Calculus Cheat Sheet All

1. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Limits
Definitions
Precise Definition : We say ()lim
xa
fxL
®
= if
for every 0e> there is a 0d>such that
whenever 0xad<-< then ()fxL e-< .

“Working” Definition : We say ()lim
xa
fxL
®
=
if we can make ()fx as close to L as we want
by taking x sufficiently close to a (on either side
of a) without letting xa=.

Right hand limit : ()lim
xa
fxL
+
®
=. This has
the same definition as the limit except it
requires xa>.

Left hand limit : ()lim
xa
fxL
-
®
=. This has the
same definition as the limit except it requires
xa<.
Limit at Infinity : We say ()lim
x
fxL
®¥
= if we
can make ()fx as close to L as we want by
taking x large enough and positive.

There is a similar definition for ()lim
x
fxL
®-¥
=
except we require x large and negative.

Infinite Limit : We say ()lim
xa
fx
®
=¥ if we
can make ()fx arbitrarily large (and positive)
by taking x sufficiently close to a (on either side
of a) without letting xa=.

There is a similar definition for ()lim
xa
fx
®
=-¥
except we make ()fx arbitrarily large and
negative.
Relationship between the limit and one-sided limits
()lim
xa
fxL
®
= Þ () ()limlim
xaxa
fxfxL
+-
®®
== () ()limlim
xaxa
fxfxL
+-
®®
== Þ ()lim
xa
fxL
®
=
() ()limlim
xaxa
fxfx
+-
®®
¹ Þ ()lim
xa
fx
®
Does Not Exist

Properties
Assume ()lim
xa
fx
®
and ()lim
xa
gx
®
both exist and c is any number then,
1. () ()limlim
xaxa
cfxcfx
®®
=éù
ëû


2. ()() () ()limlimlim
xaxaxa
fxgxfxgx
®®®
±=±éù
ëû


3. ()() () ()limlimlim
xaxaxa
fxgxfxgx
®®®
=éù
ëû

4.
()
()
()
()
lim
lim
lim
xa
xa
xa
fxfx
gxgx
®
®
®
éù
=êú
ëû
provided ()lim0
xa
gx
®
¹
5. () ()limlim
n
n
xaxa
fxfx
®®
éù
=éù
ëû ëû

6. () ()limlim
n
n
xaxa
fxfx
®®
éù =
ëû
Basic Limit Evaluations at ±¥
Note : ()sgn1a= if 0a> and ()sgn1a=- if 0a<.
1. lim
x
x®¥
=¥e & lim0
x
x®-¥
=e
2. ()limln
x
x
®¥
=¥ & ()
0
limln
x
x
-
®
=-¥
3. If 0r>thenlim0
r
x
b
x
®¥
=
4. If 0r> and
r
xis real for negative x
thenlim0
r
x
b
x
®-¥
=
5. n even : lim
n
x
x
®±¥
=¥
6. n odd : lim
n
x
x
®¥
=¥ & lim
n
x
x
®-¥
=-¥
7. n even : ()limsgn
n
x
axbxca
®±¥
+++=¥L
8. n odd : ()limsgn
n
x
axbxca
®¥
+++=¥L
9. n odd : ()limsgn
n
x
axcxda
®-¥
+++=-¥L

2. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Evaluation Techniques
Continuous Functions
If()fxis continuous at a then ()()lim
xa
fxfa
®
=

Continuous Functions and Composition
()fx is continuous at b and ()lim
xa
gxb
®
= then
()( ) ()( )()limlim
xaxa
fgxfgxfb
®®
==
Factor and Cancel
()()
()
2
2
22
2
26412
limlim
22
68
lim4
2
xx
x
xxxx
xxxx
x
x
®®
®
-++-
=
--
+
===

Rationalize Numerator/Denominator
( )( ) ()( )
()()
22
99
299
333
limlim
8181 3
91
limlim
81393
11
186108
xx
xx
xxx
xx x
x
xxxx
®®
®®
--+
=
-- +
--
==
-+++
-
==-

