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Limits Definitions Precise Definition : We say lim f ( x ) = L'if Limit at Infinity : We say lim f ( x ) = L'if we x® a

for every e > 0 there is a d'> 0 such that whenever 0 < x

- - a < d'then f ( x )

- L'< e

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

Right hand limit : lim+ f ( x ) = L'

the same definition as the limit except it requires x > a

taking x large enough and positive

- x ®-¥

can make f ( x ) arbitrarily large (and positive) by taking x sufficiently close to a (on either side of a) without letting x = a

-¥ x ®a

except we make f ( x ) arbitrarily large and x® a same definition as the limit except it requires negative

- x 0 and sgn ( a ) =
- - 1 if a < 0
- lim e x = ¥ & 2
- lim e x = 0

- lim ln ( x ) = ¥
- lim- ln ( x ) =
- b =0 x ®¥ x r 4

If r > 0 then lim

n even : lim x n = ¥ x ®± ¥

n odd : lim x n = ¥ & lim x n =

-¥ x ®¥

- x ®- ¥

n even : lim a x n + L'+ b x + c'= sgn ( a ) ¥ x ®± ¥

n odd : lim a xn + L'+ b x + c'= sgn ( a ) ¥ x ®¥

Evaluation Techniques Continuous Functions L’Hospital’s Rule f ( x) ± ¥ f ( x) 0 If f ( x ) is continuous at a then lim f ( x ) = f ( a ) x ®a = then,

If lim = or lim x® a g ( x ) x® a g ( x ) ±¥ 0 Continuous Functions and Composition f ( x) f ¢ ( x) a is a number,

- -¥ = lim lim f ( x ) is continuous at b and lim g ( x ) = b then x ®a g ( x ) x ®a g ¢ ( x ) x® a Polynomials at Infinity lim f ( g ( x ) ) = f lim g ( x ) = f ( b ) x® a x® a p ( x ) and q ( x ) are polynomials

- - 2)( x + 6) x2 + 4 x
- - 12 x®±¥ q ( x ) lim = lim 2 x ®2 x 2 ® x

- 2x x ( x

- - 2) p ( x ) and q ( x ) and then compute limit
- x+6 8 = lim = =4 x ®2 x 2 3

- 42 2 x 3

- 42 3x2

- - 4 3 x lim = lim 2 5 = lim 5 x = Rationalize Numerator/Denominator 2 x®-¥ 5 x
- - 2 x x®- ¥ x x®-¥ 2 2 2 x x 3- x 3- x 3+ x = lim 2 lim 2 Piecewise Function x®9 x
- - 81 x®9 x
- - 81 3 + x ì x 2 + 5 if x <
- -1 9- x lim g ( x ) where g ( x ) = í = lim = lim x®-2 î1
- - 3 x if x ³
- -2 x ®9 ( x 2
- - 81) 3 + x x®9 ( x + 9 ) 3 + x Compute two one sided limits,
- -1 1 lim g ( x ) = lim- x 2 + 5 = 9 = =x®-2x®-2 (18 )( 6 ) 108 lim g ( x ) = lim+ 1
- - 3 x = 7 x®-2+ x®-2 Combine Rational Expressions One sided limits are different so lim g ( x ) 1æ 1 1ö 1 æ x
- - ( x + h) ö x®-2
- - ÷ = lim çç lim ç ÷ h ®0 h x + h x ø h®0 h è x ( x + h ) ÷ø è doesn’t exist

-1 1 æ

- -h ö 1 x®-2 = lim çç =- 2 ÷ = lim and had the same value

h ®0 h x ( x + h ) ÷ h®0 x ( x + h ) x è ø

Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous

cos ( x ) and sin ( x ) for all x

- except for x’s that give 8

tan ( x ) and sec ( x ) provided division by zero

- n 3p p p 3p 3
- x (n odd) for all x
- x ¹ L,

- n x (n even) for all x ³ 0

cot ( x ) and csc ( x ) provided 5

- e x for all x
- x ¹ L',

2p ,L 6

- ln x for x > 0

b] and let M be any number between f ( a ) and f ( b )

n odd : lim a xn + L'+ c'x + d'=

- - sgn ( a ) ¥

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- x ®-¥

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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 equivalent notations for the derivative

