Calculus Cheat Sheet Limits Definitions

Calculus Cheat Sheet Limits   Definitions Limit at Infinity : We say lim Precise Definition : We say lim f x  L if f x  L if we  x a  x can make f x as close to L as we want by taking x large enough and positive.  There is a similar definition for lim f x  L  x if we can make f x as close to L as we want by taking x sufficiently close to a (on either side of a) without letting x a . except we require x large and negative.   if we Infinite Limit : We say lim f x  x   L. a can make f x arbitrarily large (and positive) by taking x sufficiently close to a (on either side of a) without letting x a . This has  a  x  lim f x Right hand limit : L  a  x  lim f x “Working” Definition : We say       0 such that for every 0 there is a L  . whenever 0  x a  thenf x  a  x   L.  There is a similar definition for lim f x  x  This has the  lim f x Left hand limit :    the same definition as the limit except it requires x a . a except we make f x arbitrarily large and negative.  lim  x a f  x    lim f x x a x a x a x lim f x  Does Not Exist x  L a a  x a   lim f x  x a a  x    same definition as the limit except it requires x a . Relationship between the limit and one-sided limits  lim f   lim f  L L x  x  lim f x  L lim f x  lim f x      n n f x   lim f  5. lim   x   x a  x a   n n 6. f x  xlim f  x   lim  a x a    f x   xlim g x 3. lim  g x  f x lim  a x a x a         g   lim f x lim 2. lim  g x f x  x  x a x a x a  g x  0 providedlim x a   f x  lim f x 4. lim   x a x a g  g x  x  lim x a   1. lim  cf x  c xlim f  x  a x a   Properties and lim g x  both exist and c is any number then, f x  Assume lim x a x a       sgn a       c x d  x 9. n odd : lim a xn  x b x c sgn a b x c sgn a  8. n odd : lim a x n   x  x 7. n even : lim a xn     x       x n 6. n odd : lim xn   & lim x      x 0 0 then lim br 0 x  x 4. If r 0 and x r is real for negative x b then lim 0 x  xr 3. If r 5. n even : lim xn      lim ln x  & 0.  x 0   2. lim ln x   x  x lim ex  1. lim ex   & 0 and sgn a   1 if a  if a   1  Basic Limit Evaluations at   Note : sgn a Calculus Cheat Sheet     a  f lim gx  f b  lim f g x  x a  lim x  Factor and Cancel x 2  x 6 x2 4 x 12 lim lim x 2 x 2 x2 2 x xx 2  Polynomials at Infinity p x and q x are polynomials. To compute of both p x and q x then compute limit. 5 x 2        x  2  lim1 3x lim g x   x  2 x  2 7 One sided limits are different so xlim2 g x       x  2   x    1 x 2 5 if x  2 1  3 x if x 2 Compute two one sided limits,  lim x2 5 9 lim g x  g x  lim where g x x 2   1   18 6 108 Combine Rational Expressions 1  1 1  1 x x h  lim    lim   h 0 h x  h x  h 0 h  xx h      1  lim 1  h    lim  h  0 h x   x  h  h 0 x xh      Piecewise Function 1 x2  doesn’t exist. If the two one sided limits had would have existed g x  been equal then lim x 2  and had the same value.         Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous. 1. Polynomials for all x. 7. cos x  and sin x for all x. 2. Rational function, except for x’s that give 8. tan x and secx provided division by zero.  3 3 3. n x (n odd) for all x. x  ,  , , , , 2 2 2 2 4. n x (n even) for all x 0 . 9. cot x and cscx provided 5. ex for all x. x  , 2 , , 0, , 2 ,  6. ln x for x 0 . Intermediate Value Theorem Suppose that f x is continuous on [a, b] and let M be any number between f a  and f b  . M  Then there exists a number c such that a c b and f c . 