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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 e > 0 there is a d > 0 such that
whenever 0 < x - a < d then f ( x ) - L < e
...
“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
by taking x sufficiently close to a (on either side
of a) without letting x = a
...
x ®a
Right hand limit : lim+ f ( x ) = L
...
Left hand limit : lim- f ( x ) = L
...
There is a similar definition for lim f ( x ) = -¥
x ®a
except we make f ( x ) arbitrarily large and
negative
...
Relationship between the limit and one-sided limits
lim f ( x ) = L Þ lim+ f ( x ) = lim- f ( x ) = L
lim+ f ( x ) = lim- f ( x ) = L Þ lim f ( x ) = L
x ®a
x ®a
x ®a
x ®a
x ®a
x ®a
lim f ( x ) ¹ lim- f ( x ) Þ lim f ( x ) Does Not Exist
x ®a +
x ®a
x ®a
Properties
Assume lim f ( x ) and lim g ( x ) both exist and c is any number then,
x ®a
x ®a
1
...
lim é f ( x ) ± g ( x ) ù = lim f ( x ) ± lim g ( x )
û x®a
x ®a ë
x ®a
3
...
lim ê
provided lim g ( x ) ¹ 0
ú=
x ®a
x ®a g ( x )
ë
û lim g ( x )
x ®a
n
n
5
...
lim é n f ( x ) ù = n lim f ( x )
û
x ®a ë
x®a
Basic Limit Evaluations at ± ¥
Note : sgn ( a ) = 1 if a > 0 and sgn ( a ) = -1 if a < 0
...
lim e x = ¥ &
x®¥
2
...
If r > 0 and x r is real for negative x
b
then lim r = 0
x ®-¥ x
3
...
n even : lim x n = ¥
x ®± ¥
6
...
n even : lim a x + L + b x + c = sgn ( a ) ¥
n
x ®± ¥
8
...
n odd : lim a x n + L + c x + d = - sgn ( a ) ¥
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...
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...
x ®-¥
© 2005 Paul Dawkins
Calculus Cheat Sheet
Evaluation Techniques
Continuous Functions
L’Hospital’s Rule
f ( x) 0
f ( x) ± ¥
If f ( x ) is continuous at a then lim f ( x ) = f ( a )
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
= lim
a is a number, ¥ or -¥
f ( x ) is continuous at b and lim g ( x ) = b then
x ®a g ( x )
x ®a g ¢ ( x )
(
)
x ®a
lim f ( g ( x ) ) = f lim g ( x ) = f ( b )
x ®a
x ®a
Polynomials at Infinity
p ( x ) and q ( x ) are polynomials
...
(
(
)
Piecewise Function
)
-1
1
=(18)( 6 ) 108
Combine Rational Expressions
1æ 1
1ö
1 æ x - ( x + h) ö
lim ç
- ÷ = lim ç
÷
h ®0 h x + h
x ø h ®0 h ç x ( x + h ) ÷
è
è
ø
1 æ -h ö
1
-1
= lim ç
=- 2
÷ = lim
h ®0 h ç x ( x + h ) ÷
h®0 x ( x + h )
x
è
ø
=
)
x 2 3 - 42
3 - 42
3x 2 - 4
3
x
lim
= lim 2 5
= lim 5 x = x ®-¥ 5 x - 2 x 2
x ®-¥ x
x ®- ¥
2
x -2
x -2
ì x 2 + 5 if x < -2
lim g ( x ) where g ( x ) = í
x ®-2
î1 - 3x if x ³ -2
Compute two one sided limits,
lim- g ( x ) = lim- x 2 + 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
...
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
...
3p p p 3p
3
...
x ¹ L , - , - , , ,L
2
2 2 2
4
...
9
...
e x for all x
...
ln x for x > 0
...
Then there exists a number c such that a < c < b and f ( c ) = M
...
math
...
edu for a complete set of Calculus notes
...
h ®0
h
If y = f ( x ) then all of the following are
equivalent notations for the derivative
...
df
dy
f ¢ ( a ) = y ¢ x =a =
=
= Df ( a )
dx x =a dx x =a
Interpretation of the Derivative
2
...
m = f ¢ ( a ) is the slope of the tangent
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 - 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
...
( f ± g )¢ = f ¢ ( x ) ± g ¢ ( x )
3
...
ç
èg
d
(c) = 0
dx
d n
6
...
f ( g ( x )) = f ¢ ( g ( x )) g¢ ( x )
dx
This is the Chain Rule
5
...
math
...
edu for a complete set of Calculus notes
...
n
n -1
d
d
1
...
cos é f ( x ) ù = - f ¢ ( x ) sin é f ( x ) ù
ë
û
ë
û
ë
û
ë
û
dx
dx
d f ( x)
d
f x
e
tan é f ( x ) ù = f ¢ ( x ) sec 2 é f ( x ) ù
2
...