Combine Rational Expressions
()
()
() ()
00
2
00
1111
limlim
111
limlim
hh
hh
xxh
hxhxhxxh
h
hxxhxxhx
®®
®®
æö-+æö
-= ç÷ç÷ ç÷
++èø èø
æö--
===- ç÷
ç÷
++
èø

L’Hospital’s Rule
If
()
()
0
lim
0
xa
fx
gx
®
= or
()
()
lim
xa
fx
gx
®
±¥
=
±¥
then,
()
()
()
()
limlim
xaxa
fxfx
gxgx
®®
¢
=
¢
a is a number, ¥ or -¥
Polynomials at Infinity
()px and ()qx are polynomials. To compute
()
()
lim
x
px
qx
®±¥
factor largest power of x out of both
()px and ()qx and then compute limit.
( )
()
2
2
2 2
2 2
4 4
55
3 3
343
limlimlim
5222 2
xxx
xx
x x
x
x
xx x
®-¥®-¥®-¥
- -
-
===-
-- -
Piecewise Function
()
2
lim
x
gx
®-
where ()
2
5if 2
13if 2
xx
gx
xx
ì+<-
=í
-³-î

Compute two one sided limits,
()
2
22
limlim59
xx
gxx
--
®-®-
=+=
()
22
limlim137
xx
gxx
++
®-®-
=-=
One sided limits are different so ()
2
lim
x
gx
®-

doesn’t exist. If the two one sided limits had
been equal then ()
2
lim
x
gx
®-
would have existed
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.
2. Rational function, except for x’s that give
division by zero.
3.
n
x(n odd) for all x.
4.
n
x(n even) for all 0x³.
5.
x
e for all x.
6. lnx for 0x>.
7. ()cosx and ()sinx for all x.
8. ()tanx and ()secx provided
33
,,,,,
2222
x
pppp
¹--LL
9. ()cotx and ()cscx provided
,2,,0,,2,x pppp¹--LL

Intermediate Value Theorem
Suppose that ()fx is continuous on [ a, b] and let M be any number between ()fa and ()fb.
Then there exists a number c such that acb<< and ()fcM=.

3. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Derivatives
Definition and Notation
If ()yfx= then the derivative is defined to be ()
()()
0
lim
h
fxhfx
fx
h
®
+-
¢= .

If ()yfx= then all of the following are
equivalent notations for the derivative.
() ()( ) ()
dfdyd
fxyfxDfx
dxdxdx
¢¢=====
If ()yfx= all of the following are equivalent
notations for derivative evaluated at xa=.
() ()
xa
xaxa
dfdy
fayDfa
dxdx
=
==
¢¢====

Interpretation of the Derivative
If ()yfx= then,
1. ()mfa¢= is the slope of the tangent
line to ()yfx= at xa=and the
equation of the tangent line at xa= is
given by ()()()yfafaxa ¢=+- .
2. ()fa¢ is the instantaneous rate of
change of ()fx at xa=.
3. If ()fx is the position of an object at
time x then ()fa¢ is the velocity of
the object at xa=.

Basic Properties and Formulas
If ()fx and ()gx are differentiable functions (the derivative exists), c and n are any real numbers,
1. () ()cfcfx
¢
¢=
2. ( ) () ()fgfxgx
¢
¢¢±=±
3. ()fgfgfg
¢
¢¢=+ – Product Rule
4.
2
ffgfg
gg
¢
¢¢æö -
=
ç÷
èø
– Quotient Rule
5. ()0
d
c
dx
=
6. ()
1nnd
xnx
dx
-
= – Power Rule
7. ()( )( ) ()( )()
d
fgxfgxgx
dx
¢¢=
This is the Chain Rule

Common Derivatives
()1
d
x
dx
=
()sincos
d
xx
dx
=
()cossin
d
xx
dx
=-
()
2
tansec
d
xx
dx
=
()secsectan
d
xxx
dx
=

()csccsccot
d
xxx
dx
=-
()
2
cotcsc
d
xx
dx
=-
( )
1
2
1
sin
1
d
x
dx x
-
=
-

( )
1
2
1
cos
1
d
x
dx x
-
=-
-

( )
1
2
1
tan
1
d
x
dxx
-
=
+

() ()ln
xxd
aaa
dx
=
()
xxd
dx
=ee
()( )
1
ln,0
d
xx
dxx
=>
()
1
ln,0
d
xx
dxx
=¹
()( )
1
log,0
ln
a
d
xx
dxxa
=>

4. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Chain Rule Variants
The chain rule applied to some specific functions.
1. ()( ) () ()
1nnd
fxnfxfx
dx
-
¢=éùéù
ëûëû

2.
()
()()
()fxfxd
fx
dx
¢=ee
3. ()( )
()
()
ln
fxd
fx
dxfx
¢
=éù
ëû

4. ()( )() ()sincos
d
fxfxfx
dx
¢=éùéù
ëûëû

5. ()( ) () ()cossin
d
fxfxfx
dx
¢=-éùéù
ëûëû

6. ()( )() ()
2
tansec
d
fxfxfx
dx
¢=éùéù
ëûëû

7. []( ) [][]()()()()secsectanfxfxfxfx
d
dx
¢=
8. ()( )
()
()
1
2
tan
1
fxd
fx
dx fx
-
¢
=éù
ëû
+éù
ëû


Higher Order Derivatives
The Second Derivative is denoted as
()
()
()
2
2
2
df
fxfx
dx
¢¢== and is defined as
() ()( )fxfx
¢
¢¢¢= , i.e. the derivative of the
first derivative, ()fx¢.

The n
th
Derivative is denoted as
()
()
n
n
n
df
fx
dx
= and is defined as
()
()
()
()( )
1nn
fxfx
- ¢
= , i.e. the derivative of
the (n-1)
st
derivative,
()
()
1n
fx
-
.
Implicit Differentiation
Find y¢ if ()
2932
sin11
xy
xyyx
-
+=+e . Remember ()yyx= 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¢.

( ) ()
()
()( )
()
29223
2922
2929223
329
3292922
2932cos11
1123
2932cos11
29cos
29cos1123
xy
xy
xyxy
xy
xyxy
yxyxyyyy
xy
yxyxyyyyy
xyy
xyyyxy
-
-
--
-
--
¢¢¢-++=+
--
¢¢¢¢-++=+Þ=
--
¢--=--
e
e
ee
e
ee


Increasing/Decreasing – Concave Up/Concave Down
Critical Points
xc= is a critical point of ()fx provided either
1. ()0fc¢= or 2. ()fc¢ doesn’t exist.

Increasing/Decreasing
1. If ()0fx¢> for all x in an interval I then
()fx is increasing on the interval I.
2. If ()0fx¢< for all x in an interval I then
()fx is decreasing on the interval I.
3. If ()0fx¢= for all x in an interval I then
()fx is constant on the interval I.

Concave Up/Concave Down
1. If ()0fx¢¢> for all x in an interval I then
()fx is concave up on the interval I.
2. If ()0fx¢¢< for all x in an interval I then
()fx is concave down on the interval I.

Inflection Points
xc= is a inflection point of ()fx if the
concavity changes at xc=.

5. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Extrema
Absolute Extrema
1. xc=is an absolute maximum of ()fx
if()()fcfx³ for all x in the domain.
2. xc= is an absolute minimum of ()fx
if()()fcfx£ for all x in the domain.

Fermat’s Theorem
If ()fx has a relative (or local) extrema at
xc=, then xc= is a critical point of ()fx.

Extreme Value Theorem
If ()fx is continuous on the closed interval
[],ab then there exist numbers c and d so that,
1. ,acdb££ , 2. ()fc is the abs. max. in
[],ab, 3. ()fd is the abs. min. in [],ab.

Finding Absolute Extrema
To find the absolute extrema of the continuous
function ()fx on the interval [],ab use the
following process.
1. Find all critical points of ()fx in [],ab.
2. Evaluate ()fx at all points found in Step 1.
3. Evaluate ()fa and ()fb.
4. Identify the abs. max. (largest function
value) and the abs. min.(smallest function
value) from the evaluations in Steps 2 & 3.

Relative (local) Extrema
1. xc= is a relative (or local) maximum of
()fx if()()fcfx³ for all x near c.
2. xc= is a relative (or local) minimum of
()fx if()()fcfx£ for all x near c.