3 2  4 x2  3   x   2 lim    lim x  4 x2  2 x2 3 3x 4 lim 2 2  x  5x 2 x x 5x     x 6 8  4 x 2 Rationalize Numerator/Denominator x 3 x 3 x 3 lim lim x  9 x2 81 x 9 x2  81 3 x 1 9 x  lim lim x 9 x 2 81 3  x x 9 x 9 3 lim x 2 p x  out factor largest power of x in q x  q x   x a  x             Evaluation Techniques L’Hospital’s Rule Continuous Functions f x  0 f x    If f x  is continuous at a then lim f x  f a x a or lim then, If lim x a g x  0 x a g x    Continuous Functions and Composition f x  f x  a is a number, or  lim lim f x is continuous at b and lim g x  b then x  a g x  x a g x  Calculus Cheat Sheet Derivatives     .      f x  If y Definition and Notation f x  h f x then the derivative is defined to bef x  lim h 0 h   a.  3. If f x is the position of an object at time x then f a  is the velocity of the object at x a .       equation of the tangent line at x a is given by y  f a f a x a .   is the instantaneous rate of change of f x at x      then, m f a is the slope of the tangent line to y f x  atx a and the  1.     Interpretation of the Derivative 2. f a f x  If y  If y f x all of the following are equivalent notations for derivative evaluated at x a . df dy y   f a  Df a  x a dx x a dx x a  If y f x then all of the following are equivalent notations for the derivative. df dy d  f  x  y   Dfx  f x  dx dx dx  f x g x        d c  0 dx d n 6. x  n xn 1 – Power Rule dx d 7. f g x  f gx g x  dx This is the Chain Rule 5.  f g x  2. cf  c f   1.  Basic Properties and Formulas If f x and g x are differentiable functions (the derivative exists), c and n are any real numbers,        d a x a x ln a  dx d x e  ex dx d lnx  1 , x 0 dx x d ln x  1 , x 0 dx x d loga x  1 , x 0 dx x ln a  Common Derivatives d csc x  csc xcot x dx d cot x  csc2 x dx d sin 1 x  1 2 dx 1 x d cos 1 x   1 2 dx 1 x d 1 1 x tan  2 dx 1 x      d x 1 dx d sin x  cos x dx d cos x  sin x dx d tan x  sec 2 x dx d sec x  sec x tan x dx   – Quotient Rule     f  f g f g 4.   g2 g   – Product Rule g  f g f g   f  3. Calculus Cheat Sheet                         Chain Rule Variants The chain rule applied to some specific functions. n n 1 d d  f x  f x   n  f   f x sin 1. 5. fx   x  f x  cos    dx dx  d ef x  d efx   2  2. 6. f x  f  x  tan f x  f x sec  dx dx d f x  d 7. secf ( x)  f ( x) secf( x)  tan f ( x)  3. ln f x  dx dx  f x  f x  d 1 d 8. tan f x  2 4. f x  f x  f  x  sin  cos  dx f x  dx 1            Higher Order Derivatives The nth Derivative is denoted as The Second Derivative is denoted as n 2 d f d f 2   n  x  f x   2 and is defined as f  f x  n and is defined as dx dx x  fn 1 x , i.e. the derivative of n 1 . x  the (n-1)st derivative, f       Implicit Differentiation if 2ex 9 y x3 y2 sin y 11 x . Remembery y x here, so products/quotients of x and y    Find y n      f   f  , i.e. the derivative of the x  first derivative, f x . x f    will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to differentiate as normal and every time you differentiate a y you tack on a y (from the chain rule). After differentiating solve for y . 2 x9 y   11 2e 2 x9 y 2 9 x 2 y2 11 2e x y 3 3 2x 9 y cos y 2x y 9e    2 2 3x y  y      y    cos y 3 2 x y y cos y y 11   3 3x y2  2 x y 9e 3 2 x y y cos y y 11 2  9y e 2 x9 y  2 9y3x2 y2 2 x9 y  2e  2x 9 y e      Increasing/Decreasing – Concave Up/Concave Down Critical Points Concave Up/Concave Down x c is a critical point of f x provided either 1. If f  x  0 for all x in an interval I then 1. f c  0 or 2.f c  doesn’t exist. f x is concave up on the interval I.  x  0   3. If f for all x in an interval I then f x is constant on the interval I. Inflection Points x c is a inflection point of f x if the concavity changes at x c .  