8 Newton's method 8 Seprting vriles 8 Averge Vlue Continuity/Differentiility Prolem Rectiliner Motion Men Vlue Theorem Studying the grph of f ( ) Trigonometric Identities Growth, DouleLife, nd HlfLife formuls Applictions of the Integrl ( Are, Volume, nd Sums) 6 7 Approimting Are ( Trpezoidl Rule, Left Endpoint, Right Endpoint, nd Midpoint) 8 st Fundmentl Theorem of Clculus 8 nd Fundmentl Theorem of Clculus 8 Integrl s n ccumultor Finding Derivtives nd Integrls given grph of f () Integrtion y Prts
2 LIMITS ** When evluting its, we re checking round the point tht we re pproching, NOT t the point. **Every time we find it, we need to check from the left nd the right hnd side (Only if there is BREAK t tht point). **Breking Points re points on the grph tht re undefined or where the grph is split into pieces. Breking Points : ) Asymptotes (when the denomintor equls ) ) Rdicls (when the rdicl equls ) ) Holes (when the numertor nd denomintor equls ) ) Piecewise functions (the # where the grph is split) + = right hnd it f f ( ) = left hnd it **If left nd right hnd its DISAGREE, then the it Does Not Eist (DNE) t tht point. **If left nd right hnd its AGREE, then the it eists t tht point s tht vlue. **Even if you cn plug in the vlue, the it might not eist t tht point. It might not eist from the left or right side or the two sides will not gree. For emple: f = for for < f = DNE ecuse + = nd f f ( ) = Note : In generl when doing its, # = # = # = LIMITS LIMITS AT NON  BREAKING POINTS Very esy. Just plug in the # E# : + = E# : + 7 = = E# : + = HOLES IN THE GRAPH E# : + Fctor nd cncel or multiply y the conjugte nd cncel, then plug in # ( ) + = = + = 7 E# : + + = + + ( + + ) ( + + ) = + ( + ) + + = 6
3 RADICALS ( You must first check tht the it eists on the side(s) you re checking) If # mkes rdicl negtive, the it will not eist t tht #. When we check t the reking point (the # tht mkes the rdicl zero) there re two possile nswers: ) if the it works from the side tht you re checking. ) DNE if the it does not work from the side tht you re checking. E # : E # : E # : + = Since the it eists from the left t we cn plug in, then = = Since the it does not eist from the right t, then + + = DNE ecuse = DNE + = + = DNE ( oth sides don't gree ). + ASYMPTOTES (# ) (Since the point DNE we hve to check point tht is close on the side we re pproching) There re three possile nswers when checking t the reking point (the # tht mkes ottom = zero) ) If we get positive nswer the it pproches ) If we get negtive nswer the it pproches ) DNE If we get positive nswer on one side nd negtive nswer on the other side, then the it DNE E # : E # : E # : E # : = Check. which gives positive nswer so = 7 = Check 8. which gives negtive nswer nd 7. which gives positive nswer so = DNE ecuse the two sides do not gree. = Check. which gives negtive nswer so + + = ecuse ( ) ( ) = = = oth sides gree ( ). TRIG. FUNCTIONS FACTS : E# : sin sin = sin tn = = sin tn cos cos = = tn = tn = = = E# : sin 8 = 8 = 8 E# : 6sin cos 6 sin cos = = 6 = 6 E# : π tn = =
4 < f ( ) = + < E # : f E # : + PIECE  WISE FUNCTIONS The reking points re nd. = (Check in > ) E # : f + = 6 (Check in < ) E # : E # : f f These net three its re not t reking points, so we just plug in the numers. E #7 : 7 + = DNE (Both sides don't gree) E #6 : = E #8 : f f Check the powers of the numertor nd denomintor. =  = 8 E # : f ( ) = (Check in > ) f ( ) = (Check in < ) = (Both sides gree) f LIMITS THAT APPROACH INFINITY ) If the denomintor (ottom) is igger power the it =. ) If the numertor (top) is igger power the it = or . ) If powers re the sme the it = E# : + = E# : + = = + = coefficient of the highest power of numertor coefficient of the highest power of denomintor E# : = E# : = E# : 6 E#6 : + 7 = E#7 : = = E#8 : = 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. f ( ) = 8 f ( ) = + + f ( ) = 7 + = ( + ) ( ) ( ) ( + ) ( ) ( + 7) vert.sym. hole vert.sym. hole vert.sym. hole vert.sym. holes = 8 none none (, 8) none none = 7 (, ), f
5 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 its s well s the vlue of the function t tht point f nd right hnd its f + f ( ). If ll three re equl then the function is continuous t. If f = f ( ) + f ( ) then the function is continuous t. If f is not equl to either one  sided it, then the function is not continuous (discontinuous) t. Continuous t Discontinuous t Derivtive t ll points f ( ) = h f(+h) f ( ) f + h h l Derivtive y Definition Derivtive t the point, f f = h f f + h h f() f(+h) f() l f(+h)=f() l +h +h Line l is secnt line h Line l is secnt line h Line l is tngent line f ( ) slope of secnt line l = f + h + h f ( ) f + h 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. The derivtive of function finds the slope of the tngent line!
