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# 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

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

### Infinite Limit : We say lim f ( x ) = ¥ if we

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

# This has x® a

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

## This has the

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

## There is a similar definition for lim f ( x ) =

-¥ 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

### x®- ¥

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

### If r > 0 and x r is real for negative x b then lim r = 0 x ®-¥ x 3

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

### To compute Factor and Cancel p ( x) lim factor largest power of x out of both ( x

• - 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

#### If the two one sided limits had been equal then lim g ( x ) would have existed

-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

## Polynomials for all x

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

### Rational function,

• 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,

### ,L 2 2 2 2 4

• 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

# Intermediate Value Theorem Suppose that f ( x ) is continuous on [a,

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

#### Then there exists a number c'such that a < c'< b and f ( c') = M

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

• - sgn ( a ) ¥

# Visit http://tutorial

edu for a complete set of Calculus notes

• x ®-¥

# Visit http://tutorial

edu for a complete set of Calculus notes

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 ) then,

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

## If f ( x ) is the position of an object at time x then f ¢ ( a ) is the velocity of

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

## ç ÷ = – Quotient Rule g2 ègø

• (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,

# Implicit Differentiation Find y¢ if e2 x

• -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,

## Chain Rule Variants The chain rule applied to some specific functions

• 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

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'

f ( x ) is constant on the interval I

Visit http://tutorial

edu for a complete set of Calculus notes

### Visit http://tutorial

edu for a complete set of Calculus notes

### Calculus Cheat Sheet

#### Absolute Extrema 1

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

### Extrema Relative (local) Extrema 1

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

## 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

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

# Find all critical points of f ( x ) in [ a,

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

Evaluate f ( a ) and f ( b )

### Identify the abs

(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

## Related Rates Sketch picture and identify known/unknown quantities

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)

# Plug in known quantities and solve for the unknown quantity

### A 15 foot ladder is resting against a wall

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

### At what rate is the distance pushed towards the wall at 14 ft/sec

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 )

## x¢ is negative because x is decreasing

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

# Plug in and solve for y¢

• - 14 ) + 176 y¢ = 0 Þ 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

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

# 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') =

• - 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

### Minimize f = d'2 = ( x

• - 0 ) + ( y
• - 2 ) and the 2

# A = y ( 500

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

#### A¢ = 500

• - 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

# Visit http://tutorial

edu for a complete set of Calculus notes

# We have q ¢ = 0

• 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

Visit http://tutorial

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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

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

## Indefinite Integral : ò f ( x ) dx = F ( x ) + c

• 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

## 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 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 ,

### F ( x ) is f ( t ) dt = u ¢ ( x ) f [ u ( x) ]

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

## F ( x ) = ò f ( x ) dx ) then ò f ( x ) dx = F ( b )

- F ( a )

• cos ( x3 ) dx

## ò 1 5x

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

• c is a constant

## ò a cf ( x ) dx = c'ò a f ( x ) dx ,

• c is a constant b

# ò a f ( x ) dx = 0

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

### If f ( x ) ³ 0 on a £ x £ b then

• f ( x ) dx ³ 0
• f ( x ) dx

#### If m £ f ( x ) £ M on a £ x £ b then m ( b

• - 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

## Choose u and dv from a

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

### ò3 ln x dx

• 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

### Strip 1 sine out and convert rest to 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

### Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) ,

• cos 2 ( x ) =

Common Integrals

### ò k dx = k x + c'n n 1 ò x dx = n+1 x + c,

• -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

### ò cos u du = sin u + c'ò sin u du =

• - 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

Visit http://tutorial

edu for a complete set of Calculus notes

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

### Strip 2 secants out and convert rest to tangents using sec 2 x = 1 + tan 2 x ,

then use the substitution u = tan x

• n odd and m even

# Use either 1

• n even and m odd

# Each integral will be dealt with differently

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

# Visit http://tutorial

edu for a complete set of Calculus notes

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

### 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

• - 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

## Use Right Triangle Trig to go back to x’s

### From 3x x 2 = x

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

#### From this we see that cot q =

• - 9x 2 = 2 cos q

-9 x2 3x

• 4 4- 9 x 2 x

Net Area :

## ò 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,

- 9x = 4

• - 4 sin q = 4 cos q = 2 cosq 2

### Applications of Integrals b

• y = f ( x) Þ A = ò

### é ù ë upper function û

#### ò Q( x) dx where the degree of P ( x ) is smaller than the degree of

• the rational expression

### Integrate the partial fraction decomposition (P

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

# Term in P

D Factor in Q ( x )

### A ax + b

• ax 2 + bx + c

#### Ax + B ax + bx + c

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

#### Term in P

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

• 7 x 2 +13 x
• dx = ò
• ( ax + b )

### Ak x + Bk A1 x + B1 +L + k ax 2 + bx + c'( ax 2 + bx + c')

• -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

### é right function ù ë û

• - éëleft function ùû dy

## Q ( x )

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

Factor in Q ( x )

### 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

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

## A = ò f ( x )

• - 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

## Rings Cylinders A = p ( outer radius ) 2

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

# 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

# Axis use f ( y ) ,

Axis use f ( y ) ,

Axis use f ( x ) ,

• g ( x ) ,

• g ( y ) ,

• g ( y ) ,

## A ( y ) and dy

• g ( x ) ,

A ( x ) and dx

Axis : y = a > 0

Axis : y = a £ 0

Axis : y = a > 0

#### Axis : y = a £ 0

- f ( x )

• outer radius: a + g ( x )
• radius : a + y

- g ( x )

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

- g ( y )

## A( x 2 +4)+ ( Bx +C ) ( x

-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

### 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 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

# Visit http://tutorial

edu for a complete set of Calculus notes

• width : f ( y )

- g ( y )

## 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

#### Visit http://tutorial

edu for a complete set of Calculus notes

### 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 =

• in a £ x £ b ,

the work done is W = ò F ( x ) dx b

• b 1 b-a a

## SA = ò 2p x ds (rotate about y-axis)

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

### 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

### Infinite Limit ¥

f ( x ) dx = lim ò f ( x ) dx t

## ò ¥ f ( x ) dx = lim ò f ( x ) dx b

• t ®-¥

f ( x ) dx provided BOTH integrals are convergent

#### Discont

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

# Discont

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

# ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent

### Discontinuity at a < c'< b :

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

# Approximating Definite Integrals For given integral

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 » 2 ëé f ( x ) + 2 f ( x ) + +2 f ( x ) + L'+ 2 f ( x ) + f ( x ) ûù b

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

#### Visit http://tutorial

edu for a complete set of Calculus notes

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