Calculus Cheat Sheet

Calculus Cheat Sheet Calculus Cheat Sheet Limits Definitions Precise Definition : We say lim f  x   L if Limit at Infinity : We say lim f  x   L if we x a x for every   0 there is a   0 such that whenever 0  x  a   then f  x   L   . can make f  x  as close to L as we want by “Working” Definition : We say lim f  x   L There is a similar definition for lim f  x   L if we can make f  x  as close to L as we want except we require x large and negative. by taking x sufficiently close to a (on either side of a) without letting x  a . Infinite Limit : We say lim f  x   if we x a Right hand limit : lim f  x   L . This has x a the same definition as the limit except it requires x  a . Left hand limit : lim f x   L . This has the x a taking x large enough and positive. x   x 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 . There is a similar definition for lim f  x    x a except we make f  x  arbitrarily large and same definition as the limit except it requires negative. x  a. Relationship between the limit and one-sided limits lim f  x   L  lim f  x  lim f  x  L lim f  x   lim f  x   L  lim f  x   L x a x a x a x a x a x a lim f  x   lim f  x   lim f x  Does Not Exist x a x a x a x a x a  x a x a 2. lim  f  x   g  x   lim f  x   lim g  x  x a x a x a 3. lim  f  x  g  x   lim f  x  lim g  x  x a x a x a f x   f  x   lim x a provided lim g  x  0 4. lim   x a x a g x g lim     x  a x  n n 5. lim  f  x     lim f  x  x a  x a     n lim f x  n f x   6. lim x a  x a  Basic Limit Evaluations at   Note : sgn a  1 if a  0 and sgn  a   1 if a  0 . 1. lim e x   & x 2. lim ln  x    x  lim e x  0 x   & x 0 b 0 x xr r 4. If r  0 and x is real for negative x b then lim r  0 x   x 3. If r  0 then lim  5. n even : lim x n     x lim ln  x     n 6. n odd : lim x n   & lim x      x   lim f  g  x   f lim g  x   f  b  x a x a Polynomials at Infinity p  x  and q x  are polynomials. To compute Factor and Cancel x 2  4x 12  x  2  x  6  lim  lim 2 x 2 x 2 x  2x x  x 2 lim x  x 6 8  4 x 2 Rationalize Numerator/Denominator 3 x 3 x 3 x  lim 2 lim 2 x9 x  81 x9 x  81 3  x 9x 1  lim  lim x9 x 2  81  3  x x9  x  9  3  x  lim x 2    x p  x q  x factor largest power of x in q  x  out of both p x and q  x  then compute limit. lim x    x 2 3  42 3  42 3 x 2 4 3 x x  lim  lim  5x  2x 2 x    x 2  5x  2  x    5x  2 2 Piecewise Function  1 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   Properties Assume lim f  x  and lim g  x both exist and c is any number then, 1. lim  cf  x   clim f  x Evaluation Techniques Continuous Functions L’Hospital’s Rule f  x 0 f  x   f  x   f a  If f  x  is continuous at a then lim x a If lim  or lim  then, x a g  x  x a g  x  0  Continuous Functions and Composition f  x f   x  lim a is a number,  or  lim f  x  is continuous at b and lim g  x  b then x a g  x  x a g  x x a 1  h  1 1  lim     lim h 0 h  x x  h    h 0 x  x  h  x 2    x2  5 if x   2 lim2 g x  where g  x   x  1 3 x if x   2 Compute two one sided limits, 2 lim g  x  lim x  5  9 x  2 x  2 x  2 x  2 lim g  x  lim 1  3 x  7 One sided limits are different so lim g  x x 2 doesn’t exist. If the two one sided limits had been equal then lim2 g  x would have existed x  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 x 5. e for all x. x   , 2 ,  , 0, , 2 , 6. ln x for x  0 . 7. n even : lim a x n    b x  c  sgn  a     Intermediate Value Theorem Suppose that f  x  is continuous on [a, b] and let M be any number between f  a  and f  b  . 