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typed by Sean Bird, Covenant Christian High School
                updated August 15, 2009
                                                                    AP CALCULUS
                                                              Stuff you MUST know Cold                                               * means topic only on BC
    Curve sketching and analysis                                  Differentiation Rules                               Approx. Methods for Integration
 y = f(x) must be continuous at each:                       Chain Rule                                              Trapezoidal Rule
                dy                                           d                      du    dy dy du                      b
critical point:    = 0 or undefined
                dx and look out for endpoints                dx
                                                                [ f (u )] = f '(u ) dx OR dx = du dx                ∫a
                                                                                                                            f ( x)dx =           1 b− a
                                                                                                                                                 2 n      [ f ( x0 ) + 2 f ( x1 ) + ...
local minimum:                                                                                                                                                                + 2 f ( xn −1 ) + f ( xn )]
 dy goes (–,0,+) or (–,und,+) or d 2 y >0                   Product Rule                                            Simpson’s Rule
 dx                              dx 2                                                                                   b

local maximum:
                                                              d
                                                              dx
                                                                 (uv) =
                                                                        du
                                                                        dx
                                                                           v+u
                                                                               dv
                                                                               dx
                                                                                  OR u ' v + uv '                   ∫a
                                                                                                                            f ( x)dx =
                                                      2                                                                     1
  dy goes (+,0,–) or (+,und,–) or d y <0                                                                                    3   ∆x[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + ...
  dx                              dx 2                                                                                                        2 f ( xn −2 ) + 4 f ( xn−1 ) + f ( xn )]
point of inflection: concavity changes
                                                            Quotient Rule
 d 2 y goes from (+,0,–), (–,0,+),
                                                               d u           du
                                                                               dx   v −u   dv
                                                                                           dx        u ' v − uv '
                                                                   =
                                                               dx  v               v2
                                                                                                OR
                                                                                                          v2
 dx 2          (+,und,–), or (–,und,+)                                                                                           Theorem of the Mean Value
                                                                                                                                  i.e. AVERAGE VALUE
                                                                                                                    If the function f(x) is continuous on [a, b]
              Basic Derivatives
                                                          “PLUS A CONSTANT”                                         and the first derivative exists on the
                  d n                                                                                               interval (a, b), then there exists a number
                  dx
                           ( )
                        x = nx n−1                            The Fundamental Theorem of                            x = c on (a, b) such that
                                                                       Calculus                                                                                          b
                d
                    ( sin x ) = cos x                                                                                                                f (c ) =
                                                                                                                                                              ∫          a
                                                                                                                                                                             f ( x)dx
               dx                                                       b
                                                                                                                                            (b − a)
              d
                   ( cos x ) = − sin x
                                                                    ∫a
                                                                            f ( x)dx = F (b) − F (a)
                                                                                                                    This value f(c) is the “average value” of
              dx                                                            where F '( x) = f ( x)                  the function on the interval [a, b].
               d
                   ( tan x ) = sec2 x
              dx
                                                                         Corollary to FTC                              Solids of Revolution and friends
              d
                  ( cot x ) = − csc2 x                                                                              Disk Method
             dx                                                d b( x)                                                                  x =b
                                                                                                                         V =π∫
                                                                                                                                                               2
                                                                                                                                               [ R ( x)]
                                                               dx ∫a ( x )
            d                                                              f (t )dt =                                                                              dx
                ( sec x ) = sec x tan x                                                                                                 x=a

            dx                                                                                                      Washer Method
           d                                                       f (b( x))b '( x) − f (a ( x))a '( x)
           dx
              ( csc x ) = − csc x cot x                                                                                  V =π∫
                                                                                                                                        b

                                                                                                                                        a
                                                                                                                                            ([ R( x)] − [r ( x)] ) dx
                                                                                                                                                             2                     2


                                                              Intermediate Value Theorem                            General volume equation (not rotated)
                 d              1 du
                     ( ln u ) =                           If the function f(x) is continuous on [a, b],                             b
                dx              u dx                      and y is a number between f(a) and f(b),                       V = ∫ Area ( x) dx
                                                                                                                                    a
                  d u            du                       then there exists at least one number x= c
                 dx
                           ( )
                        e = eu
                                  dx                      in the open interval (a, b) such that                     *Arc Length L = ∫ 1 + [ f '( x)] dx
                                                                                                                                                                   b                       2