for all x in an interval I then    x  0 f x is decreasing on the interval I. for all x in an interval I then f x is concave down on the interval I. f x is increasing on the interval I. 2. If f 0  2. If f  x  Increasing/Decreasing 1. If f x  0 for all x in an interval I then Calculus Cheat Sheet     x  0  to the right ofx x  0 c. to the left   c is to the left  to the right ofx c .   ofx c and f a ,b then there exist numbers c and d so that, 1. a c , d b , 2. f c is the abs. max. in a ,b ,3. f d is the abs. min. in a ,b .  Extreme Value Theorem If f x is continuous on the closed interval x  0  x  0   of x c and f 2. a rel. min. of f x if f   c is a critical point of f x  .    c , then x 1. a rel. max. of f x if f  1st Derivative Test If x c is a critical point of f x then x Fermat’s Theorem If f x has a relative (or local) extrema at x     x for all x in the domain.  f 2. x c is a relative (or local) minimum of f x if f c  f x for all x near c.    2. x c is an absolute minimum of f x if f c  x for all x in the domain.   f  if f c Extrema Relative (local) Extrema 1. x c is a relative (or local) maximum of f x if f c  f x for all x near c.   Absolute Extrema 1. x c is an absolute maximum of f x  is   3. not a relative extrema of f x if f x the same sign on both sides of x c .   Finding Absolute Extrema To find the absolute extrema of the continuous function f x on the interval a , b  use the following process. 1. Find all critical points of f x in a , b .   2nd Derivative Test If x c is a critical point of f x such that f c  0 thenx c c  0 1. is a relative maximum of f x if f  3. Evaluate f a and f b  . Finding Relative Extrema and/or Classify Critical Points 1. Find all critical points of f x  . 2. Use the 1st derivative test or the 2nd derivative test on each critical point. 4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3. b a .  a  f     f b  Mean Value Theorem If f x is continuous on the closed interval a , b  and differentiable on the open interval a , b then there is a number a c b such that f c 0 then (n+1)st guess is xn 1  xn  f xn f xn provided f    If xn is the n guess for the root/solution of f x  Newton’s Method th xn  exists. .   2. Evaluate f x at all points found in Step 1. 0  c 2. is a relative minimum of f x if f  3. may be a relative maximum, relative c  0 . minimum, or neither if f     . Calculus Cheat Sheet    Related Rates Sketch picture and identify known/unknown quantities. Write down equation relating quantities and differentiate with respect to t using implicit differentiation (i.e. add on a derivative every time you differentiate a function of t). Plug in known quantities and solve for the unknown quantity. Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one The bottom is initially 10 ft away and is being starts walking north. The angle changes at pushed towards the wall at 41 ft/sec. How fast 0.01 rad/min. At what rate is the distance between them changing when 0.5 rad? is the top moving after 12 sec?             We have 0.01 rad/min. and want to find x . We can use various trig fcns but easiest is, x x  sec tan sec  50 50 and solve. We know 0.5 so plug in x  0.01  sec0.5  tan 0.5 50 x 0.3112 ft/min Remember to have calculator in radians!   7 y  ft/sec 4 176    714  176 y 0       176 . Plug in and solve  so y  15 2 7 2 for y .      