8. n odd : lim a x    b x  c  sgn a   Then there exists a number c such that a  c  b and f  c   M . x n x  9. n odd : lim a x n    c x  d   sgn a   x   Calculus Cheat Sheet Calculus Cheat Sheet Derivatives Definition and Notation f  x  h  f  x If y  f  x  then the derivative is defined to be f   x   lim . h 0 h If y  f  x  then all of the following are If y  f  x  all of the following are equivalent equivalent notations for the derivative. df dy d f   x   y     f  x   Df x  dx dx dx notations for derivative evaluated at x  a . df dy f a   y  x a    Df a  dx x a dx x a If y  f  x  then, Interpretation of the Derivative 2. f  a  is the instantaneous rate of 1. m  f   a  is the slope of the tangent change of f  x  at x  a . line to y  f  x  at x  a and the equation of the tangent line at x  a is given by y  f  a   f   a x  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 . Basic Properties and Formulas If f  x  and g  x  are differentiable functions (the derivative exists), c and n are any real numbers, 1.  c f   c f   x  3.  f g   d  c  0 dx d n 6.  x   n xn 1 – Power Rule dx d f  g  x    f   g x  g  x  7. dx This is the Chain Rule 5. 2.  f  g   f   x   g  x  f  g  f g – Product Rule  f  f  g  f g  4.    – Quotient Rule g2 g    Chain Rule Variants The chain rule applied to some specific functions. n n 1 d d  f  x    n  f  x   f   x  cos  f  x    f   x  sin  f  x   1. 5. dx dx  d d f  x f  x tan  f  x   f  xsec 2  f  x 2. 6. e  f   x e dx dx d d f  x  7. ln f  x    sec  f ( x )   f  ( x) sec f ( x ) tan  f ( x ) 3. dx dx f x  f   x d d tan 1  f  x   8. 2 sin f  x  f   x  cos f  x  4. dx 1  f  x   dx    sin x   cos x  cosx    sinx  tanx   sec2 x sec x   sec x tan x d  csc x    csc x cot x dx d 2  cot x    csc x dx d sin  1 x   1 2 dx 1 x d 1 cos 1 x     dx 1 x 2 d 1 tan 1 x   1 x2 dx d  a x   a x ln a dx d x  e   ex dx d  ln x   1 , x  0 dx x d 1  ln x   , x  0 dx x d 1 loga  x   , x0  dx x ln a          Higher Order Derivatives The nth Derivative is denoted as The Second Derivative is denoted as 2 d f dn f 2  f  x  f  x  2 and is defined as f  n  x   and is defined as dx dx n   1  n n    f  x   f  x , i.e. the derivative of the f  x  f  x , i.e. the derivative of first derivative, f   x  .   the (n-1)st derivative, f  n 1  x . Implicit Differentiation Find y if e2x 9y  x 3 y 2  sin y  11x . Remember y  y  x  here, so products/quotients of x and y 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 . e 2x  9y  2  9 y   3 x 2y 2  2 x 3y y  cos  y  y 11 2 x 9 y  9 y e 2 x 9 y  3 x 2 y 2  2 x 3 y y   cos  y  y   11 2e 3 x   1  2 x y 9 e x Common Derivatives d dx d dx d dx d dx d dx  2 9 y cos  y  y 11  2 e2 x 9 y 3 x 2 y 2  y  2 x 9 y  3 x 2y 2 11  2 e 2 x3 y  9e 2x  9 y  cos  y 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. Increasing/Decreasing 1. If f  x  0 for all x in an interval I then f  x is increasing on the interval I. 2. If f  x  0 for all x in an interval I then f  x is decreasing on the interval I. 3. If f  x  0 for all x in an interval I then f  x is constant on the interval I. 2. If f  x   0 for all x in an interval I then f  x  is concave down on the interval I. Inflection Points x  c is a inflection point of f  x  if the concavity changes at x  c . Calculus Cheat Sheet Absolute Extrema 1. x  c is an absolute maximum of f x  if f c   f x for all x in the domain. 2. x  c is an absolute minimum of f  x if f c   f x for all x in the domain. Calculus Cheat Sheet 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. 2. x  c is a relative (or local) minimum of f  x  if f  c   f  x  for all x near c. Fermat’s Theorem If f  x  has a relative (or local) extrema at x  c , then x  c is a critical point of f  x  . Extreme Value Theorem If f  x  is continuous on the closed interval 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  . 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 . 2. Evaluate f  x  at all points found in Step 1. 3. Evaluate f  a  and f  b  . 