                                                                                                                                                                 a
          where u is a function of x,                                        f (c ) = y .                                                                            b
                                                                                                                                                          =∫
                                                                                                                                                                                       2           2
             and a is a constant.                                                                                                                                  a
                                                                                                                                                                             [ x '(t )] + [ y '(t )]   dt
              More Derivatives                                                                                      Distance, Velocity, and Acceleration
         d  −1 u                         1     du                Mean Value Theorem                               velocity = d (position)
              sin    =
         dx 
                 a
                                         2
                                         a −u  2 dx                                                                                              dt
                                                          If the function f(x) is continuous on [a, b],                                                   d (velocity)
            d                               −1                                                                      acceleration =
                                                          AND the first derivative exists on the
            dx
                   (
                cos −1 x =       )                        interval (a, b), then there is at least one
                                                                                                                                                          dt
                              1 − x2                                                                                                                                   dx dy
                                                          number x = c in (a, b) such that                          *velocity vector =                                   ,
         d  −1 u            a       du                                                                                                                               dt dt
              tan        =          ⋅                                           f (b) − f (a )
        dx         a  a 2 + u 2 dx
                                                                      f '(c) =                .
                                                                                                                    speed = v = ( x ')2 + ( y ') 2 *
                                                                                    b−a
              d                −1
             dx
                       (
                  cot −1 x =
                             1 + x2
                                     )                                                                              displacement =                      ∫t
                                                                                                                                                          tf
                                                                                                                                                                 v dt
                                                                                                                                                          o

      d  −1 u               a          du                                 Rolle’s Theorem                                                      final time

      dx 
         
           sec        =
                  a  u u 2 − a 2 dx
                    
                                       ⋅                                                                            distance =              ∫initial time v dt
                                                          If the function f(x) is continuous on [a, b],                                        tf
           d
               (
              csc −1 x =
                           −1
                             )                            AND the first derivative exists on the                                            ∫t   o
                                                                                                                                                       ( x ') 2 + ( y ') 2 dt *
           dx            x x2 − 1                         interval (a, b), AND f(a) = f(b), then there              average velocity =
                                                          is at least one number x = c in (a, b) such                         final position − initial position
           d u( x)                     du
           dx
               ( )
               a      = a ( ) ln a ⋅
                            u x

                                       dx
                                                          that
                                                                             f '(c) = 0 .
                                                                                                                            =
                                                                                                                                          total time
                                                                                                                               ∆x
              d                    1                                                                                        =
                 ( log a x ) =                                                                                                 ∆t
              dx                x ln a
BC TOPICS and important TRIG identities and values
              l’Hôpital’s Rule                           Slope of a Parametric equation                                     Values of Trigonometric
        f (a) 0            ∞                        Given a x(t) and a y(t) the slope is                                  Functions for Common Angles
   If          = or = ,
        g (b) 0            ∞                                           dy dy
                                                                           = dt                                           θ        sin θ      cos θ      tan θ
            f ( x)         f '( x)                                     dx dx
then lim           = lim                                                      dt
      x →a g ( x)    x → a g '( x )                                                                                       0°         0          1          0
             Euler’s Method                                                  Polar Curve                                   π         1           3          3
If given that dy = f ( x, y ) and that              For a polar curve r(θ), the                                            6         2          2          3
                dx
                                                    AREA inside a “leaf” is
the solution passes through (xo, yo),                                        θ2               2
                                                                                                                           π          2          2
 y ( xo ) = yo                                                             ∫θ1
                                                                                  1
                                                                                  2
                                                                                       r (θ )  dθ
                                                                                                                         4         2          2
                                                                                                                                                           1