x is negative because x is decreasing. Using Pythagorean Theorem and differentiating, 2 2 2 x  y 15 2 x x 2 y y 0 After 12 sec we have x 10 12 14  7 and Optimization Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for one of the two variables and plug into first equation. Find critical points of equation in range of variables and verify that they are min/max as needed. Ex. We’re enclosing a rectangular field with Ex. Determine point(s) on y x2 1 that are 500 ft of fence material and one side of the closest to (0,2). field is a building. Determine dimensions that will maximize the enclosed area.       y 1 y 2  y 3 y 3 Differentiate and find critical point(s). f  2 y 3 y 32 By the 2nd derivative test this is a rel. min. and so all we need to do is find x value(s). 1 x2 32  1 12 x 2    2   2     The 2 points are then  , and  1 2 3 2  The dimensions are then 250 x 125. 2 x and plug into the function. 2 x2  y 1  f x2 y 2       2 500y  2y Differentiate and find critical point(s). A  500 4y y 125 By 2nd deriv. test this is a rel. max. and so is the answer we’re after. Finally, find x. x 500 2  125  250 constraint is y x 1 . Solve constraint for  A  y500 2 y   x 500 2y and the 2   into area. 2 y 2   2  2 Minimize f d x 0 Maximize A xy subject to constraint of x 2 y 500 . Solve constraint for x and plug 1 2 , 3 2 . Calculus Cheat Sheet Integrals Definitions f x is continuous Anti-Derivative : An anti-derivative of f x   f x . f x  dx F x  c Indefinite Integral : where F x is an anti-derivative of f x . on a , b .Divide a , b into n subintervals of  x and choose xi * from each interval.   width , such that F x is a function, F x   Definite Integral: Suppose f x  dx  lim  a n n x . f xi*  i 1   Then  b                   Fundamental Theorem of Calculus Variants of Part I : Part I : If f x is continuous on a , b then d u x  x  u x f u x  f t dt a ,b  g x  f t dt is also continuous on dx a a d b d x   v x f v  f t dt x  f  t dt f x . and g x  dx vx  dx a   u x d Part II : f x  is continuous ona , b ,F x is  u x f u( x) v x f v( x) f t dt dx vx  x dx  ) an anti-derivative of f x  (i.e. F x  f  .  Properties     cf x dx b a cf x dx  f x dx  g x dx  f x g x dx   b b b f x  g x dx  f x     a  dx  a g x dx  a b  b a dx f x  b c   g x on a x b b then  f x dx a b  a a g x dx b dx 0 on a  x b then f x  a  0 M b a Common Integrals cos u du sin u c    sec u du ln sec u tan u a2 u sec u tan u du sec u c c 1   a2 u 2  csc u cot udu  csc u c    csc2 u du  cot u c    edu  u ec du sin  dx a1 ln ax b b ln u du u ln u u c u  a1 tan 2 du 1 au c   1   sec2 u du tan u c c   ax  1 dx ln x x 1   x 1 dx  1 ua c    tan u du ln sec u c sin u du  cos u c 1  c, n  1  x n dx  n1 1 xn   k dx k x c   f x dx  b  a  m b a If m  f x M on a x b then  a  If f x  b  f x dx  f x dxfor  any value of c.  If f x c f x dx, c is a constant  b a a c dx  c b a f x dx a f x dx  bf x dx f x dx a b c b a    f x  dx 0 a  a c  f  x dx , c is a constant   f x  dx  F b F a   a  b then c  Calculus Cheat Sheet Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. g b g a     using  f u du  g x  dx . For indefinite integrals drop the limits of integration.  2 1 3 du    x 1  u 13 1 :: x 2  u 2 3  cos u du  53 sinu  35 sin 8 sin 1  1 8 8 Integration by Parts : u dv uv  v du  and b b u dv uv  a a  x dx 2 8 5 1 3   du 3 x dx 3 3 dx cos x  2  u x 2 5x  1  5 x2 cos x3  dx b  v du. Choose u and dv from a dv .  