4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3. 1st Derivative Test If x  c is a critical point of f  x then x  c is 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? 1. a rel. max. of f  x  if f  x   0 to the left of x  c and f   x  0 to the right of x  c . 2. a rel. min. of f  x if f   x  0 to the left of x  c and f  x  0 to the right of x  c . 3. not a relative extrema of f  x  if f   x  is the same sign on both sides of x  c . 2nd Derivative Test If x  c is a critical point of f  x such that f  c   0 then x  c 1. is a relative maximum of f  x  if f   c   0 . 2. is a relative minimum of f  x  if f c  0. 3. may be a relative maximum, relative minimum, or neither if f   c   0 . Finding Relative Extrema and/or Classify Critical Points 1. Find all critical points of f  x  . x is negative because x is decreasing. Using Pythagorean Theorem and differentiating, x2  y2  152  2 x x  2 y y  0 After 12 sec we have x  10  12  14   7 and so y  15 2  7 2  176 . Plug in and solve for y . 7 7   14   176 y  0  y  ft/sec 4 176 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  x 2  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. 2. Use the 1st derivative test or the 2nd derivative test on each critical point. Mean Value Theorem If f x  is continuous on the closed interval  a , b and differentiable on the open interval a , b  f  b  f  a . then there is a number a  c  b such that f  c  ba Newton’s Method If xn is the nth guess for the root/solution of f x  0 then (n+1)st guess is xn  1  xn  provided f   xn  exists. f  xn  f   xn  We have    0.01 rad/min. and want to find x . We can use various trig fcns but easiest is, x x sec   sec tan    50 50 We know  0.5 so plug in  and solve. x sec  0.5 tan  0.5  0.01  50 x  0.3112 ft/min Remember to have calculator in radians! Minimize f  d 2   x  0    y  2  and the 2 Maximize A  xy subject to constraint of x  2 y  500 . Solve constraint for x and plug into area. A  y  500 2y  x  500  2 y   500 y  2y 2 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 The dimensions are then 250 x 125. 2 constraint is y  x 2  1. Solve constraint for 2 x and plug into the function. 2 2 2 x  y 1  f  x   y  2   y  1   y  2   y2  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). x2  32  1  12  x   12 2 The 2 points are then  1 2   , 32 and  1 2  , 32 . Calculus Cheat Sheet Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose f x  is continuous Anti-Derivative : An anti-derivative of f  x on a , b  . Divide a , b  into n subintervals of is a function, F  x  , such that F  x   f x  . width  x and choose x i* from each interval. Indefinite Integral :  f x dx  F x   c Then n b  * i Ex. 1 2 1 5 x ux Fundamental Theorem of Calculus Variants of Part I : Part I : If f  x  is continuous on  a , b then d u x  x f t  dt  u  x  f u x  g  x   f  t dt is also continuous on a , b  dx a a d b d x f t dt  v  x  f v  x  f  t dt  f  x . and g x  dx v x dx a d u x  Part II : f x  is continuous on  a , b , F x  is f t  dt  u  x  f u ( x )  v  x  f dx v x an anti-derivative of f  x  (i.e. F  x    f  x  dx ) 2 cos  x 3  dx 1 5 x cos  x  dx  1 2 2 2 v (x )  b c dx  c b  a   a cf  x dx  c  a f  x dx , c is a constant a  a b  b  f x dx   f x dx   f x dx for any value of c. If f x   g x  on a  x  b then  f  x  dx   g  x  dx b a a f  x  dx    f  x  dx b c a a a f  x  dx   b a b b If f x   0 on a  x  b then b a b  f  x  dx  0 a If m  f x  M on a  x  b then m b  a    a f  x  dx  M b  a  b Common Integrals  k dx  k x  c n n  x dx  n11 x  c, n  1 x dx dx  x    ln x  c  a x b dx  a ln ax b c  ln u du  u ln u   u  c  e du  e  c 1 1 1 1 u 1 u cos u du  sin u  c sin u du   cos u  c sec u du  tan u  c sec u tan u du  sec u  c csc u