6 E # : f = = Find f ( ) Use f ( ) = f + h h h h = h + 6h + h h f ( ) f + h h = h 6h + h h = h 6 + h = 6 E # : f = = Find f Use f = f + h h h h = h + 8h + h + h h f f + h h = h 8 + h + h = 8 tngent line to curve The derivtive finds the slope of the tngent line. The norml line is perpendiculr to the tngent line. E # : f = f norml line to curve = Find eqution of the tngent line nd norml line t =. + h h h = h + h + h h Eqution of Line (point  slope form) : y y = m = nd f = f Eqution of the tngent line : y = ( ) Eqution of the norml line : y = = h + h = ( slope of the tngent line) Questions from AP Test When you see these prolems, you need to tke derivtive of the given eqution. sin( + h) sin E# : = cos E# : = h h h h Eqution: sin Derivtive: cos Eqution: Derivtive: 6 ( + h) When you see these prolems, you need to tke derivtive of the given eqution nd plug in #. ( + h) E# : = 6 E# : h h h Eqution: Eqution: ( + h) = h Derivtive: Derivtive t = : 6 Derivtive: Derivtive t = :
7 Derivtive Formuls *Power Rule y = n y = n n E# : y = y = E# : y = y = y = y = *Product Rule y = f () g() y = f ()g() + g () f () E # : y = sin y = ( ) ( sin ) + ( cos ) *Quotient Rule y = f () g() y = f ()g() g () f () g() E # : y = sin y = cos sin = 6 *Chin Rule y = f () n OR y = f g( ) y = n( f ()) n f () y = f g( ) g = sin + cos cos sin E # : y = ( +) y = ( +) = 6( +) *Implicit Differentition function in terms of s nd y s must write dy E# : y + y + = derivtive y + dy dy d + y d + = dy dy d + y d = y dy d + y *Trig. Functions d dy Now solve for d. = y everytime you tke deriv. of y dy y = d + y ( Tke the derivtive of the trig. function times the derivtive of the ngle) Function Derivtive Emple sin cos cos sin tn sec csc csc cot sec sec tn cot csc d d cos d sin( ) = cos( ) = cos( ) d = cos d csc d sin( ) = 8 cos( )sin d tn ( ) d = ( sec ) = sec ( ) = csc d d d cot d sec sin cot( ) = csc( )cot = sec sin = csc tn ( sin ) cos = csc 7
8 *Nturl Log E# : y = ln( +) y = y = ln( f ()) y = f () f () + = + E# : y = ln(sin ) y = cos = cot sin E# : y = log chnge of se y = ln ln y = ln = ln *Constnt Vrile y = f () y = f () f () ln ( steps: itself, derivtive of eponent, ln of se) E# : y = E# : y = E# : y = e y = ln y = ln y = e ln e = e *Vrile Vrile ( tke ln of oth sides) ln y = g( )ln f ( ) y = f () g() dy y d = g ( )ln f + f ( ) f ( ) g( ) dy d = f ( ( ) )g g E# : y = sin tke ln of oth sides then tke derivtive lny = sin ln dy y d = cos ln + sin *Vrile Vrile y = f () g() ( lternte wy) dy d = sin ln f + f ( ) f ( ) g( ) cos ln + sin Need to chnge y = f ( ) g( ) to y = e ln f ()g() y = e g()ln f () then tke derivtive. E# : y = sin sin ln y = e y = e sin ln cos ln + sin = sin cos ln + sin *Inverse Trig. Functions y = rcsin f ( ) y = rctn f ( ) y = rcsec f ( ) y = f ( ) f ( ) y = + f ( ) f ( ) y = f f ( ) E# : y = rcsin E# : y = rctn E# : y = rcsece y = y = 8 y = y = y = y = 6 e e f e e ( ) 8
9 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 is d, derivtive of y is y dy, derivtive of r dr is r, derivtive of t is t = t dv V mens volume ; mens rte of chnge of volume (how fst the volume is chnging) dr r mens rdius ; mens rte of chnge of rdius (how fst the rdius is chnging) d is how fst is chnging; dy is how fst y is chnging Volume of sphere Surfce Are of sphere Are of circle Circumference of circle V = πr A = πr A = πr C = πr dv = πr dr da = 8πr dr da = πr dr dc = π dr 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. E # : The rdius of sphericl lloon is incresing t the rte of ft / min. How fst is the surfce re of the lloon chnging when the rdius is ft.? A = πr da dr = 8πr da da = 8π = 6π ft /min. The surfce re of the lloon is incresing t 6π ft /min. E # : Wter is poured into cylinder with rdius t the rte of in / s. How fst is the height of the wter chnging when the height is 6 in? V = πr h r = V = πh dv = π dh = π dh dh = π in / s The height is incresing t.7 in / s.
