Quadratic Formula
Given: ax 2  bx  c  0
 b  b 2  4ac
Solve for x: x 
2a
Square of a Sum
(a  b) 2  a 2  2ab  b 2 and
(a  b  c) 2  a 2  b 2  c 2  2ab  2bc  2ac
Series Formulas
Geometric Series: a, ar, ar2, ar3, …  a + ar + ar2+ ar3 + …
Sum of first n terms (note: First n terms starts at r0 and goes to rn-1)
1 rn
a + ar + ar2+ ar3 + … + arn-1= a(1 + r + r2+ r3 + … + rn-1)= a 
1 r
Infinite Convergent Series (-1 < r < 1)
1
a + ar + ar2+ ar3 + … = a(1 + r + r2+ r3 + …)= a 
1 r
Arithmetic Series: a, a+d, a+2d, a+3d3, …  a + (a+d) + (a+2d) + (a+3d) + …
Sum of first n terms (note: First n terms starts at 0d and goes to (n-1)d)
n(n  1)
a + (a+d) + (a+2d) + (a+3d) + … + (a+(n-1)d) = na  d 
2
Special Case for first n integers
n(n  1)
1+2+3+…+n=
2
Definition of ea

an
 ea

n 0 n!
Also – dividing both sides by ea gives
n

a a
e
 1 (This is the pdf for the Poisson distribution)

n!
n 0
Limit Theorems: Let n be a positive integer, k be a constant, and f and g be functions
which have limits at c then:
lim k  k
x c
lim x  c
x c
lim kf ( x)  k lim f ( x)
x c
x c
lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)
x c
x c
x c
x c
x c
x c
x c
x c
lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)
lim  f ( x)  g ( x)  lim f ( x)  lim g ( x)
x c
lim  f ( x)  g ( x)  lim f ( x)  lim g ( x) Provided lim g ( x)  0
x c

x c
lim  f ( x)  lim f ( x)
n
x c
lim
x c
x c
n

x c
x c
n
f ( x)  n lim f ( x) Provided lim f ( x)  0 when n is even
x c
x c
Substitution Theorem: if f is a polynomial or rational function, and if rational
denominator at c ≠ 0
lim f ( x)  f (c)
x c
Derivative
 f (c  h )  f (c ) 
f ' (c)  lim 
 Provided this limit exists
h 0
h

Alternate Form
 f ( x )  f (c ) 
f ' (c)  lim 

x c
xc

Rules of Differentiation
f (x)
Power Rule
f ' ( x)
c (a constant)
0
cx n n  
cnx n 1
g ( x )  h( x )
g ' ( x )  h' ( x )
Product Rule g ( x)  h( x)
g ' ( x )  h( x )  g ( x )  h' ( x )
u ( x)v( x) w( x)
u' vw  uv' w  uvw'
Quotient Rule
g ( x)
h( x )
h( x ) g ' ( x )  g ( x ) h ' ( x )
Chain Rule
g (h( x))
g ' (h( x))  h' ( x)
e g ( x)
g ' ( x)  e g ( x )
ln( g ( x))
g ' ( x)
g ( x)
a x a  0
a x ln( a)
ex
ex
ln x
1
x
log b x
1
x  ln( b)
sin x
cos x
cos x
 sin x
cf (x) 
cf (x)
 f ( x) 
p f ( x) 
BPP Power Rule
h( x)2
p
p 1
 f ( x)
L’Hopital’s Rule
If lim f ( x)  0 & lim g ( x)  0 or lim f ( x)   & lim g ( x)  
x c
x c
x c
x c
Then
 f ( x) 
 f ( x) 
Provided lim g ' ( x)  0
lim 
 lim 
x c
x c g ( x ) 
x c g ( x ) 




Note: Can be applied multiple times.
Antiderivatives (Integrals) of Some Frequently Used Functions
f (x)
 f ( x)dx
g ( x )  h( x )
 g ( x)dx   h( x)dx  c
x n  1
x n 1
c
n 1
1
x
ln( x )  c
ex
ex  c
a x a  0
ax
c
ln a
xe ax
xeax e ax
 2 c
a
a
n
e  ax or
1
e ax
1  ax
1
e or ax
a
ae
sin x
 cos x  c
cos x
sin x  c
Additional integration rules:
For integer n  0 and real number c  0


0
x n e cx dx 
n!
c n 1
Integration by Substitution
To find ∫f(x)dx substitute u=g(x), g being differentiable then du =g′(x)dx then we
can rewrite ∫f(x)dx as an integral with respect to the variable u. So if this changes
f(x)dx to h(u)du and H is the antiderivative of h, then
 f ( x)dx   h(u)du  H (u)  c  H ( g ( x))  c
Example: Find ∫(x3-1)4/3x2dx – Substitue u= x3-1, so that du=3x2dx or ⅓du= x2dx
7
7
1 u 3 1 3
 u (c) now
now the integral can be written as ∫u ∙⅓du = ⅓∫u du = 
3 73 7
substitute back the u= x3-1 gives the final result ∫(x3-1)4/3x2dx = 1/7(x3-1)7/3(+c)
4/3
4/3
Integration by Parts
 f ( x)  g ( x)dx then do integration by parts
If you have an integral of the form
 f ( x)  g ( x)dx  f ( x) g ( x)   f ( x)  g ( x)dx
Use when
 f ( x)  g ( x)dx is easier than  f ( x)  g ( x)dx
For a definite integral

b
a
f ( x)  g ( x)dx   f ( x) g ( x)a   f ( x)  g ( x)dx
b
b
a