cot udu   csc u  c csc u du   cot u  c 2 2  tan u du  ln sec u  c  sec u du  ln sec u  tan u  c u  a  u du  a tan  a  c 1 u  a u du  sin  a  c 1 2 1 1 2 1 2 2 5 3 cos  u  du  53 sin u  1 8 b  a u dv  uv b a 5 3  sin 8  sin 1  b   v du . Choose u and dv from a Ex.  x e ux  xe x x dx Ex. x dv  e  du  dx v  e x dx  xe  x   e  x dx  xe  x e  x  c 5  3 ln x dx u  ln x dv  dx  du  x1 dx v  x  3 ln x dx  x ln x 3   3 dx   x ln  x   x  3 5 5 5 5  5ln  5  3ln 3  2 Products and (some) Quotients of Trig Functions For  sinn x cosm x dx we have the following : For  tann x secm x dx we have the following : 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using cosines using sin 2 x  1  cos 2 x , then use the substitution u  cos x . tan2 x  sec2 x  1 , then use the substitution u  sec x . 2. m odd. Strip 1 cosine out and convert rest 2 2 2. m even. Strip 2 secants out and convert rest to sines using cos x  1  sin x , then use to tangents using sec2 x  1 tan2 x , then the substitution u  sin x . 3. n and m both odd. Use either 1. or 2. use the substitution u  tan x . 4. n and m both even. Use double angle 3. n odd and m even. Use either 1. or 2. and/or half angle formulas to reduce the 4. n even and m odd. Each integral will be integral into a form that can be integrated. dealt with differently. Trig Formulas : sin 2x   2 sin  x cos  x  , cos2  x   12  1  cos  2 x  , sin 2  x   21 1 cos  2x   f  x  dx c a using integral and compute du by differentiating u and compute v using v   dv . Properties  cf  x dx  c  f  x dx , c is a constant  a f  x dx  0 8 3 1 3 a b 2  du  3x dx  x dx  du 3 Integration by Parts :  u dv  uv   v du and then  f  x  dx  F  b  F  a  . b b x  1  u  13  1 :: x  2  u  23  8 b  f x  g x dx   f x dx   g x dx b b b  a f  x   g  x  dx   a f  x  dx   a g  x  dx g b   a f  g x  g  x dx   g a  f u  du u Substitution : The substitution u  g  x  will convert du  g  x  dx . For indefinite integrals drop the limits of integration. where F x  is an anti-derivative of f  x  . f x   x .  a f  x  dx  nlim i Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. Ex.  tan 3 x sec 5 x dx  tan 3 Ex. x sec xdx   tan x sec x tan x sec xdx 5 2 4  sec x  1 sec   u  1 u du  2 2 4 x tan x sec xdx 4  sec x  sec 5 x  c 1 7 7 1 5  u  sec x  sin 5 x  cos x dx 3 sin5 x (sin 2 x) 2sin x sin4 x sinx  cos x dx   cos x dx   cos x dx sin x  dx   (1 cos x )  u  cos x cos x (1  u )   du   1 2 u  u du u u 3 3 3 2 2 3 2 2 3 2 3 4  sec x  2 ln cos x  cos2 x  c 1 2 2 1 2 Calculus Cheat Sheet 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. 2 2 2 a  b x  x  ab sin  2 2 2 b x  a  x  ab sec cos   1  sin  2 Ex. x 16 2 4 9x 2 x  32 sin  2 2 2 a  b x  x  ab tan  2 2 tan   sec  1 2   dx dx  32 cos d 2 2 sec  1  tan  16 4 sin2   2 cos  9  23 cos   d   sin122 Recall 2 2 Use Right Triangle Trig to go back to x’s. From 3x x 2  x . Because we have an indefinite substitution we have sin  2 so, 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 From this we see that cot   x In this case we have 4  9x  2 cos  . 2 16 2 4 9 x 2 dx   4 4 9x2 3x  a f  x dx represents the net area between f  x and the x-axis with area above x-axis positive and area below x-axis negative. Area Between Curves : The general formulas for the two main cases for each are,   12 csc d    12 cot   c 2 4  9x  4  4sin   4 cos   2 cos 2 d Applications of Integrals b Net Area : y  f  x  A   b a  upper function  c P x  Q x dx where the degree of P  x  is smaller than the degree 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. Term in P.F.D Factor in Q  x  A ax  b ax  b Ax  B 2 ax  bx  c ax2  bx  c Ex. 7x 2  13x  x 1 x ( ) ( 2 4) k  ax 2 (x 1 )(x2 4 )  x7x1 x13x4 dx   x41  x x 4 dx   x4 1  x3 x 4  x16 4 dx  4 ln x 1  ln  x  4   8 tan  x  )( 2 k A1 x  B1 Ak x  Bk   k ax2  bx  c  ax2  bx  c 7 x2  13 x dx 2 (  bx  c Term in P.F.D A1 A2 Ak   k ax  b  ax  b 2  ax  b   ax  b  3 16 2 ) 2 2 3 2 2 1 2 Here is partial fraction form and recombined.  