10 E # : Wter is leking out of cone with dimeter inches nd height inches t the rte of 7 in / s. How fst is the height of the wter chnging when the height is in? r h = r = h V = πr h V = π h dv = dh 8 πh 7 = 8 π dh The height is decresing t.8877 in / s h V = πh dh = 67 6π in / s E # : 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 8 ft. from the wll? + y = 7 y = when = 8 d dy dy + y = 8 + = dy = 8 ft. per second. ) How fst is the re of the tringle formed chnging t this time? A = y da = d y + dy da = da = 6 ft. per second. θ 7 y c) How fst is the ngle etween the ottom of the ldder nd the floor chnging t this time? sinθ = y cosθ dθ 7 = 7 dy dθ = rdins per second E # : 8 7 dθ = 7 8 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? d = ( speed of the mn wlking ) dy Use similr tringles: + y = 6 y =? ( speed of the length of shdow ) y = 6 + 6y 7y = 6 y = 6 7 dy = 6 7 d dy = 6 7 dy = feet per second 7 ) At wht rte, in feet per second, is the tip of the shdow chnging? Tip of shdow is + y, so speed of tip is d + dy = + 7 = 6 7 feet per second 6 y
11 Properties of Derivtives Derivtive is rte of chnge; it finds the chnge in y over the chnge in, dy, which is slope. d st derivtive m. nd min., incresing nd decresing, slope of the tngent line to the curve, nd velocity. nd derivtive inflection points, concvity, nd ccelertion. E # : Slope of the tngent line to the curve Given f ( ) = Find eqution of the tngent line nd norml line t =. = f ( ) = 6 = 8 f = f f Eqution of Line (point  slope form) : y y = m Eqution of the tngent line : y 8 = Eqution of the norml line : y 8 = incresing : slopes of tngent lines re positive f decresing : slopes of tngent lines re negtive f Properties of First Derivtive ( ) > ( ) < ( = ) f ( ) = mimum point : Slopes switch from positive to negtive. found y setting f minimum point : Slopes switch from negtive to positive. found y setting concve up : slopes of tngent lines re incresing. f concve down : slopes of tngent lines re decresing. f Properties of Second Derivtive ( ) > ( ) < ( = ) inflection points : points where the grph switches concvity. found y setting f slopes of tngent line switch from incresing to decresing or vice vers. 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 Mimum; 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 Mimum; m is minimum; I is n inflection point
12 Appliction of Derivtives To find rel. m., rel. min., where the grph is incresing nd decresing, we set the first derivtive = E# : y = 6 + ) Plug #'s in ech intervl on the numer line into first derivtive. y = f ( ) > mens grph is incresing on tht intervl. = 6( 6) f ( ) < mens grph is decresing on tht intervl. = 6( )( + ) f ( ) switches from + to, then point is reltive mimum. = nd = f ( ) switches from to +, then point is reltive minimum. ) To find the Y vlue of m. nd min. plug into originl eqution. +  f'() +  m min incresing decresing (, 6) (, 7), (, ) (, ) To find inflection points, where the grph is concve up nd concve down, we set the second derivtive = y = 6 = 6( ) = ) Plug #'s in ech intervl on the numer line into second derivtive. f ( ) > mens grph is concve up on tht intervl. f ( ) < mens grph is concve down on tht intervl. An inflection point occurs t the points where f ( ) switches from + to or from to +. ) To find the Y vlue of inf. pt. plug into originl eqution. f''()  +! inflection pt. concve down concve up,,, M I 68 m
13 Optimiztion Prolems ) Drw nd lel picture. ) Write eqution sed on fct given nd write eqution for wht you need to mimize or minimize. ) Plug in fct eqution into the eqution you wnt to mimize or minimize. ) Tke derivtive nd set equl to zero. ) Find remining informtion. E# : An open o of mimum volume is to e mde from squre piece of mteril, 8 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 mimum volume of your o? V = (8 ) V = + = V = 7 + V = + 7 Cutting off mkes no sense (minimum o). = We need to cut off inches to hve o with = mimum volume. The mimum volume is V = )( =, = V = (8 ) V = in E# : A frmer plns to fence rectngulr psture djcent to river. The frmer hs 8 feet of fence in which to enclose the psture. Wht dimensions should e used so tht the enclosed re will e mimum? Wht is the mimum Are? P = y + A = y A = 8 y = A = 8 = y + A = ( 8 y) y y = A = 88 ft = 8 y A = 8y y = 8 E# : = A frmer plns to fence two equl rectngulr psture djcent to river. The frmer hs feet of fence in which to enclose the pstures. Wht dimensions should e used so tht the enclosed re will e mimum? Wht is the mimum Are? = + y A = y A = 6y = A = y = A = 6 y y y = A = ft 6 y = A = y y = 6 = Y Y RIVER RIVER Y Y Y
14 E# : A crte, open t the top, hs verticl sides, squre ottom nd volume of ft. Wht dimensions give us minimum surfce re? Wht is the surfce re? V = y A = + y A = = A = + = y A = + A = = A = ft = y A = + = = y = y = Y E# : A rectngle is ounded y the is nd the semicircle y = 8. Wht length nd wih should the rectngle hve so tht its re is mimum? y = 8 A = y A = 8 A = 8 A = 8 + ( 8 ) ( 8 ) ( 8 ) 8 A = A = 6 = ( 8 ) y = 8 6 = y = = ( Bowtie) ( ) A = A = 8 Y= 8  Y