                                                    where θ1 and θ2 are the “first” two times that r =                     π          3         1
                                                    0.                                                                                                      3
 y ( xn ) = y ( xn−1 ) + f ( xn−1 , yn−1 ) ⋅ ∆x                                                                            3         2          2
                                                    The SLOPE of r(θ) at a given θ is
In other words:                                                          dy dy / dθ                                        π
                                                                           =                                                         1          0        “∞ ”
                xnew = xold + ∆x                                         dx dx / dθ                                        2
                          dy                                                                                               π         0         −1          0
                                                                              dθ  r (θ ) sin θ 
                                                                               d
                                                                           = d                 
       ynew = yold +                         ⋅ ∆x                                                                     Know both the inverse trig and the trig
                          dx ( xold , yold )
                                                                              dθ  r (θ ) cos θ 
                                                                                                                    values. E.g. tan(π/4)=1 & tan-1(1)= π/4
          Integration by Parts                                               Ratio Test                                           Trig Identities
                                                                   ∞
                                                                                                                      Double Argument
             ∫   udv = uv − ∫ vdu                   The series ∑ ak converges if
                                                                  k =0
                                                                                                                      sin 2 x = 2sin x cos x
Integral of Log                                                                                                       cos 2 x = cos 2 x − sin 2 x = 1 − 2sin 2 x
Use IBP and let u = ln x (Recall                                                        ak +1
                                                                                 lim          <1                                1
u=LIPET)                                                                         k →∞    ak                           cos 2 x = (1 + cos 2 x )
                                                                                                                                2
         ∫   ln x dx = x ln x − x + C               If the limit equal 1, you know nothing.

               Taylor Series                                                                                                    1
If the function f is “smooth” at x =                             Lagrange Error Bound                                 sin 2 x =    (1 − cos 2 x )
                                                                                                                                 2
a, then it can be approximated by                   If Pn ( x) is the nth degree Taylor polynomial                    Pythagorean
the nth degree polynomial
                                                    of f(x) about c and f ( n+1) (t ) ≤ M for all t                   sin 2 x + cos 2 x = 1
  f ( x) ≈ f (a ) + f '(a )( x − a)                                                                                   (others are easily derivable by
                                                    between x and c, then
                   +
                       f ''(a)
                                 ( x − a)2 + …                                                                        dividing by sin2x or cos2x)
                                                                                             M          n+1
                         2!                                  f ( x) − Pn ( x) ≤                     x−c               1 + tan 2 x = sec 2 x
                       f ( n ) (a)
                                                                                          ( n + 1)!
                   +               ( x − a)n .                                                                        cot 2 x + 1 = csc 2 x
                           n!                                                                                         Reciprocal
             Maclaurin Series                                                                                                     1
A Taylor Series about x = 0 is                            Alternating Series Error Bound                              sec x =         or cos x sec x = 1
                                                                                                                               cos x
called Maclaurin.
                                                                                                                                 1
                        x 2 x3                                N                                                       csc x =         or sin x csc x = 1
                                                    If S N = ∑ ( −1) an is the Nth partial sum of a
                                                                            n
         ex = 1 + x + + +…                                                                                                     sin x
                        2! 3!                                k =1                                                     Odd-Even
                     2
                   x       x4                       convergent alternating series, then                               sin(–x) = – sin x (odd)
      cos x = 1 − + − …
                   2! 4!                                           S∞ − S N ≤ aN +1                                   cos(–x) = cos x        (even)
                    x3 x5                                                                                             Some more handy INTEGRALS:
       sin x = x − + − …

         1
                    3! 5!                                                 Geometric Series
                                                                                                           ∞          ∫   tan x dx = ln sec x + C
             = 1 + x + x2 + x3 + …                  a + ar + ar 2 + ar 3 + … + ar n −1 + … = ∑ ar n −1                             = − ln cos x + C
      1− x                                                                                                n =1

                    x 2 x3 x 4                                                                         a
 ln( x + 1) = x − + − + …
                     2         3 4
                                                    diverges if |r|≥1; converges to
                                                                                                      1− r
                                                                                                           if |r|<1   ∫   sec x dx = ln sec x + tan x + C

This is available at http://covenantchristian.org/bird/Smart/Calc1/StuffMUSTknowColdNew.htm

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Stuff You Must Know Cold for the AP Calculus BC Exam!