ë
û
ë
û
dx
dx
d
f ¢( x)
d
7
...
ln é f ( x ) ù =
ë
û
dx
dx
f ( x)
f ¢( x)
d
d
tan -1 é f ( x ) ù =
8
...
sin é f ( x ) ù = f ¢ ( x ) cos é f ( x ) ù
ë
û
ë
û
ë
û
dx
1 + é f ( x )ù
dx
ë
û
)
(
(
(
(
)
(
)
)
)
(
(
)
)
Higher Order Derivatives
The Second Derivative is denoted as
The nth Derivative is denoted as
d2 f
dn f
f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as
f ( n ) ( x ) = n and is defined as
dx
dx
¢
f ¢¢ ( x ) = ( f ¢ ( x ) )¢ , i
...
the derivative of the
f ( n ) ( x ) = f ( n -1) ( x ) , i
...
the derivative of
first derivative, f ¢ ( x )
...
(
)
( )
Implicit Differentiation
¢ if e 2 x -9 y + x3 y 2 = sin ( y ) + 11x
...
The “trick” is to
differentiate as normal and every time you differentiate a y you tack on a y¢ (from the chain rule)
...
e 2 x -9 y ( 2 - 9 y¢ ) + 3 x 2 y 2 + 2 x3 y y¢ = cos ( y ) y¢ + 11
2e
2 x -9 y
- 9 y¢e
( 2 x y - 9e x
3
2 x -9 y
2 -9 y
+ 3x y + 2 x y y¢ = cos ( y ) y¢ + 11
2
2
3
- cos ( y ) ) y¢ = 11 - 2e2 x -9 y - 3x 2 y 2
Þ
11 - 2e 2 x -9 y - 3x 2 y 2
y¢ = 3
2 x y - 9e2 x -9 y - cos ( y )
Increasing/Decreasing – Concave Up/Concave Down
Critical Points
x = c is a critical point of f ( x ) provided either
1
...
f ¢ ( c ) doesn’t exist
...
If f ¢ ( x ) > 0 for all x in an interval I then
f ( x ) is increasing on the interval I
...
If f ¢ ( x ) < 0 for all x in an interval I then
f ( x ) is decreasing on the interval I
...
If f ¢ ( x ) = 0 for all x in an interval I then
Concave Up/Concave Down
1
...
2
...
Inflection Points
x = c is a inflection point of f ( x ) if the
concavity changes at x = c
...
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...
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...
© 2005 Paul Dawkins
Calculus Cheat Sheet
Absolute Extrema
1
...
Extrema
Relative (local) Extrema
1
...
2
...
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 )
...
a £ c, d £ b , 2
...
max
...
f ( d ) is the abs
...
in [ a, b]
...
1
...
2
...
3
...
4
...
max
...
min
...
2
...
1st Derivative Test
If x = c is a critical point of f ( x ) then x = c is
1
...
max
...
2
...
min
...
3
...
2nd Derivative Test
If x = c is a critical point of f ( x ) such that
f ¢ ( c ) = 0 then x = c
1
...
2
...
3
...
Finding Relative Extrema and/or
Classify Critical Points
1
...
2
...
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 (b) - f ( a )
...
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...
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...
© 2005 Paul Dawkins
Calculus Cheat Sheet
Related Rates
Sketch picture and identify known/unknown quantities
...
e
...
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
...
01 rad/min
...
How fast
4
between them changing when q = 0
...
Using
Pythagorean Theorem and differentiating,
x 2 + y 2 = 152 Þ 2 x x¢ + 2 y y¢ = 0
After 12 sec we have x = 10 - 12 ( 1 ) = 7 and
4
so y = 152 - 7 2 = 176
...
7
7 ( - 1 ) + 176 y¢ = 0 Þ y¢ =
ft/sec
4
4 176
We have q ¢ = 0
...
and want to find
x¢
...
05 so plug in q ¢ and solve
...
5 ) tan ( 0
...
01) =
50
x¢ = 0
...
Solve constraint for
one of the two variables and plug into first equation
...
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)
...
Determine dimensions that
will maximize the enclosed area
...
Solve constraint for x and plug
into area
...
A¢ = 500 - 4 y Þ y = 125
nd
By 2 deriv
...
max
...
Finally, find x
...
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...
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...
2
constraint is y = x 2 + 1
...
2
x2 = y - 1 Þ f = x2 + ( y - 2)
= y -1 + ( y - 2) = y 2 - 3 y + 3
Differentiate and find critical point(s)
...
min
...
x 2 = 3 - 1 = 1 Þ x = ± 12
2
2
2
The 2 points are then
(
1
2
)
(
, 3 and 2
1
2
)
,3
...