15 *Integrl of constnt Integrtion Formuls d = + C E# : d = + C E# : π d = π + C *Polynomils n d = n+ n + + C E# : + 8 d = + + C *Frctions ( Bring up denomintor, then tke integrl) E# : *Constnt Vrile E# : E# : d d = d = e d = *Trig Functions d = ln + C + C = + C ( steps : itself, divided y deriv. of eponent, divided y ln of se ) ln + C E# : d = e ln e + C = e + C E# : e d = ( Alwys divide y derivtive of the ngle) ln + C e ln e + C = e + C sin d = cos + C cos d = sin + C tn d = ln cos + C csc d = ln csc + cot + C sec d = ln sec + tn + C cot d = ln sin + C E# : cos d = sin + C E# : sin 6 d = cos6 + C 6 E# : tn d = ln cos + C E# : sec 7 d = ln sec 7 + tn 7 + C 7 E# : csc d = ln csc + cot + C E#6 : cot d = ln sin + C f ( ) *Nturl Log : d = ln f ( ) + C top is the derivtive of the ottom f E# : E# : d = ln + C E# : sin cos d = ln cos + C E# : d = ln + C d + = d = + ln ( +) + C *Integrl ( top is higher or sme power thn ottom) ( Must divide ottom eqution into top eqution). E# : + d Long Division = d = + ln + + C E# : + d = + d = + ln + C
16 *Inverse Trig Functions d = rcsin + C d = + rctn + C d = rcsec + C Find vrile v nd constnt. The top MUST e the derivtive of the vrile v. E# : E# : E# : d = rcsin + C E# : d = v = = v = = d = 6 + d = rctn + C E# : d = +6 d = +6 v = = v = = = d = rcsec + C E#6 : 6 d = 6 v = = v = 7 = = rcsec 7 + C rcsin + C rctn + C rctn + C d 6
17 *Sustitution When integrting we usully let u = the prt in the prenthesis, the prt under the rdicl, the denomintor, the eponent, or the ngle of the trig. function. E# : + d = E# : u = + du = d ( +) d = d ( u) du = u + C = ( +) + C d ( + ) d = + = = u du 6 = 6 u + C u = + du = 6 d = 8 + E# : + d = u u du = (u u )du = u u + C E# : u = + = u = ( + ) ( + ) + C du = d cos d = u = du = d E# : + d = u du = u cos d = cosu du = sinu + C = sin + C = = u = + You must switch everything from to u. Including the #'s. du = 6 d u = + = u = + = Properties of Logrithms + C logrithmic form eponentil form : y = ln e y = Log Lws : y = ln y = ln ln + ln y = ln y ln ln y = ln y ln 8 = ln = ln ln + ln = ln ln 7 ln = ln 7 Chnge of Bse Lw : y = log y = ln ln Fct : You cn t tke ln/log of negtive # or zero. We use logrithms to solve ny prolem tht hs vrile in the eponent. E# : e = ln e = ln ln e = ln = E# : ln = e ln = e = e 7 Memorize : ln e = ln = log = log = ln
18 *Newton s Method ( Used to pproimte the zero of the function) c f ( c) where c is the pproimtion for the zero. f ( c) E# : If Newton s method is used to pproimte the rel root of + =, then first pproimtion = would led to third pproimtion of = f () = + f () = + Plug in to find f () f () = or.7 = Plug in to find f / f / = 86 or.686 = DIFFERENTIAL EQUATIONS ( Seprting Vriles) (used when you re given the derivtive nd you need to find the originl eqution. We seprte the 's nd y's nd tke the integrl). E# : Find the generl solution given dy d = y dy d = y y dy = d y dy = d E# : Find the prticulr solution y = f () for E# given (, ) y = + C y = + C y = + C = 8 + C 7 = C y = + 7 y = + 7 E# : Find the prticulr solution y = f () given dy d dy dy = 6 d y = 6y nd (, ) = 6 d y ln y = + C y = e +C y = C e = C y = e If the rte of growth of something is proportionl to itself ( y = ky), then it is the growth formul. Proof : y = ky dy dy = ky y = k e ln y = e kt +C y = e kt e C y = C e kt dy = k y ln y = kt + C *Averge Vlue (use this when you re sked to find the verge of nything) f () d E# : Find the verge vlue of f () = from [, ] Avg. vlue = d = = 6 = 7 = 7 8
19 *Continuity / Differentiility Prolem E# : f ( ) = At f ( ) =, < = 6, 6 = At f ( ) = f ( ) = f ( ) =. Therefore f () is continuous. + f () is continuous iff oth hlves of the function hve the sme nswer t the reking point. =, < f 6, At oth hlves of the derivtive = 6. Therefore the function is differentile. Since oth sides of f At f () = = 6 6 = 6 f () is differentile if nd only if the derivtive of oth hlves of the function hve the sme nswer t the reking point. nd f ( ) gree t, then f ( ) is continuous nd differentile t =. *Rectiliner Motion (Position, Velocity, Accelertion Prolems) We designte position s ( t), y( t),or s( t). The derivtive of position, ( t) = v( t) velocity. The derivtive of velocity, v ( t) = ( t) ccelertion. We often tlk out position, velocity, nd ccelertion when we re discussing prticles moving long the is or yis. A prticle is t rest or is chnging direction when v t A prticle is moving to the right or up when v t To find the verge velocity of prticle =. > nd to the left or down when v( t) <. v(t) To find the mimum or minimum ccelertion of prticle set t =, then check the vlues on numer line to see if nd how they switch signs. Speed is the solute vlue of velocity, v t If v( t) nd ( t) gree If v( t) nd ( t) disgree Distnce trveled is v( t) v( t) ( t) + + v t ( t) + +. speed is incresing. speed is decresing.