df dy d'f ¢ ( x ) = y¢ = = = ( f ( x) ) = Df ( x ) dx dx dx

If y = f ( x ) all of the following are equivalent notations for derivative evaluated at x = a

df dy f ¢ ( a ) = y ¢ x= a = = = Df ( a ) dx x =a dx x =a

Interpretation of the Derivative 2

f ¢ ( a ) is the instantaneous rate of

m = f ¢ ( a ) is the slope of the tangent line to y = f ( x ) at x = a and the

- change of f ( x ) at x = a

equation of the tangent line at x = a is given by y = f ( a ) + f ¢ ( a )( x

- the object at x = a
- ( c'f )¢ = c'f ¢ ( x )
- ( f g )¢ =

± g )¢ = f ¢ ( x ) ± g ¢ ( x ) f ¢ g + f g ¢ – Product Rule

- f g ¢ 4

- (c) = 0 dx d'n 6

( x ) = n x n-1 – Power Rule dx d'f ( g ( x) ) = f ¢ ( g ( x) ) g¢ ( x) 7

- dx This is the Chain Rule

Higher Order Derivatives The Second Derivative is denoted as The nth Derivative is denoted as 2 d'f dn f f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n) ( x ) = n and is defined as dx dx ¢ ¢ n n-1 ( ) f ¢¢ ( x ) = ( f ¢ ( x ) ) ,

the derivative of the f ( x ) = f ( ) ( x ) ,

the derivative of first derivative,

- f ¢ ( x )
- the (n-1)st derivative,

- -9 y + x3 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¢

- e2 x-9 y ( 2
- - 9 y¢ ) + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 2e 2 x
- - 9 y¢e 2x
- - 9 y + 3x 2 y 2 + 2 x 3 y y¢ = cos ( y ) y¢ + 11 3
- d ( csc x ) =
- - csc x cot x dx d'( cot x ) =
- - csc2 x dx d'( sin
- -1 x ) = 1 2 dx 1- x d
- -1 ( cos x ) =
- - 1 2 dx 1- x d'1 tan
- -1 x ) = ( dx 1 + x2
- ( 2 x y

Common Derivatives 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

Basic Properties and Formulas If f ( x ) and g ( x ) are differentiable functions (the derivative exists),

- c and n are any real numbers,

- n n-1 d'd 1

é f ( x )ùû = n éë f ( x ) ùû f ¢ ( x ) 5

- cos éë f ( x ) ùû =
- - f ¢ ( x ) sin éë f ( x ) ùû dx ë dx d'f ( x) d'2
- e = f ¢ ( x ) e f ( x) 6

tan éë f ( x )ùû = f ¢ ( x ) sec2 éë f ( x ) ùû dx dx d'f ¢( x) d'7

( sec [ f ( x)]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )] 3

ln ëé f ( x ) ûù = dx dx f ( x) f ¢( x ) d'd 8

- -1 ëé f ( x ) ûù = sin éë f ( x ) ùû = f ¢ ( x ) cos éë f ( x ) ùû 2 4
- dx 1 + ëé f ( x )ûù dx

d x ( a ) = a x ln ( a ) dx d'x ( e ) = ex dx d'( ln ( x ) ) = 1x ,

- x > 0 dx d'( ln x ) = 1x ,

x ¹ 0 dx d'( log a ( x )) = x ln1 a ,

- x > 0 dx
- - cos ( y ) ) y ¢ = 11

- 2e2 x

- - 3 x 2 y 2

- 2e 2 x

- - 3 x 2 y 2 2 x3 y

- 9e2 x

- - 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

If f ¢¢ ( x ) < 0 for all x in an interval I then f ( x ) is concave down on the interval I

f ( x ) is constant on the interval I

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x = c'is an absolute maximum of f ( x ) if f ( c') ³ f ( x ) for all x in the domain

x = c'is a relative (or local) maximum of f ( x ) if f ( c') ³ f ( x ) for all x near c

x = c'is an absolute minimum of f ( x ) if f ( c') £ f ( x ) for all x in the domain

then x = c'is a critical point of f ( x )

b] then there exist numbers c'and d'so that,

- a £ c,
- d £ b ,
- f ( c') is the abs
- f ( d') is the abs

Finding Absolute Extrema To find the absolute extrema of the continuous function f ( x ) on the interval [ a ,

- b ] use the following process

Evaluate f ( x ) at all points found in Step 1

Evaluate f ( a ) and f ( b )