integral and compute du by differentiating u and compute v using v   1  2 Ex.   du b a gx will convert  f g x  g x  dx  Substitution : The substitution u 3 ln x dx u ln x  x c dv dx 5 du 1 5 5 x dx v  e x 5 xln x  x 3 3 ln x dx xln x 3 3dx   x    dx v   x dx   xe x e e    du     Ex.  dv  ex x x   x dx x e e  ux  5  x x e dx Ex.  5ln 5  3ln 3  2 For tann x secm x dx we have the following :   Products and (some) Quotients of Trig Functions For sinn x cosm x dx we have the following : odd. Strip 1 sine out and convert rest to 1. odd. Strip 1 tangent and 1 secant out and convert the rest to secants using cosines using sin 2 x 1 cos2 x , then use tan 2 x sec 2 x  1 , then use the substitution the substitution u cos x . u sec x . 2. odd. Strip 1 cosine out and convert rest 2 2 2. even. Strip 2 secants out and convert rest to sines using cos x 1 sin x , then use the substitution u sin x . to tangents using sec 2 x 1 tan 2 x , then 3. and both odd. Use either 1. or 2. use the substitution u tan x . 3. odd and even. Use either 1. or 2. 4. and both even. Use double angle 4. even and odd. Each integral will be and/or half angle formulas to reduce the dealt with differently. integral into a form that can be integrated. 2 , sin 2 x  21  1 cos 2x   x ,cos x 12 1 cos 2x   Trig Formulas : sin 2 x 2 sin x cos 4  sec x 1  sec xtan xsec xdx 4  u 2 1  u 4 du u   17 sec 7 x 15 sec 5 x c   2 sec x  sin 5 cos x 3x dx dx   sin 4 xsin x cos 3x dx  (1 cos 2 x) 2 sin x cos3 x (1 u 2 )2 u3   (sin 2 x ) 2 sin x   xsec xdx 2   tan  xsec xtan xsec xdx 5  tan 3 sin5 x cos3 x  Ex.  tan3 x sec5 x dx  Ex.        1. dx du  cos 3 x u cos x 2 4  1 2u 3 u u  dx  du  12 sec2 x 2 ln cos x 12 cos2 x c d  d   12 sin2  12 csc d   12 cot  c    2     2 3 cos 2 cos  d 4 sin2 2 cos   3 9   dx  x  23 sin 2 2 sec  1 tan  16  dx x 2 4 9 x2  2 2 tan  sec 1  16 Ex.      2 2 cos  1 sin   Calculus Cheat Sheet Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. b2 x2 a 2  x ab sec a 2 b 2x 2  x ab tan a 2 b 2x 2  x ab sin 2 4 9x2  4 4 sin         x2 2 cos Use Right Triangle Trig to go back to x’s. From 3 x so, x . Because we have an indefinite substitution we have sin 2  Recall 4 cos2 Px  Partial Fractions : If integrating Q x . So,   c   dx  x2 4 9 x2  .   2 cos  4 9x 2 2 4 4 x 9 x    In this case we have 16 4 9 x2 3x   From this we see that cot    integral we’ll assume positive and drop absolute value bars. If we had a definite integral we’d need to compute ’s and remove absolute value bars based on that and, x if x 0 x  x if x  0  dx  where the degree ofP x is smaller than the degree of Q x  . Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition (P.F.D.). For each factor in the denominator we get term(s) in the decomposition according to the following table. Factor in Q x b     A( x2  C 4   x 4)  ) ( Bx C) ( x 1 (x  1 ) ( x2 4) 2 4 8 tan x  1 2 Here is partial fraction form and recombined.   Set coefficients equal to get a system and solve to get constants. A B 7 C  B 13 4 A C 0 A 4 B 3 C 16  dx ln x    16 x2  4  3 2   4 ln x 1  x 4   32 x  k  2  4 x 1     Set numerators equal and collect like terms. 7 x2  13 x  A B x2 C B x 4 A C  3xx 164 dx 4 x 1 xA1 Bx2  ax2 bx c  dx  k Ak x Bk   7 x2 13 x 2 (x 1 ) (x  4 )  13 x 2 ) ( x 4 )  bx c dx  (  2 x7x1 2 A1 x B1  ax 2 bx c k     ( x 1) ( x2 4) ax  7 x2 13 x Ex. Term in P.F.D Ak A1  A2  2 ax b   b ax b  ax  ax  k  ax 2 bx c Ax B 2 ax bx c  ax b A ax b   Term in P.F.D  Factor in Qx        An alternate method that sometimes works to find constants. Start with setting numerators equal in 2 13 x Ax2 4  Bx  C x 1 . Chose nice values of x and plug in. previous example : 7 x  For example if x 1 we get 20 5A which gives A 4 . This won’t always work easily. Calculus Cheat Sheet Applications of Integrals b  Net Area : a f x  dx represents the net area betweenf x and the x-axis with area above x-axis positive and area below x-axis negative.    