Divide [ a, b ] into n subintervals of
is a function, F ( x ) , such that F ¢ ( x ) = f ( x )
...
Indefinite Integral : ò f ( x ) dx = F ( x ) + c
*
i
Then
¥
where F ( x ) is an anti-derivative of f ( x )
...
i
b
®¥
=1
*
i
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 éu ( x ) ù
ë
û
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 év ( x ) ù
and g ¢ ( x ) =
f ( t ) dt = f ( x )
...
e
...
a
ò f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx
b
b
b
f ( x ) ± g ( x ) dx = ò f ( x ) dx ± ò g ( x ) dx
òa
a
a
a
Properties
ò cf ( x ) dx = c ò f ( x ) dx , c is a constant
b
b
cf ( x ) dx = c ò f ( x ) dx , c is a constant
òa
a
b
f ( x ) dx = 0
òa
b
òa
a
ò a f ( x ) dx = -òb f ( x ) dx
a
b
b
a
a
ò f ( x ) dx £ ò
If f ( x ) ³ g ( x ) on a £ x £ b then
If f ( x ) ³ 0 on a £ x £ b then
b
f ( x ) dx = ò f ( t ) dt
b
a
a
f ( x ) dx
b
ò f ( x ) dx ³ ò g ( x ) dx
b
ò f ( x ) dx ³ 0
a
b
If m £ f ( x ) £ M on a £ x £ b then m ( b - a ) £ ò f ( x ) dx £ M ( b - a )
a
ò k dx = k x + c
n
n
1
ò x dx = n+1 x + c, n ¹ -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
+1
-1
1
1
u
1
u
Common Integrals
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 = sec u + 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
u
1
ò a - u du = sin ( a ) + c
1
2
1
2
-1
-1
2
2
2
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...
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...
© 2005 Paul Dawkins
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
...
For indefinite integrals drop the limits of integration
...
2
ò 1 5x
2
cos ( x3 ) dx
2
ò 1 5x
2
cos ( x3 ) dx = ò
u = x 3 Þ du = 3x 2 dx Þ x 2 dx = 1 du
3
5
cos
1 3
8
= 5 sin ( u ) 1 =
3
x = 1 Þ u = 1 = 1 :: x = 2 Þ u = 2 = 8
3
8
3
Integration by Parts : ò u dv = uv - ò v du and
b
ò a u dv = uv
b
a
5
3
( u ) du
( sin (8) - sin (1) )
b
- ò v du
...
Ex
...
dv = e- x Þ
du = dx v = -e - x
dx = - xe + ò e dx = - xe - e
-x
-x
-x
-x
+c
5
ò3 ln x dx
u = ln x
5
ò3
dv = dx Þ du = 1 dx v = x
x
ln x dx = x ln x 3 - ò dx = ( x ln ( x ) - x )
5
5
3
5
3
= 5ln ( 5) - 3ln ( 3) - 2
Products and (some) Quotients of Trig Functions
For ò sin n x cos m x dx we have the following :
For ò tan n x sec m x dx we have the following :
1
...
Strip 1 sine out and convert rest to
1
...
2
...
Strip 1 cosine out and convert rest
2
...
3
...
Use either 1
...
4
...
Use double angle
3
...
integral into a form that can be integrated
...
ò tan 3 x sec5 x dx
ò tan
3
sin5 x
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
= - ò (1-u ) du = - ò 1-2u +u du
u
u
Ex
...
Strip 1 tangent and 1 secant out and
convert the rest to secants using
tan 2 x = sec 2 x - 1 , then use the substitution
u = sec x
...
Strip 2 secants out and convert rest
to tangents using sec2 x = 1 + tan 2 x , then
use the substitution u = tan x
...
Use either 1
...
n even and m odd
...
2
1
1
2 (1 + cos ( 2 x ) ) , sin ( x ) = 2 (1 - cos ( 2 x ) )
( u = sec x )
= 1 sec7 x - 1 sec5 x + c
7
5
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...
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...
3
5
3
2
4
3
2
3
2
2
3
2 2
2
3
3
4
= 1 sec2 x + 2 ln cos x - 1 cos 2 x + c
2
2
© 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
...
òx
16
2
4 -9 x 2
tan 2 q = sec 2 q - 1
ó
õ
dx
x = 2 sin q Þ dx = 2 cos q dq
3
3
sec2 q = 1 + tan 2 q
16
4 sin 2 q ( 2cosq )
9
Recall x 2 = x
...
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
Use Right Triangle Trig to go back to x’s
...