(largest function value) and the abs

(smallest function value) from the evaluations in Steps 2 & 3

x = c'is a relative (or local) minimum of f ( x ) if f ( c') £ f ( x ) for all x near c

- 1st Derivative Test If x = c'is a critical point of f ( x ) then x = c'is

Write down equation relating quantities and differentiate with respect to t using implicit differentiation (i

add on a derivative every time you differentiate a function of t)

Two people are 50 ft apart when one The bottom is initially 10 ft away and is being starts walking north

The angle q changes at 0

01 rad/min

How fast between them changing when q = 0

? is the top moving after 12 sec

of f ( x ) if f ¢ ( x ) > 0 to the left of x = c'and f ¢ ( x ) < 0 to the right of x = c'

of f ( x ) if f ¢ ( x ) < 0 to the left of x = c'and f ¢ ( x ) > 0 to the right of x = c'

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

is a relative minimum of f ( x ) if f ¢¢ ( c') > 0

- 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 )

Using Pythagorean Theorem and differentiating,

x 2 + y 2 = 15 2 Þ 2 x x¢ + 2 y y¢ = 0 After 12 sec we have x = 10

- - 12 ( 14 ) = 7 and so y = 152
- - 7 2 = 176

- - 14 ) + 176 y¢ = 0 Þ y¢ = ft/sec 4 176

write down equation to be optimized and constraint

Solve constraint for one of the two variables and plug into first equation

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

b ] and differentiable on the open interval ( a ,

b ) then there is a number a < c'< b such that f ¢ ( c') =

- - f ( a) b-a

Newton’s Method If xn is the n guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn+1 = xn th

- f ( xn ) f ¢ ( xn )
- provided f ¢ ( xn ) exists

- - 0 ) + ( y
- - 2 ) and the 2

- - 2 y ) x = 500
- - 2 y Þ = 500 y
- - 2 y 2 Differentiate and find critical point(s)

- - 4 y Þ y = 125 By 2nd deriv
- test this is a rel

and so is the answer where after

Finally,

- x = 500
- - 2 (125 ) = 250 The dimensions are then 250 x 125

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01 rad/min

- and want to find x¢

We can use various trig fcns but easiest is,

x x¢ sec q = Þ sec q tan q q ¢ = 50 50 We know q = 0

- 05 so plug in q ¢ and solve
- x¢ sec ( 0
- 5 ) tan ( 0
- 01) = 50 x¢ = 0
- 3112 ft/sec Remember to have calculator in radians

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- constraint is y = x 2 + 1

Solve constraint for x 2 and plug into the function

2 x2 = y

- -1 Þ f = x2 + ( y

- 2 ) = y

- 1+ ( y

- 2) = y2

- - 3 y + 3 Differentiate and find critical point(s)
- f ¢ = 2y
- - 3 Þ y = 32 nd By the 2 derivative test this is a rel

and so all we need to do is find x value(s)

- x 2 = 32
- - 1 = 12 Þ x = ± 12 2

Calculus Cheat Sheet

Integrals Definitions Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x )

Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class

( ) ò a f ( g ( x )) g ¢ ( x ) dx = ò g (a ) f ( u ) du

- on [ a ,

Divide [ a ,

- b ] into n subintervals of
- is a function,

F ( x ) ,

such that F ¢ ( x ) = f ( x )

u Substitution : The substitution u = g ( x ) will convert

width D'x and choose x from each interval

- du = g ¢ ( x ) dx

For indefinite integrals drop the limits of integration

where F ( x ) is an anti-derivative of f ( x )

- f (x )D x

ò 1 5x

b ] then d'u(x) x f ( t ) dt = u ¢ ( x ) f éëu ( x ) ùû g ( x ) = ò f ( t ) dt is also continuous on [ a ,

b ] dx ò a a d'b d'x f ( t ) dt =

- - v¢ ( x ) f éë v ( x )ùû and g ¢ ( x ) = f ( t ) dt = f ( x )

dx ò v( x) dx ò a d'u( x) Part II : f ( x ) is continuous on [ a ,

- - v¢ ( x ) f [ v( x ) ] dx ò v( x) an anti-derivative of f ( x ) (i

- F ( a )