1
st
Derivative Test
If xc= is a critical point of ()fx then xc= is
1. a rel. max. of ()fx if()0fx¢> to the left
of xc= and()0fx¢< to the right of xc=.
2. a rel. min. of ()fx if()0fx¢< to the left
ofxc=and()0fx¢>to the right of xc=.
3. not a relative extrema of ()fx if()fx¢ is
the same sign on both sides of xc=.

2
nd
Derivative Test
If xc= is a critical point of ()fx such that
()0fc¢= then xc=
1. is a relative maximum of ()fx if()0fc¢¢<.
2. is a relative minimum of ()fx if()0fc¢¢>.
3. may be a relative maximum, relative
minimum, or neither if ()0fc¢¢=.

Finding Relative Extrema and/or
Classify Critical Points
1. Find all critical points of ()fx.
2. Use the 1
st
derivative test or the 2
nd

derivative test on each critical point.

Mean Value Theorem
If ()fx is continuous on the closed interval [],ab and differentiable on the open interval (),ab
then there is a number acb<< such that ()
()()fbfa
fc
ba
-
¢=
-
.

Newton’s Method
If
n
x is the n
th
guess for the root/solution of ()0fx= then (n+1)
st
guess is
()
()
1
n
nn
n
fx
xx
fx
+
=-
¢

provided ()
n
fx¢ exists.

6. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
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.
The bottom is initially 10 ft away and is being
pushed towards the wall at
1
4
ft/sec. How fast
is the top moving after 12 sec?

x¢is negative because x is decreasing. Using
Pythagorean Theorem and differentiating,
222
15220xyxxyy ¢¢+=Þ+=
After 12 sec we have ()
1
4
10127x=-= and
so
22
157176y=-= . Plug in and solve
for y¢.
()
1
4
7
71760 ft/sec
4176
yy¢¢-+=Þ=
Ex. Two people are 50 ft apart when one
starts walking north. The angle q changes at
0.01 rad/min. At what rate is the distance
between them changing when 0.5q= rad?

We have 0.01q¢= rad/min. and want to find
x¢. We can use various trig fcns but easiest is,
secsectan
5050
xx
qqqq
¢
¢=Þ=
We know 0.05q= so plug in q¢ and solve.
()()()sec0.5tan0.50.01
50
0.3112 ft/sec
x
x
¢
=
¢=

Remember to have calculator in radians!

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
500 ft of fence material and one side of the
field is a building. Determine dimensions that
will maximize the enclosed area.

Maximize Axy= subject to constraint of
2500xy+= . Solve constraint for x and plug
into area.
( )
2
5002
5002
5002
Ayy
xy
yy
=-
=-Þ
=-

Differentiate and find critical point(s).
5004125Ayy¢=-Þ=
By 2
nd
deriv. test this is a rel. max. and so is
the answer we’re after. Finally, find x.
()5002125250x=-=
The dimensions are then 250 x 125.
Ex. Determine point(s) on
2
1yx=+ that are
closest to (0,2).

Minimize ()()
22
2
02fdxy==-+- and the
constraint is
2
1yx=+ . Solve constraint for
2
x and plug into the function.
()
()
2
22
2
2
12
1233
xyfxy
yyyy
=-Þ=+-
=-+-=-+
Differentiate and find critical point(s).
3
2
23fyy¢=-Þ=
By the 2
nd
derivative test this is a rel. min. and
so all we need to do is find x value(s).
2 3 11
22 2
1xx=-=Þ=±
The 2 points are then ()
31
22
, and ( )
31
22
,- .

7. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Integrals
Definitions
Definite Integral: Suppose ()fx is continuous
on [],ab. Divide [],ab into n subintervals of
width xD and choose
*
i
x from each interval.
Then () ()
*
1
lim
i
b
a n
i
fxdxfxx
®¥
=
¥
=Dåò
.
Anti-Derivative : An anti-derivative of ()fx
is a function, ()Fx, such that ()()Fxfx¢= .
Indefinite Integral : () ()fxdxFxc =+ò

where ()Fx is an anti-derivative of ()fx.