20 *MenVlue Theorem (Only pplies if the function is continuous nd differentile) f() Slope of tngent line = slope of line etween two points f (c) = f () f () 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() f() *When given the grph of f (), it is like you re looking t # line. This is the grph of f (). Where f () = (int) is where the possile m. nd min. re. Signs re sed on if the grph is ove incresing or elow the  is (decresing). c E : Grph of f incresing when f decresing when f m. occurs when f min. occurs when f ( ) from 6 6 ( ) > ( ) < ( ) switches from + to. ( ) switches from to +. The f ( ) is the slope of the tngent line of f ( ). concve up when slope is positive. concve down when slope is negtive. inf. pts. occur when slopes switch from + to or to +. f'() rel.min rel. m incresing decresing =, = (, ) (,6] [ 6, ) (, ) f''() 66 inf. pts. concve up concve down =,,, ( 6, ) (, ) (, 6) (, ) (, ) Grph of f '() 6
21 *Doule  Life Formul y = C () t d *Hlf  Life Formul y = C *Growth Formul y = C e kt ( Comes from y = ky ) y = ending mount, C= intil mount, t = time, d = doulelife time, h = hlflife time, k = growth constnt *Trigonometric Identities *Reciprocl Identities Identities *Doule Angle Formuls *Hlf Angle Formuls cosθ = secθ sinθ = cscθ *Pythgoren Identities sinθ cosθ = tnθ sin = sin cos cos sin = cosθ sinθ = cotθ cos = cos sin cos + cos = sin + cos = + tn = sec + cot = csc t h
22 *Are Applictions of Integrls (Are,Volume, Sums) A = [top equtionottom eqution] d f() f() g() g() h() c c Are from to c Are from to c c A = [ f () g()] d + [g() f ()] d A = [ f () h()] d + [g() h()] d Find Are of y = sin from to π From to π, sin is on top of is (y = ) sin π π c From π to π, is (y = ) is on top of sin sin π π Are = sin d + sin d = cos + cos π π = + = π π *Volume V = A( ) d where A( ) is the Are of the cross section. Volume if cross section rotted is circle( A = π r ) Volume if cross section rotted is cylinder ( A = πrh) ( ottom function) V = π top function d V = π top function ottom function [ ] d rdius rdius height *Volume rotted out : Verticl cross section the  is the y  is ( g() ) [ ] π f () d π f () g() d the line y = k the line = c ( g() + k) π f () + k d π + c the line y = m the line = d ( m f ()) π m g() d π d [ f () g() ] [ f () g() ] d d y=m =c y=k f() g() =d
23 *Volume rotted out : ( Verticl cross section) the  is the y  is π d π d = 8π 8. = 8π. the line y = the line = π ( + ) + d π + d 7π = 7. = π.66 the line y = the line = 6 π d π 6 = 6π d 87.7 = 6π 7. y = y = = = 6 y = *Volume rotted out : ( Horizontl cross section) the  is the y  is π y y dy π y = 8π 8. = 8π. the line y = the line = π ( y + ) y dy π y + dy + 7π = 7. = π.66 the line y = the line = 6 π ( y) y dy π 6 = 6π 6 y 87.7 = 6π 7. dy dy For horizontl cross sections we must switch everything from to y. y to to y = = ± y
24 *Volume (Region is not rotted) V = A( ) is the Are of the cross section. d where A  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. E# : Let R e the region in the first qudrnt elow f nd ove g( ) from = to =. Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the is re : Squres( A = s ) f ( ) g( ) d f ( ) g( ) V = f ( ) g( ) Equilterl Δ's A = s f() g() f ( ) g( ) Semicircle A = πr V = ( f ( ) g( ) ) d V = ( f ( ) g( ) ) d f ( ) g( ) = dimeter V = f ( ) g Rectngle with h = A = h f ( ) g( ) f d π f ( ) g V = π ( f ( ) g( ) ) 8 d = rdius ( g( ) ) V = f ( ) g( ) Regulr Hegon A = p f ( g( ) )d ( g( ) ) V = f d f ( ) g( ) V = tn6 f ( ) g ( ) ( 6( f ( ) g( ) )) d V = ( f ( ) g( ) ) d tn6 f ( ) g( ) = pothem 6 f ( g( ) ) = perimeter
25 E# : Let R e the region in the first qudrnt under the grph of y = for. Find the volume of the solid whose se is the region R nd whose cross sections cut y plnes perpendiculr to the is re : Squres A = s V = Equilterl Δ's A = s d = / / 6 R 7 8 d = ln = ln ln = ln = ln.8 Semicircle A = πr V = d = d = ln = ln. = dimeter V = π d = = rdius Rectngle with h = A = h V = Regulr Hegon A = p V = d = 6 d = = pothem 6 = perimeter Multipliers for other figures : π d = π 8 8 d = π 8 ln = π ln.8 d = d = ln = ln. d = d = ln = ln.76( SL): 6( LL): 6( HYP) : 8 ( LEG): ( HYP): Regulr Octgon : tn 67. or tn π 8