- cos ( x3 ) dx

Properties ò cf ( x) dx = c'ò f ( x) dx ,

- c is a constant

- c is a constant b

ò a f ( x ) dx = ò a f ( t ) dt

ò a f ( x) dx =

- - òb f ( x) dx a

ò f ( x ) dx £ ò b

- f ( x ) dx ³ 0
- f ( x ) dx

- - a ) £ ò f ( x ) dx £ M ( b

- a ) b

8 5 1 3

- cos ( u ) dx

= 53 sin ( u ) 1 = 53 ( sin (8 )

- - sin (1) ) 8

x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 2 3 = 8 Integration by Parts : ò u dv = uv

- - ò v du and

- ò v du

integral and compute du by differentiating u and compute v using v = ò dv

- du = dx v =
- -x ò xe dx =
- - xe + ò e dx =

- e + c

- u = ln x
- dv = dx Þ du = 1x dx v = x

ò3 ln x dx = x ln x 3

- - ò3 dx = ( x ln ( x )

- x ) 3 5

Products and (some) Quotients of Trig Functions For ò tan n x sec m x dx we have the following : For ò sin n x cos m x dx we have the following : 1

- cosines using sin 2 x = 1
- - cos 2 x ,

then use the substitution u = cos x

Strip 1 cosine out and convert rest 2

- to sines using cos2 x = 1
- - sin 2 x ,

then use the substitution u = sin x

- n and m both odd

Use either 1

- n and m both even

Use double angle 3

and/or half angle formulas to reduce the 4

integral into a form that can be integrated

- cos 2 ( x ) =

Common Integrals

- -1 ò x dx = ò x dx = ln x + c'ò a x + b dx = a ln ax + b + c'ò ln u du = u ln ( u )
- - u + c'ò e du = e + c
- cos ( x3 ) dx = ò
- = 5ln ( 5 )

- 3ln ( 3)

ò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx b b b ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx

Þ du = 3 x dx Þ x dx = du 2

- - cos u + c'ò sec u du = tan u + c'ò sec u tan u du = secu + c'ò csc u cot udu =
- - csc u + c'ò csc u du =
- - cot u + c'2

ò tan u du = ln sec u + c'ò sec u du = ln sec u + tan u + c'u ò a + u du = a tan ( a ) + c'1 u ò a

- - u du = sin ( a ) + c'1
- x sec5 x dx = ò ( sec2 x
- - 1) sec 4 x tan x sec xdx
- ( u = sec x )
- = 17 sec 7 x
- - 15 sec5 x + c

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ò cos x dx (sin x) sin x sin x sin x sin x ò cos x dx = ò cos x dx = ò cos x dx (1- cos x ) sin x =ò dx ( u = cos x ) cos x (1-u ) 1 2 u + u du =

-ò du =

- - ò u u sin 5 x
- 3 5 2 4 ò tan x sec xdx = ò tan x sec x tan x sec xdx
- = ò ( u 2
- - 1) u 4 du

Strip 1 tangent and 1 secant out and convert the rest to secants using tan 2 x = sec 2 x

then use the substitution u = sec x

then use the substitution u = tan x

- n odd and m even

- n even and m odd

- 2 1 1 2 (1 + cos ( 2 x ) ) ,
- sin ( x ) = 2 (1
- - cos ( 2 x ) )

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- = 12 sec 2 x + 2 ln cos x
- - 12 cos2 x + c'© 2005 Paul Dawkins

Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions

- - b 2 x 2 Þ x = ab sin q
- - a 2 Þ x = ab sec q
- cos 2 q = 1
- - sin 2 q Ex

4- 9 x2

- tan 2 q = sec 2 q

x = sin q Þ dx = cos q d'q 2 3

- a 2 + b 2 x 2 Þ x = ab tan q
- sec 2 q = 1 + tan 2 q
- 16 4 sin 2 q ( 2cosq ) 9

( 23 cos q ) dq = ò sin122 q dq

- = ò 12 csc dq =
- -12 cot q + c

Because we have an indefinite substitution we have sin q = 2 so,

integral we’ll assume positive and drop absolute value bars

If we had a definite integral we’d need to compute q ’s and remove absolute value bars based on that and,

ì x if x ³ 0 x =í î- x if x < 0

- - 9x 2 = 2 cos q

-9 x2 3x

- 4 4- 9 x 2 x

Net Area :