x A 1  Bx C x2 4  dx & x  f  y   A   d c right function   left function  dy . So, 4 9 x 2  x Q x . Factor denominator as completely as possible and find the partial fraction decomposition of Factor in Q x function 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. b A   f  x   g  x  dx a Partial Fractions : If integrating   lower A  d c f  y  g y dy c b A   f  x  g  x dx   g  x  f  x dx a c Volumes of Revolution : The two main formulas are V   A x dx and V   A  y  dy . Here is some general information about each method of computing and some examples. Rings Cylinders 2 2 A   outer radius   inner radius A  2 radius  width / height    Limits: x/y of right/bo t ring to x/y of left/top ring Horz. Axis use f  x  , Vert. Axis use f  y  , Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use f  y , Vert. Axis use f x  , g  x  , A x and dx. g  y  , A y  and dy. g  y  , A y and dy. g  x  , A x  and dx. Ex. Axis : y  a  0 Ex. Axis : y  a  0 Ex. Axis : y  a  0 Ex. Axis : y  a  0 outer radius : a  f  x  outer radius: a  g  x  inner radius: a  f x  radius : a  y width : f  y  g  y radius : a  y inner radius : a  g  x  A (x 2 4) ( Bx C) ( x 1) (x 1 ) (x2 4 ) Set numerators equal and collect like terms. 7 x 2  13 x  A  B  x 2  C  B x  4 A  C Set coefficients equal to get a system and solve to get constants. A B  7 C  B  13 4A  C  0 A 4 B 3 C  16 An alternate method that sometimes works to find constants. Start with setting numerators equal in 2 2 previous example : 7 x  13 x  A  x  4   Bx  C  x  1 . Chose nice values of x and plug in. For example if x  1 we get 20  5 A which gives A  4 . This won’t always work easily. width : f  y   g  y  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 Work : If a force of F  x moves an object Average Function Value : The average value of f  x  on a  x  b is f avg  b in a  x  b , the work done is W   F  x  dx a 1 b a b  a f x dx Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, b b b SA   2 y ds (rotate about x-axis) L   ds a SA   2  x ds (rotate about y-axis) a a where ds is dependent upon the form of the function being worked with as follows.   1   ds  1  ds  dxdt  dy dx 2 dx if y  f  x , a  x  b ds  dx dy 2 dy if x  f  y , a  y  b ds  r2   2   dt dy dt  dr d  2 2 if x  f  t , y  g  t , a  t  b d  if r  f  , a    b 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  t f  x  dx  lim  f  x  dx 1.  3.   f  x dx    f  x dx   t  a 2. a  c    c  b b f x dx  lim  f x dx  t   t f  x  dx provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at a:  f  x  dx  lim b a t a  f  x  dx 2. Discont. at b :  f  x  dx  lim b b t a b c b a a c t b  f  x  dx t a 3. Discontinuity at a  c  b :  f  x  dx   f  x  dx   f  x  dx provided both are convergent. Comparison Test for Improper Integrals : If f  x  g  x  0 on  a ,   then, 1. If   a   f  x  dx conv. then  a g  x  dx conv. Useful fact : If a  0 then  a 1 xp  2. If  g  x  dx divg. then  f  x  dx divg. a a dx converges if p 1 and diverges for p  1 . Approximating Definite Integrals For given integral b a f x dx and a n (must be even for Simpson’s Rule) define x  b a n and divide a , b  into n subintervals  x0 , x1  , x1 , x2 , … ,  xn 1 , xn  with x0  a and xn  b then, Midpoint Rule :  f  x  dx  x f x  f x   f x  , xi b * 1 a * 2 * n * is midpoint  xi  1 , xi  x  f  x 0   2 f  x1    2 f  x 2     2 f  xn 1   f  x n  Trapezoid Rule :  f  x  dx  a 2 b x  f  x 0   4 f  x 1   2 f  x 2    2 f  xn  2   4 f  xn  1   f  xn  Simpson’s Rule :  f  x  dx  a 3 b