26 Approimting Are We pproimte Are using rectngles (left, right, nd midpoint) nd trpezoids. *Riemnn Sums ) Left edge Rectngles f () = + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = Totl Are=.7 8 ) Right edge Rectngles f () = + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = Totl Are = c) Midpoint Rectngles f () = + from [, ] using sudivisions (Find re of ech rectngle nd dd together) A = Totl Are = 8.6 d d) Actul Are = + = + = =.667 *Trpezoidl Rule (used to pproimte re under curve, using trpezoids). Are f ( ) + f ( ) + f ( ) + f ( )... f ( n ) + f ( n ) n where n is the numer of sudivisions. [ ] E# : f () = + Approimte the re under the curve from [, ] using the trpezoidl rule with sudivisions. A = f () + f + f () + f + f () = + + () = = 76 6 = =.7 All you re doing is finding the re of the trpezoids nd dding them together!  6
27 *Approimting Are when given only ( no eqution given) To estimte the re of plot of lnd, surveyor tkes severl mesurements. The mesurements re tken every feet for the ft. long plot of lnd, where y represents the distnce cross the lnd t ech ft. increment. 6 7 y ) Estimte using Trpezoidl Rule ) Estimte using Midpoint sudivisions A 8 f + f ( ) f ( ) + f A [ ] A [ ] A A 78. A 78 f ( ) + f ( ) + f ( 7) + f ( ) c) Estimte Avg. vlue using Trpezoidl Rule d) Wht re you finding in prt c? 6.7 The verge distnce cross the lnd. Avg.Vlue 78. e) Estimte using Left Endpoint f) Estimte using Right Endpoint A f + f ( ) + f f ( ) 8 A 8 A f ( ) + f + f ( )...+ f [ ] A [ ] A 778 A 7 *Approimting Are when given only ( no eqution given) Unequl sudivisions: You must find ech Are seprtely. y ) Estimte using Trpezoids A = ( + )h A ( + ) + ( +) + ( + ) ) Estimte using Left Endpoint ( A = wih left height) A + + Trpezoids shown c) Estimte using Right Endpoint A = wih right height A + + 7
28 st Fundmentl Theorem of Clculus Just plug in the top # minus the ottom #. E# : d = = 8 = 8 E# : π π sin d = cos π π = 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. E# : E# : d d d d d d f t = f t = = E# : t + = + + = + + Integrl s n ccumultor A definite integrl finds the chnge in the eqution ove it. The integrl of velocity from to is the chnge in position (distnce trvelled) from to. The integrl of ccelertion from to is the chnge in velocity from time to time. The integrl of f () is the chnge in f (). d d t = 6 = 7 E# : If f = then find f. f ' Since f () d = f f it finds the chnge in f from to. Since the re under f from to = this will help find f. f = f + f () d = + = 7 E# : If f = then: f Integrls going left re negtive. Integrls going right re positive. = + f ( ) d f  = + f ( ) d E# : Given v v( ) = v + ( t) . = +. =. f = = 6 f 8 = 8 nd ( t) = sin t find v = + f ( ) d = + f ( ) d 8 Grph of f'() π 6 = + = 7 8 = + π = 6 + π = = (The integrl of ccelertion finds the chnge in velocity)
29 Finding Derivtives nd Integrls given grph of f() Grph of f() 7 8 The derivtive is the slope of ech line. = f ( 7) = DNE = DNE f ( 8) = = f ( ) = DNE = f (.) = f f f f The integrl finds the totl re etween f() nd the  is. f d = 7 + = 8 f d = f d = = f d = = 8 ( f ( ) + ) d = 68 f ( ) d = Grph of f() velocity of runner in meters per second velocity of runner in meters per second E# : Find the velocity of the runner t t = nd t = 7 seconds. v = = v( 7) = Find the ccelertion of the runner t t = nd t = 7 seconds Since v ( t) = ( t), you find ccelertion y finding the derivtive (slope) of velocity. = ( 7) = Find the distnce trvelled y the runner from t = nd t = seconds E# : Given () = find () nd (). () = + v(t) = + = 6 () = + v(t) = + 8 = Distnce trvelled = v t v t = 8
30 *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 is in the prolem it 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 ( ) g ( )d = f ( )g g( ) f ( ) du more simply u dv = uv v du E# : e d = e e d Cse : E# : ln d = = e e + C = u = dv = e d u = ln dv = d ln d ln + C du = d v = e du = v = *Tulr method E# : cos d = sin + cos sin + C Deriv. Integrl + cos sin + cos sin *Specil cse E# : Deriv. d = Integrl + ln ( ln ) ln ( ln ) + C ( neither function's derivtive goes wy so we use integrtion y prts twice nd dd integrl to the other side) st time u = e du = e d nd time u = e du = e d dv = sin v = cos dv = cos v = sin E# : e sin d = e + e cos d = e cos + e sin e sin d e sin d = e cos + e sin e sin d = e cos + e sin + C
Calculus AB. For a function f(x), the derivative would be f '(
lculus AB Derivtive Formuls Derivtive Nottion: For function f(), the derivtive would e f '( ) Leiniz's Nottion: For the derivtive of y in terms of, we write d For the second derivtive using Leiniz's Nottion:
More information( ) Same as above but m = f x = f x  symmetric to yaxis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.
AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More information( ) where f ( x ) is a. AB/BC Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB/ Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 Find the intersection of f ( x) nd g( x). A3 Show tht f ( x) is even. A4 Show tht
More informationCalculus Notes BC CH.7 Applications of Integration CH. 8 Integration Techniques CH. 9 Infinite Series
PG. Topic Clculus Notes BC PG. Topic 6 CH. Limits / Continuity  Find Limits Grphiclly nd Numericlly  Evluting Limits Anlyticlly  Continuity nd Onesided Limits 5 Infinite Limits/ 5 Limits tht Approch
More informationInstantaneous Rate of Change of at a :
AP Clculus AB Formuls & Justiictions Averge Rte o Chnge o on [, ]:.r.c. = ( ) ( ) (lger slope o Deinition o the Derivtive: y ) (slope o secnt line) ( h) ( ) ( ) ( ) '( ) lim lim h0 h 0 3 ( ) ( ) '( ) lim
More informationLoudoun Valley High School Calculus Summertime Fun Packet
Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!
More informationMA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations
LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll
More informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewiseliner function f, for 4, is shown below.
More informationKeys to Success. 1. MC Calculator Usually only 5 out of 17 questions actually require calculators.
Keys to Success Aout the Test:. MC Clcultor Usully only 5 out of 7 questions ctully require clcultors.. FreeResponse Tips. You get ooklets write ll work in the nswer ooklet (it is white on the insie)
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationcritical number where f '(x) = 0 or f '(x) is undef (where denom. of f '(x) = 0)
Decoding AB Clculus Voculry solute mx/min x f(x) (sometimes do sign digrm line lso) Edpts C.N. ccelertion rte of chnge in velocity or x''(t) = v'(t) = (t) AROC Slope of secnt line, f () f () verge vlue
More informationDERIVATIVES NOTES HARRIS MATH CAMP Introduction
f DERIVATIVES NOTES HARRIS MATH CAMP 208. Introduction Reding: Section 2. The derivtive of function t point is the slope of the tngent line to the function t tht point. Wht does this men, nd how do we
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationTime in Seconds Speed in ft/sec (a) Sketch a possible graph for this function.
4. Are under Curve A cr is trveling so tht its speed is never decresing during 1second intervl. The speed t vrious moments in time is listed in the tle elow. Time in Seconds 3 6 9 1 Speed in t/sec 3 37
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationAB Calculus Path to a Five Problems
AB Clculus Pth to Five Problems # Topic Completed Definition of Limit OneSided Limits 3 Horizontl Asymptotes & Limits t Infinity 4 Verticl Asymptotes & Infinite Limits 5 The Weird Limits 6 Continuity
More informationA. Limits  L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. 1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by
More informationlim f(x) does not exist, such that reducing a common factor between p(x) and q(x) results in the agreeable function k(x), then
AP Clculus AB/BC Formul nd Concept Chet Sheet Limit of Continuous Function If f(x) is continuous function for ll rel numers, then lim f(x) = f(c) Limits of Rtionl Functions A. If f(x) is rtionl function
More informationCh AP Problems
Ch. 7.7. AP Prolems. Willy nd his friends decided to rce ech other one fternoon. Willy volunteered to rce first. His position is descried y the function f(t). Joe, his friend from school, rced ginst him,
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationReview Exercises for Chapter 4
_R.qd // : PM Pge CHAPTER Integrtion Review Eercises for Chpter In Eercises nd, use the grph of to sketch grph of f. To print n enlrged cop of the grph, go to the wesite www.mthgrphs.com... In Eercises
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationEdexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks
Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1
More informationTrigonometric Functions
Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationAP Calculus AB First Semester Final Review
P Clculus B This review is esigne to give the stuent BSIC outline of wht nees to e reviewe for the P Clculus B First Semester Finl m. It is up to the iniviul stuent to etermine how much etr work is require
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions
Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationSample Problems for the Final of Math 121, Fall, 2005
Smple Problems for the Finl of Mth, Fll, 5 The following is collection of vrious types of smple problems covering sections.8,.,.5, nd.8 6.5 of the text which constitute only prt of the common Mth Finl.
More informationTO: Next Year s AP Calculus Students
TO: Net Yer s AP Clculus Students As you probbly know, the students who tke AP Clculus AB nd pss the Advnced Plcement Test will plce out of one semester of college Clculus; those who tke AP Clculus BC
More informationChapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More informationWhat Is Calculus? 42 CHAPTER 1 Limits and Their Properties
60_00.qd //0 : PM Pge CHAPTER Limits nd Their Properties The Mistress Fellows, Girton College, Cmridge Section. STUDY TIP As ou progress through this course, rememer tht lerning clculus is just one of
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More informationCalculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION
lculus Section I Prt LULTOR MY NOT US ON THIS PRT OF TH XMINTION In this test: Unless otherwise specified, the domin of function f is ssumed to e the set of ll rel numers for which f () is rel numer..
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
More informationSection 4: Integration ECO4112F 2011
Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic
More informationTopics for final
Topics for 161.01 finl.1 The tngent nd velocity problems. Estimting limits from tbles. Instntneous velocity is limit of verge velocity. Slope of tngent line is limit of slope of secnt lines.. The limit
More informationKEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a
KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider
More informationFINALTERM EXAMINATION 9 (Session  ) Clculus & Anlyticl GeometryI Question No: ( Mrs: )  Plese choose one f ( x) x According to PowerRule of differentition, if d [ x n ] n x n n x n n x + ( n ) x n+
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More information[ ( ) ( )] Section 6.1 Area of Regions between two Curves. Goals: 1. To find the area between two curves
Gols: 1. To find the re etween two curves Section 6.1 Are of Regions etween two Curves I. Are of Region Between Two Curves A. Grphicl Represention = _ B. Integrl Represention [ ( ) ( )] f x g x dx = C.