  • 1. typed by Sean Bird, Covenant Christian High School updated August 15, 2009 AP CALCULUS Stuff you MUST know Cold * means topic only on BC Curve sketching and analysis Differentiation Rules Approx. Methods for Integration y = f(x) must be continuous at each: Chain Rule Trapezoidal Rule dy d du dy dy du b critical point: = 0 or undefined dx and look out for endpoints dx [ f (u )] = f '(u ) dx OR dx = du dx ∫a f ( x)dx = 1 b− a 2 n [ f ( x0 ) + 2 f ( x1 ) + ... local minimum: + 2 f ( xn −1 ) + f ( xn )] dy goes (–,0,+) or (–,und,+) or d 2 y >0 Product Rule Simpson’s Rule dx dx 2 b local maximum: d dx (uv) = du dx v+u dv dx OR u ' v + uv ' ∫a f ( x)dx = 2 1 dy goes (+,0,–) or (+,und,–) or d y <0 3 ∆x[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + ... dx dx 2 2 f ( xn −2 ) + 4 f ( xn−1 ) + f ( xn )] point of inflection: concavity changes Quotient Rule d 2 y goes from (+,0,–), (–,0,+), d u du dx v −u dv dx u ' v − uv '  = dx  v  v2 OR v2 dx 2 (+,und,–), or (–,und,+) Theorem of the Mean Value i.e. AVERAGE VALUE If the function f(x) is continuous on [a, b] Basic Derivatives “PLUS A CONSTANT” and the first derivative exists on the d n interval (a, b), then there exists a number dx ( ) x = nx n−1 The Fundamental Theorem of x = c on (a, b) such that Calculus b d ( sin x ) = cos x f (c ) = ∫ a f ( x)dx dx b (b − a) d ( cos x ) = − sin x ∫a f ( x)dx = F (b) − F (a) This value f(c) is the “average value” of dx where F '( x) = f ( x) the function on the interval [a, b]. d ( tan x ) = sec2 x dx Corollary to FTC Solids of Revolution and friends d ( cot x ) = − csc2 x Disk Method dx d b( x) x =b V =π∫ 2 [ R ( x)] dx ∫a ( x ) d f (t )dt = dx ( sec x ) = sec x tan x x=a dx Washer Method d f (b( x))b '( x) − f (a ( x))a '( x) dx ( csc x ) = − csc x cot x V =π∫ b a ([ R( x)] − [r ( x)] ) dx 2 2 Intermediate Value Theorem General volume equation (not rotated) d 1 du ( ln u ) = If the function f(x) is continuous on [a, b], b dx u dx and y is a number between f(a) and f(b), V = ∫ Area ( x) dx a d u du then there exists at least one number x= c dx ( ) e = eu dx in the open interval (a, b) such that *Arc Length L = ∫ 1 + [ f '( x)] dx b 2 a where u is a function of x, f (c ) = y . b =∫ 2 2 and a is a constant. a [ x '(t )] + [ y '(t )] dt More Derivatives Distance, Velocity, and Acceleration d  −1 u  1 du Mean Value Theorem velocity = d (position) sin = dx   a  2 a −u 2 dx dt If the function f(x) is continuous on [a, b], d (velocity) d −1 acceleration = AND the first derivative exists on the dx ( cos −1 x = ) interval (a, b), then there is at least one dt 1 − x2 dx dy number x = c in (a, b) such that *velocity vector = , d  −1 u  a du dt dt tan = ⋅ f (b) − f (a ) dx  a  a 2 + u 2 dx  f '(c) = . speed = v = ( x ')2 + ( y ') 2 * b−a d −1 dx ( cot −1 x = 1 + x2 ) displacement = ∫t tf v dt o d  −1 u  a du Rolle’s Theorem final time dx   sec = a  u u 2 − a 2 dx  ⋅ distance = ∫initial time v dt If the function f(x) is continuous on [a, b], tf d ( csc −1 x = −1 ) AND the first derivative exists on the ∫t o ( x ') 2 + ( y ') 2 dt * dx x x2 − 1 interval (a, b), AND f(a) = f(b), then there average velocity = is at least one number x = c in (a, b) such final position − initial position d u( x) du dx ( ) a = a ( ) ln a ⋅ u x dx that f '(c) = 0 . = total time ∆x d 1 = ( log a x ) = ∆t dx x ln a