2
Partial Fractions : If integrating
12
( 2 cos q ) dq = ò sin2 q dq
3
= ò 12 csc 2 dq = -12 cot q + c
2
2
4 - 9x 2 = 4 - 4sin q = 4 cos q = 2 cos q
In this case we have
a 2 + b 2 x 2 Þ x = a tan q
b
16
2
4 -9 x 2
dx = - 4
4 -9 x 2
3x
4 -9 x 2
x
...
Factor denominator as completely as possible and find the partial fraction decomposition of
the rational expression
...
F
...
For each factor in the
denominator we get term(s) in the decomposition according to the following table
...
F
...
ò
ò
A
ax + b
Ax + B
2
ax + bx + c
( ax + b )
( ax
2
( x -1)( x
7 x2 +13 x
( x -1)( x2 + 4 )
2
Ak x + Bk
A1 x + B1
+L +
k
2
ax + bx + c
( ax 2 + bx + c )
k
7 x2 +13 x
dx
+4)
2
( x -1)( x + 4 )
dx = ò x4 1 + 3xx2+16 dx
+4
= ò x41 +
-
Ak
A1
A2
+
+L +
2
k
ax + b ( ax + b )
( ax + b )
k
+ bx + c )
7 x2 +13 x
Term in P
...
D
3x
x2 + 4
+
16
x2 + 4
dx
3
= 4 ln x - 1 + 2 ln ( x 2 + 4 ) + 8 tan -1 ( x )
2
Here is partial fraction form and recombined
...
7 x 2 + 13x = ( A + B ) x 2 + ( C - B ) x + 4 A - C
Set coefficients equal to get a system and solve
to get constants
...
Start with setting numerators equal in
previous example : 7 x 2 + 13x = A ( x 2 + 4 ) + ( Bx + C ) ( x - 1)
...
For example if x = 1 we get 20 = 5A which gives A = 4
...
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...
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...
© 2005 Paul Dawkins
Calculus Cheat Sheet
Applications of Integrals
Net Area :
b
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
...
Here are some
sketches of a couple possible situations and formulas for a couple of possible cases
...
Here is
some general information about each method of computing and some examples
...
to x/y of outer cyl
...
Axis use f ( x ) ,
Vert
...
Axis use f ( y ) ,
Vert
...
g ( y ) , A ( y ) and dy
...
g ( x ) , A ( x ) and dx
...
Axis : y = a > 0
Ex
...
Axis : y = a > 0
Ex
...
If axis of rotation is the x-axis use the
y = a £ 0 case with a = 0
...
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© 2005 Paul Dawkins
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 = ò F ( x ) dx
a
b
1
b-a a
ò f ( x ) dx
Arc Length Surface Area : Note that this is often a Calc II topic
...
( )
1+ ( )
ds = 1 +
dy
dx
ds =
dx
dy
( dx )
dt
( )
2
dx if y = f ( x ) , a £ x £ b
ds =
2
dy if x = f ( y ) , a £ y £ b
dr
ds = r 2 + ( dq ) dq if r = f (q ) , a £ q £ b
2
+
dy
dt
2
dt if x = f ( t ) , y = g ( t ) , a £ t £ b
2
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
...
This is typically a Calc II topic
...
3
...
a
¥
c
¥
-
-
c
ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò
b
f ( x ) dx = lim
ò¥
-
t ®-¥
b
ò f ( x ) dx
t
f ( x ) dx provided BOTH integrals are convergent
...
Discont
...
Discontinuity at a < c < b :
t
2
...
at b : ò f ( x ) dx = lim ò f ( x ) dx
-
t
a
b
c
a
a
t ®b
a
b
c
ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent
...
If ò g ( x ) dx divg
...
a
¥
¥
a
1
...
then ò g ( x ) dx conv
...
Approximating Definite Integrals
For given integral
b
n
ò a f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b-a
and
divide [ a, b] into n subintervals [ x0 , x1 ] , [ x1 , x2 ] , … , [ xn -1 , xn ] with x0 = a and xn = b then,
Midpoint Rule :
Trapezoid Rule :
Simpson’s Rule :
ò f ( x ) dx » Dx é f ( x ) + f ( x ) + L + f ( x )ù , xi
ë
û
b
*
1
a
*
2
*
n
*
is midpoint [ xi -1 , xi ]
Dx
ë
û
ò f ( x ) dx » 2 é f ( x ) + 2 f ( x ) + +2 f ( x ) + L + 2 f ( x ) + f ( x )ù
b
a
0
1
2
n -1
n
Dx
ë
û
ò f ( x ) dx » 3 é f ( x ) + 4 f ( x ) + 2 f ( x ) + L + 2 f ( x ) + 4 f ( x ) + f ( x )ù
b
a
0
1
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...
2
n-2
n -1
n
© 2005 Paul Dawkins