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
More informationUnit Six AP Calculus Unit 6 Review Definite Integrals. Name Period Date NONCALCULATOR SECTION
Unit Six AP Clculus Unit 6 Review Definite Integrls Nme Period Dte NONCALCULATOR SECTION Voculry: Directions Define ech word nd give n exmple. 1. Definite Integrl. Men Vlue Theorem (for definite integrls)
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re openended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnkoutnnswer problems! (There re plenty of those in the
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
More informationMath 154B Elementary Algebra2 nd Half Spring 2015
Mth 154B Elementry Alger nd Hlf Spring 015 Study Guide for Exm 4, Chpter 9 Exm 4 is scheduled for Thursdy, April rd. You my use " x 5" note crd (oth sides) nd scientific clcultor. You re expected to know
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationOptimization Lecture 1 Review of Differential Calculus for Functions of Single Variable.
Optimiztion Lecture 1 Review of Differentil Clculus for Functions of Single Vrible http://users.encs.concordi.c/~luisrod, Jnury 14 Outline Optimiztion Problems Rel Numbers nd Rel Vectors Open, Closed nd
More information7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?
7.1 Integrl s Net Chnge Clculus 7.1 INTEGRAL AS NET CHANGE Distnce versus Displcement We hve lredy seen how the position of n oject cn e found y finding the integrl of the velocity function. The chnge
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationUnit 5. Integration techniques
18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationA. Limits  L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.
A. Limits  L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More informationChapter 7: Applications of Integrals
Chpter 7: Applictions of Integrls 78 Chpter 7 Overview: Applictions of Integrls Clculus, like most mthemticl fields, egn with tring to solve everd prolems. The theor nd opertions were formlized lter. As
More informationAP Calculus AB Summer Packet
AP Clculus AB Summer Pcket Nme: Welcome to AP Clculus AB! Congrtultions! You hve mde it to one of the most dvnced mth course in high school! It s quite n ccomplishment nd you should e proud of yourself
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationThe Fundamental Theorem of Calculus, Particle Motion, and Average Value
The Fundmentl Theorem of Clculus, Prticle Motion, nd Averge Vlue b Three Things to Alwys Keep In Mind: (1) v( dt p( b) p( ), where v( represents the velocity nd p( represents the position. b (2) v ( dt
More informationMath 259 Winter Solutions to Homework #9
Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges 658659 (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier
More informationCHAPTER 10 PARAMETRIC, VECTOR, AND POLAR FUNCTIONS. dy dx
CHAPTER 0 PARAMETRIC, VECTOR, AND POLAR FUNCTIONS 0.. PARAMETRIC FUNCTIONS A) Recll tht for prmetric equtions,. B) If the equtions x f(t), nd y g(t) define y s twicedifferentile function of x, then t
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More informationAPPLICATIONS OF DEFINITE INTEGRALS
Chpter 6 APPICATIONS OF DEFINITE INTEGRAS OVERVIEW In Chpter 5 we discovered the connection etween Riemnn sums ssocited with prtition P of the finite closed intervl [, ] nd the process of integrtion. We
More informationMat 210 Updated on April 28, 2013
Mt Brief Clculus Mt Updted on April 8, Alger: m n / / m n m n / mn n m n m n n ( ) ( )( ) n terms n n n n n n ( )( ) Common denomintor: ( ) ( )( ) ( )( ) ( )( ) ( )( ) Prctice prolems: Simplify using common
More informationShape and measurement
C H A P T E R 5 Shpe nd mesurement Wht is Pythgors theorem? How do we use Pythgors theorem? How do we find the perimeter of shpe? How do we find the re of shpe? How do we find the volume of shpe? How do
More informationSection 13.1 Right Triangles
Section 13.1 Right Tringles Ojectives: 1. To find vlues of trigonometric functions for cute ngles. 2. To solve tringles involving right ngles. Review   1. SOH sin = Reciprocl csc = 2. H cos = Reciprocl
More informationLogarithmic Functions
Logrithmic Functions Definition: Let > 0,. Then log is the number to which you rise to get. Logrithms re in essence eponents. Their domins re powers of the bse nd their rnges re the eponents needed to
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationFinal Exam Review. Exam 1 Material
Lessons 24: Limits Limit Solving Strtegy for Finl Exm Review Exm 1 Mteril For piecewise functions, you lwys nee to look t the left n right its! If f(x) is not piecewise function, plug c into f(x), i.e.,
More information5.2 Volumes: Disks and Washers
4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict
More informationMAT137 Calculus! Lecture 20
officil website http://uoft.me/mat137 MAT137 Clculus! Lecture 20 Tody: 4.6 Concvity 4.7 Asypmtotes Net: 4.8 Curve Sketching 4.5 More Optimiztion Problems MVT Applictions Emple 1 Let f () = 3 27 20. 1 Find
More informationFirst midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009
Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No
More information1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE
ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check
More informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationCONIC SECTIONS. Chapter 11
CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
More informationAPPM 1360 Exam 2 Spring 2016
APPM 6 Em Spring 6. 8 pts, 7 pts ech For ech of the following prts, let f + nd g 4. For prts, b, nd c, set up, but do not evlute, the integrl needed to find the requested informtion. The volume of the
More information