- x-axis with area above x-axis positive and area below x-axis negative

- 9x = 4

- - 4 sin q = 4 cos q = 2 cosq 2

- y = f ( x) Þ A = ò

- the rational expression

For each factor in the denominator we get term(s) in the decomposition according to the following table

D Factor in Q ( x )

- ax 2 + bx + c

- 7 x2 +13 x ( x
- -1)( x 2 + 4 )
- 7 x2 +13 x ( x
- -1)( x 2 + 4 )
- + bx + c')

D A1 A2 Ak + +L+ k ax + b ( ax + b )2 ( ax + b )

- 7 x 2 +13 x
- dx = ò
- ( ax + b )

- -1)( x 2 + 4 )

3x x2 +4

- 3 x +16 x2 + 4

16 x2 +4

- = 4 ln x
- - 1 + 32 ln ( x 2 + 4 ) + 8 tan
- -1 ( x2 ) Here is partial fraction form and recombined

- - éëleft function ùû dy

Factor denominator as completely as possible and find the partial fraction decomposition of

Factor in Q ( x )

A = ò f ( x )

- - g ( x ) dx b

Partial Fractions : If integrating

- - éëlower function ùû dx & x = f ( y ) Þ A = ò

A = ò f ( y )

- - g ( y ) dy d

- - g ( x ) dx + ò g ( x )
- - f ( x ) dx

Volumes of Revolution : The two main formulas are V = ò A ( x ) dx and V = ò A ( y ) dy

Here is some general information about each method of computing and some examples

- - ( inner radius ) 2 A = 2p ( radius ) ( width / height )

- to x/y of outer cyl

Axis use f ( y ) ,

Axis use f ( x ) ,

- g ( x ) ,

- g ( y ) ,

- g ( y ) ,

- g ( x ) ,

A ( x ) and dx

Axis : y = a > 0

Axis : y = a £ 0

Axis : y = a > 0

- outer radius : a

- f ( x )

- outer radius: a + g ( x )
- radius : a + y
- inner radius : a

- g ( x )

- inner radius: a + f ( x )
- radius : a
- - y width : f ( y )

- g ( y )

-1) ( x

- -1)( x 2 + 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

- - B = 13 4A- C = 0 A=4 B=3 C = 16

Start with setting numerators equal in previous example : 7 x 2 + 13 x = A ( x 2 + 4 ) + ( Bx + C ) ( x

Chose nice values of x and plug in

For example if x = 1 we get 20 = 5A which gives A = 4

This won’t always work easily

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- width : f ( y )

- g ( y )

For vertical axis of rotation ( x = a > 0 and x = a £ 0 ) interchange x and y to get appropriate formulas

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Work : If a force of F ( x ) moves an object

- in a £ x £ b ,

the work done is W = ò F ( x ) dx b

- b 1 b-a a

where ds is dependent upon the form of the function being worked with as follows

- ( ) 1+ ( )
- ds = 1 + ds =
- dy dx dx dy
- ( dxdt )
- dx if y = f ( x ) ,
- a £ x £ b
- dy if x = f ( y ) ,
- a £ y £ b

ds = r 2 + ( ddrq ) d'q if r = f (q ) ,

- a £ q £ b
- dt if x = f ( t ) ,
- y = g ( t ) ,
- a £ t £ 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

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

f ( x ) dx = lim ò f ( x ) dx t

- t ®-¥

f ( x ) dx provided BOTH integrals are convergent

at a: ò f ( x ) dx = lim+ ò f ( x ) dx b

at b : ò f ( x ) dx = lim- ò f ( x ) dx

Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a,

¥ ) then,

- f ( x ) dx conv
- then ò g ( x ) dx conv

Useful fact : If a > 0 then

- f ( x ) dx divg

dx converges if p > 1 and diverges for p £ 1

If ò g ( x ) dx divg

- then ò

f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b

-n a and

- divide [ a ,

b ] into n subintervals [ x0 ,

xn ] with x0 = a and xn = b then,

Midpoint Rule : Trapezoid Rule : Simpson’s Rule :

ò f ( x ) dx » Dx éë f ( x ) + f ( x ) + L'+ f ( x ) ùû ,

- is midpoint [ xi

Dx ò f ( x ) dx » 3 ëé f ( x ) + 4 f ( x ) + 2 f ( x ) +L + 2 f ( x ) + 4 f ( x ) + f ( x )ûù b

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