Fundamental Theorem of Calculus
Part I : If ()fx is continuous on [],ab then
() ()
x
a
gxftdt=ò
is also continuous on [],ab
and () () ()
x
a
d
gxftdtfx
dx
¢==ò
.
Part II : ()fxis continuous on [],ab, ()Fx is
an anti-derivative of ()fx(i.e. () ()Fxfxdx=ò
)
then () () ()
b
a
fxdxFbFa =-ò
.
Variants of Part I :
()
()
() ()
ux
a
d
ftdtuxfux
dx
¢= éù
ëûò

()
()
() ()
b
vx
d
ftdtvxfvx
dx
¢=- éù
ëûò

()
()
()
()[]()[]()()
ux
vx
uxvx
d
ftdtuxfvxf
dx
¢¢=-ò


Properties
()() () ()fxgxdxfxdxgxdx±=±òòò

()() () ()
bbb
aaa
fxgxdxfxdxgxdx±=±òòò

() 0
a
a
fxdx =ò

() ()
ba
ab
fxdxfxdx =-òò

() ()cfxdxcfxdx =òò
, c is a constant
() ()
bb
aa
cfxdxcfxdx =òò
, c is a constant
() ()
bb
aa
fxdxftdt =òò

() ()
bb
aa
fxdxfxdx £òò
If ()()fxgx³ onaxb££then () ()
ba
ab
fxdxgxdx ³òò

If()0fx³ on axb££ then () 0
b
a
fxdx ³ò

If ()mfxM££ on axb££ then () () ()
b
a
mbafxdxMba-££-ò


Common Integrals
kdxkxc=+ò

11
1
,1
nn
n
xdxxcn
+
+
=+¹-ò

1 1
ln
x
xdxdxxc
-
==+òò

11
ln
a
axb
dxaxbc
+
=++ò

()lnlnuduuuuc=-+ò

uu
duc=+ò
ee
cossinuduuc=+ò

sincosuduuc=-+ò

2
sectanuduuc=+ò

sectansecuuduuc =+ò

csccotcscuuduuc =-+ò

2
csccotuduuc=-+ò

tanlnsecuduuc=+ò

seclnsectanuduuuc=++ò

()
11 1
22
tan
u
aa
au
duc
-
+
=+ò

()
1
22
1
sin
u
a
au
duc
-
-
=+ò

8. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Standard Integration Techniques
Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class.

u Substitution : The substitution ()ugx= will convert ()( )() ()
()
()bgb
aga
fgxgxdxfudu ¢ =òò
using
()dugxdx¢= . For indefinite integrals drop the limits of integration.
Ex. ()
23
2
1
5cosxxdxò

322 1
3
3uxduxdxxdxdu=Þ=Þ=
33
111::228xuxu=Þ===Þ==
() ()
() () ()( )
23
28
5
3
11
8
55
33 1
5coscos
sinsin8sin1
xxdxudu
u
=
==-
òò


Integration by Parts : udvuvvdu=-òò
and
bb b
aaa
udvuvvdu=-òò
. Choose u and dv from
integral and compute du by differentiating u and compute v using vdv=ò
.
Ex.
x
xdx
-
ò
e
xx
uxdvdudxv
--
==Þ==- ee
xxxxx
xdxxdxxc
-----
=-+=--+òò
eeeee
Ex.
5
3
lnxdxò

1
ln
x
uxdvdxdudxvx==Þ==
()( )
() ()
55 55
3 333
lnlnln
5ln53ln32
xdxxxdxxxx=-=-
=--
òò