a    Area Between Curves : The general formulas for the two main cases for each are, b d upper function    lower function  dx right function   &x  f y  A y  f x  A   c left function    dy If the curves intersect then the area of each portion must be found individually. Here are some sketches of a couple possible situations and formulas for a couple of possible cases. c A   f x  g x dx a  Volumes of Revolution : The two main formulas are V A x  dx and V b g x  f x dx   f y g  y dy   c A y  dy some general information about each method of computing and some examples. Rings Cylinders  radius 2  A  outer radius 2 inner A 2 radius width / height       . Here is  d c  A   f x g  x dx   b a   A  Limits: x/ y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use f x  , Vert. Axis use f y , Horz. Axis use f y  , Vert. Axis use f x  ,  outer radius:  inner radius: a a g x f x Ex. Axis : y  a 0 radius : a y  radius : a y  width : f y g y  width : f y  g y    inner radius : a g x  f x  outer radius : a Ex. Axis : y  a 0  Ex. Axis : y  a 0 g x  , Ax and dx.  Ex. Axis : y  a 0 g y  , Ay and dy.  g y  , Ay and dy.  g x  , Ax and dx.  These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the y  a 0 case with a 0 . For vertical axis of rotation ( x  a 0 and x  a 0 ) interchange x and y to get appropriate formulas. Calculus Cheat Sheet moves an object Work : If a force of F x  Average Function Value : The average value b  1 b a  of f x on a  x b isfavg F x  dx   a  in a  x b , the work done isW b a f x  dx Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,   a 2y ds (rotate about x-axis) SA b 2x ds  b SA  ds  a   b  L a (rotate about y-axis)    ,a f b b  r g  t , a t   d  if f t  , y  ds  r if x 2  dr d 2  y b   f y , a dt 2 2 ds   x dxdt  dydt x b  dx dy f x , a  ds  1 2 y  2 dx   dx if   dy if ds  1 dy  where ds is dependent upon the form of the function being worked with as follows. With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute. Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic. Infinite Limit f x dx  c    c f x dx  f x dx   b  f x dx 2. b lim  t a t  t  t   lim   a  3.  f x dx  1. f x dx  f x dxprovided BOTH integrals are convergent. Discontinuous Integrand b  t b f x  dx lim 2. Discont. at b :  a t  f x dx t b c a a b  a   t  b f x  dx lim 1. Discont. at a:  a a f x dx b 3. Discontinuity at a c b :f x  dx   f x dx  fx dxprovided  both are convergent. f x  g x  0 ona , a dx p x    divg. then f x  dx  2. If g x  dx conv. a  a divg. converges if p 1 and diverges for p 1 .  1  a then,   0 then  Useful fact : If a conv. then gx  dx   f x dx  a   1. If  Comparison Test for Improper Integrals : If  c Approximating Definite Integrals b f x  dx and a n (must be even for Simpson’s Rule) definex  nb aand  a  a and x n b then,   f n x         4 f nx1    f nx 2  f nx    2 f nx 1     2  2 f x2     2 f x2     0   with x * f xn*  xi 1 , xi i is midpoint  ,x  Simpson’s Rule :   b x x0 2 f x1 f  2 x  f x  dx  f  x0 4 f x1 3 f x  dx  a   Trapezoid Rule : * 2  b  f x    a *   x  f x dx f x1  b Midpoint Rule :   a divide a , b  into n subintervals x0 , x1 , x1 , x2 , … , xn 1 , xn   For given integral 