  • 2. BC TOPICS and important TRIG identities and values l’Hôpital’s Rule Slope of a Parametric equation Values of Trigonometric f (a) 0 ∞ Given a x(t) and a y(t) the slope is Functions for Common Angles If = or = , g (b) 0 ∞ dy dy = dt θ sin θ cos θ tan θ f ( x) f '( x) dx dx then lim = lim dt x →a g ( x) x → a g '( x ) 0° 0 1 0 Euler’s Method Polar Curve π 1 3 3 If given that dy = f ( x, y ) and that For a polar curve r(θ), the 6 2 2 3 dx AREA inside a “leaf” is the solution passes through (xo, yo), θ2 2 π 2 2 y ( xo ) = yo ∫θ1 1 2  r (θ )  dθ   4 2 2 1 where θ1 and θ2 are the “first” two times that r = π 3 1 0. 3 y ( xn ) = y ( xn−1 ) + f ( xn−1 , yn−1 ) ⋅ ∆x 3 2 2 The SLOPE of r(θ) at a given θ is In other words: dy dy / dθ π = 1 0 “∞ ” xnew = xold + ∆x dx dx / dθ 2 dy π 0 −1 0 dθ  r (θ ) sin θ  d = d   ynew = yold + ⋅ ∆x Know both the inverse trig and the trig dx ( xold , yold ) dθ  r (θ ) cos θ    values. E.g. tan(π/4)=1 & tan-1(1)= π/4 Integration by Parts Ratio Test Trig Identities ∞ Double Argument ∫ udv = uv − ∫ vdu The series ∑ ak converges if k =0 sin 2 x = 2sin x cos x Integral of Log cos 2 x = cos 2 x − sin 2 x = 1 − 2sin 2 x Use IBP and let u = ln x (Recall ak +1 lim <1 1 u=LIPET) k →∞ ak cos 2 x = (1 + cos 2 x ) 2 ∫ ln x dx = x ln x − x + C If the limit equal 1, you know nothing. Taylor Series 1 If the function f is “smooth” at x = Lagrange Error Bound sin 2 x = (1 − cos 2 x ) 2 a, then it can be approximated by If Pn ( x) is the nth degree Taylor polynomial Pythagorean the nth degree polynomial of f(x) about c and f ( n+1) (t ) ≤ M for all t sin 2 x + cos 2 x = 1 f ( x) ≈ f (a ) + f '(a )( x − a) (others are easily derivable by between x and c, then + f ''(a) ( x − a)2 + … dividing by sin2x or cos2x) M n+1 2! f ( x) − Pn ( x) ≤ x−c 1 + tan 2 x = sec 2 x f ( n ) (a) ( n + 1)! + ( x − a)n . cot 2 x + 1 = csc 2 x n! Reciprocal Maclaurin Series 1 A Taylor Series about x = 0 is Alternating Series Error Bound sec x = or cos x sec x = 1 cos x called Maclaurin. 1 x 2 x3 N csc x = or sin x csc x = 1 If S N = ∑ ( −1) an is the Nth partial sum of a n ex = 1 + x + + +… sin x 2! 3! k =1 Odd-Even 2 x x4 convergent alternating series, then sin(–x) = – sin x (odd) cos x = 1 − + − … 2! 4! S∞ − S N ≤ aN +1 cos(–x) = cos x (even) x3 x5 Some more handy INTEGRALS: sin x = x − + − … 1 3! 5! Geometric Series ∞ ∫ tan x dx = ln sec x + C = 1 + x + x2 + x3 + … a + ar + ar 2 + ar 3 + … + ar n −1 + … = ∑ ar n −1 = − ln cos x + C 1− x n =1 x 2 x3 x 4 a ln( x + 1) = x − + − + … 2 3 4 diverges if |r|≥1; converges to 1− r if |r|<1 ∫ sec x dx = ln sec x + tan x + C This is available at http://covenantchristian.org/bird/Smart/Calc1/StuffMUSTknowColdNew.htm