Products and (some) Quotients of Trig Functions
For sincos
nm
xxdxò
we have the following :
1. n odd. Strip 1 sine out and convert rest to
cosines using
22
sin1cosxx=- , then use
the substitution cosux= .
2. m odd. Strip 1 cosine out and convert rest
to sines using
22
cos1sinxx=- , then use
the substitution sinux= .
3. n and m both odd. Use either 1. or 2.
4. n and m both even. Use double angle
and/or half angle formulas to reduce the
integral into a form that can be integrated.
Fortansec
nm
xxdxò
we have the following :
1. n odd. Strip 1 tangent and 1 secant out and
convert the rest to secants using
22
tansec1xx=- , then use the substitution
secux= .
2. m even. Strip 2 secants out and convert rest
to tangents using
22
sec1tanxx=+ , then
use the substitution tanux= .
3. n odd and m even. Use either 1. or 2.
4. n even and m odd. Each integral will be
dealt with differently.
Trig Formulas : () ()()sin22sincosxxx= , () ()( )
2 1
2
cos1cos2xx=+ , () ()( )
2 1
2
sin1cos2xx=-

Ex.
35
tansecxxdxò

( )
( ) ( )
3524
24
24
7511
75
tansectansectansec
sec1sectansec
1sec
secsec
xxdxxxxxdx
xxxxdx
uuduux
xxc
=
=-
=-=
=-+
òò
ò
ò

Ex.
5
3
sin
cos
x
x
dxò

( )
2211
22
2254
333
22
3
22
24
33
sin(sin)sinsinsin
coscoscos
sin(1cos)
cos
(1) 12
cos
sec2lncoscos
xxxxx
xxx
xx
x
u uu
uu
dxdxdx
dxux
dudu
xxxc
-
- -+
==
==
=-=-
=+-+
òòò
ò
òò

9. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 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.
222
sin
a
b
abxx q-Þ=
22
cos1sinqq=-
222
sec
a
b
bxax q-Þ=
22
tansec1qq=-
222
tan
a
b
abxx q+Þ=
22
sec1tanqq=+

Ex.
2 2
16
49x x
dx
-ò

22
33
sincosxdxdqqq=Þ=
222
44sin4cos2cos49x qqq=-==-
Recall
2
xx=. Because we have an indefinite
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,
if 0
if 0
xx
x
xx
³ì
=í
-<î

In this case we have
2
2cos49x q=- .
( )
( )
2
3
sin2cos
2
224
9
16 12
sin
cos
12csc12cot
dd
dc
qq q
qqq
qq
=
==-+
ó
õ ò
ò

Use Right Triangle Trig to go back to x’s. From
substitution we have
3
2
sin
x
q= so,

From this we see that
2
49
3
cot
x
x
q
-
= . So,
2
2 2
16 449
49
x
x
x x
dxc
-
-
=-+ò


Partial Fractions : If integrating
()
()
Px
Qx
dxò
where the degree of ()Px is smaller than the degree of
()Qx. 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 ()Qx Term in P.F.D Factor in ()Qx Term in P.F.D
axb+
A
axb+
( )
k
axb+
( ) ( )
12
2
k
k
AAA
axb axbaxb
+++
+ ++
L
2
axbxc++
2
AxB
axbxc
+
++
( )
2
k
axbxc++
( )
11
2
2
kk
k
AxBAxB
axbxc
axbxc
++
++
++
++
L

Ex.
2
()()
2
14
713
xx
xx
dx
-+
+
ò

( ) ()
2 2
22
()()
213
22
2
3164
114 4
3164
1 44
713
4ln1ln48tan
x
xxx x
x
x xx
xx
x
dxdx
dx
xx
-
+
--+ +
- ++
+
=+
=++
=-+++
òò
ò

Here is partial fraction form and recombined.
2
222
4)()()
()()()()
2 1
114414
(713 BxCx
xxxxxx
AxBxCAxx
++ +-
--++-+
++
=+=
Set numerators equal and collect like terms.
( )( )
22
7134xxABxCBxAC+=++-+-
Set coefficients equal to get a system and solve
to get constants.
71340
4316
ABCBAC
ABC
+=-=-=
===


An alternate method that sometimes works to find constants. Start with setting numerators equal in
previous example : ( )( )()
22
71341xxAxBxCx+=+++- . Chose nice values of x and plug in.
For example if 1x= we get 205A= which gives 4A=. This won’t always work easily.

10. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Applications of Integrals
Net Area : ()
b
a
fxdxò
represents the net area between ()fx 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,
() upper functionlower function
b
a
yfxAdx
éù éù
ëûëû=Þ=- ò
& () right functionleft function
d
c
xfyAdy
éù éù
ëûëû=Þ=- ò

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
Afxgxdx=-ò


() ()
d
c
Afygydy=-ò


()() () ()
cb
ac
Afxgxdxgxfxdx=-+-òò


Volumes of Revolution : The two main formulas are ()VAxdx=ò
and ()VAydy=ò
. Here is
some general information about each method of computing and some examples.
Rings Cylinders
( ) ( )( )
22
outer radiusinner radiusAp=- ( )( )radiuswidth / height2Ap=
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 ()fx,
()gx,()Ax and dx.
Vert. Axis use ()fy,
()gy,()Ay and dy.
Horz. Axis use ()fy,
()gy,()Ay and dy.
Vert. Axis use ()fx,
()gx,()Ax and dx.

Ex. Axis : 0ya=> Ex. Axis : 0ya=£ Ex. Axis : 0ya=> Ex. Axis : 0ya=£

outer radius : ()afx-
inner radius : ()agx-
outer radius: ()agx+
inner radius: ()afx+
radius :ay-
width : ()()fygy-
radius :ay+
width : ()()fygy-

These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the
0ya=£ case with 0a=. For vertical axis of rotation ( 0xa=> and 0xa=£) interchange x and
y to get appropriate formulas.

11. Calculus Cheat Sheet
Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins
Work : If a force of ()Fxmoves an object
inaxb££, the work done is ()
b
a
WFxdx=ò

Average Function Value : The average value
of ()fx on axb££ is ()
1
b
avg
aba
ffxdx
-
=ò


Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are,
b
a
Lds=ò
2
b
a
SAyds p=ò
(rotate about x-axis) 2
b
a
SAxds p=ò
(rotate about y-axis)
where ds is dependent upon the form of the function being worked with as follows.
() ()
2
1 if ,
dy
dx
dsdxyfxaxb=+=££
() ()
2
1 if ,
dx
dy
dsdyxfyayb=+=££
()() () ()
2
2
if ,,
dydx
dtdt
dsdtxftygtatb=+==££
() ()
2
2
if ,
dr
d
dsrdrfab
q
qqq=+=££
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
1. () ()lim
t
aa t
fxdxfxdx
®¥
¥
=òò
2. () ()lim
bb
tt
fxdxfxdx
- ®-¥¥
=òò
3. () () ()
c
c
fxdxfxdxfxdx
--
¥¥
¥¥
=+òòò
provided BOTH integrals are convergent.
Discontinuous Integrand
1. Discont. at a:() ()lim
bb
at ta
fxdxfxdx
+
®
=òò
2. Discont. at b :() ()lim
bt
aa tb
fxdxfxdx
-
®
=òò
3. Discontinuity at acb<< : () () ()
bcb
aac
fxdxfxdxfxdx =+òòò
provided both are convergent.

Comparison Test for Improper Integrals : If ()()0fxgx³³ on [),a¥ then,
1. If()
a
fxdx
¥
ò
conv. then ()
a
gxdx
¥
ò
conv. 2. If()
a
gxdx
¥
ò
divg. then ()
a
fxdx
¥
ò
divg.
Useful fact : If 0a> then
1
a
p
x
dx
¥
ò
converges if 1p> and diverges for 1p£.

Approximating Definite Integrals
For given integral ()
b
a
fxdxò
and a n (must be even for Simpson’s Rule) define
ba
n
x
-
D= and
divide [],ab into n subintervals []
01
,xx, []
12
,xx, … , [ ]
1
,
nn
xx
-
with
0
xa= and
n
xb= then,
Midpoint Rule : () ()() ()
***
12
b
n
a
fxdxxfxfxfx éù»D+++
ëûò
L ,
*
i
x is midpoint [ ]
1
,
ii
xx
-

Trapezoid Rule : () () () () ()()
0121
222
2
b
nn
a
x
fxdxfxfxfxfxfx
-
D
»++++++éù
ëûò
L
Simpson’s Rule : () () () () () ()()
01221
4224
3
b
nnn
a
x
fxdxfxfxfxfxfxfx
--
D
»++++++éù
ëûò
L