Calculus Formula and Improper Integrals:

Data Overview Z ∞
1
dx (9)
1 xp
Calculus - Period 1 This function is convergent for p > 1 and divergent
for p ≤ 1.

Differentiation and Integration Comparison Theorem:

If f (x) ≥ g(x) ≥ 0 for x ≥ a then:
Chain Rule: R∞ R∞
• If a f (x)dx is convergent, then a g(x)dx
(f (g(x)))0 = f 0 (g(x))g 0 (x) (1) is convergent.
R∞ R∞
• If a g(x)dx is divergent, then a f (x)dx is
Implicit Differentiation: divergent.
When applying implicit differentiation for a func-
tion y of x, every term with a y should, after
normal differentiation (often involving the product
rule), be multiplied by y 0 because of the chain rule. Complex Numbers
After that, the equation should be solved for y 0 .
Complex Number Notations:
Lineair Approximations:
i2 = −1 (10)
0
f (x) − f (a) ≈ f (a)(x − a) (2)
z = a + bi = r(cos θ + i sin θ) = reiθ (11)
a = r cos θ and b = r sin θ (12)
Mean Value Theorem: √
If f is continuous on [a, b] and differentiable on |z| = r = a2 + b2
(13)
(a, b), then there is a c in (a, b) such that: θ = arctan ab or θ = arctan ab + π

f (b) − f (a) = f 0 (c)(b − a) (3) eiθ = cos θ + i sin θ (14)

Integration: Complex Number Calculation:

Z b
f (x)dx = F (b) − F (a) (4) (a + bi) + (c + di) = (a + c) + (b + d)i
(15)
a (a + bi)(c + di) = (ac − bd) + (ad + bc)i
Where F is any antiderivative/primitive function
z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 ))
of f , that is, F 0 = f . (16)
r1 eiθ1 · r2 eiθ2 = r1 r2 ei(θ1 +θ2 )
Substitution Rule:
If u = g(x) then: Complex Conjugates:
Z Z
f (g(x))g 0 (x)dx = f (u)du (5) z = a + bi ⇒ z = a − bi (17)
z+w =z+w
Z b Z g(b)
zw = z w
f (g(x))g 0 (x)dx = f (u)du (6) (18)
a g(a) zn = zn
zz = |z|2
Integration By Parts:
Z Z
f (x)g (x)dx = f (x)g(x) − f 0 (x)g(x)dx (7)
0
Differential Equations
Z b Z b Separable Differential Equations:
b
f (x)g 0 (x)dx = [f (x)g(x)]a − f 0 (x)g(x)dx
a a dy
(8) Form : = y 0 = P (x)Q(y)
dx

1
 Z Z
1 number, the series is divergent. Be careful not to
Solution : dy = P (x)dx (19)
Q(y)
P
confuse the series an with the series an = s.

First-Order Differential Equations Monotonic Sequence Theorem

If a sequence is either increasing (an+1 > an for all
Form : y 0 + P (x)y = Q(x) n ≥ 1) or decreasing (an+1 < an for all n ≥ 1), it
is called a monotonic sequence. If there are c1 and
Let Υ(x) be any integral of P (x). Solution is: c2 such that c1 < an < c2 for all n ≥ 1, it is called
Z  bounded. Every bounded monotonic sequence is
−Υ(x) Υ(x)
y=e e Q(x)dx + C (20) convergent.

Test for divergence:

Homogeneous second-order linear differen- If limn→∞ an does not exist, or if limn→∞ 6= 0,
tial equations: then the series sn is divergent.

Form : ay 00 + by 0 + cy = 0 Integral test:

If f is a continuous positive decreasing function
• b2 − 4ac > 0: on [1, ∞) and an = f (n) for integer n, then the
Define r such that ar2 + br + c = 0. series
R ∞ sn is convergent if, and only if, the integral
1
f (x)dx is convergent.
Solution : y = c1 er1 x + c2 er2 x (21)
Comparison test:
• b2 − 4ac = 0: Suppose an and bn are series with positive terms
Define r such that ar2 + br + c = 0. and an ≤ bn for all n, then:
P P
Solution : y = c1 erx + c2 xerx (22) • If bn is convergent, then an is conver-
gent.
• b2 − 4ac < 0:
P P
√
• If an is divergent, then bn is divergent.
b 4ac−b2
Define α = − 2a and β = 2a . Solution:

y = eαx (c1 cos βx + c2 sin βx) (23) Limit comparison test:

Suppose an and bn are series with positive terms.
If limn→∞ abnn = c and 0 < c 6= ∞, then either both
Nonhomogeneous second-order linear differ- series are convergent or divergent.
ential equations:
Alternating series test:
Form : ay 00 + by 0 + cy = P (x) If the alternating series
∞
First solve ayc00 + byc0 + cyc = 0. Then use an auxil- X
(−1)n−1 an = a1 − a2 + a3 − a4 + a5 − a6 . . .
iary equation to find one solution yp for the given
n=1
differential equation. The solution is:
satisfies an+1 ≤ an for all n and limn→∞ an = 0,
y = yc + y p (24) then the series is convergent.

Absolute P convergence:
Calculus - Period 2 A series
P an is called absolutely convergent
P if the
series |an | is convergent. A series an is called
conditionally convergent if it is convergent but not
Testing Series
P
absolutely convergent. If a series an is abso-
lutely convergent, then it is convergent.
Convergence/Divergence:
Suppose Ratio test:
Pn a is a series of numbers a1 , a2 , . . ., and
sn = k=1 ak . A series sn converges if limn→∞ sn =  
• If limn→∞  aan+1  = L < 1, then the series
 
s exists as a real number. The limit s is then called P n

the sum of series a. If s doesn’t exist as a finite an is absolutely convergent.

2
  
• If limn→∞  aan+1  = L > 1, then the series The second way to represent functions as power
 
n
P
an is divergent. series goes as follows. Let f (n) (x) be the n’th
derivative of f (x). Supposing the function f (x)
has a power series (this sometimes still has to be
Root test: proven), the following function must be true:
p
• P
If limn→∞ n |an | = L < 1, then the series X∞
f (n) (a)
an is absolutely convergent. f (x) = (x − a)n (29)
p n=0
n!
• P
If limn→∞ n |an | = L > 1, then the series
an is divergent. This representation is called the Taylor series of
f (x) at a. For the special case that a = 0, it is
called the Maclaurin series.
Power Series Binomial series:
If k is any real number and |x| < 1, the power
Radius of convergence:
function representation of (1 + x)k is:
Power series are written as
∞ ∞  
X k
X k
f (x) = cn (x − a) n
(25) (1 + x) = xn (30)
n
n=0 n=0

where x is a variable and the cn ’s are constant co-  

k k(k − 1) . . . (k − n + 1)
efficients of the series. When tested for converges, where = (31)
n n!
there are only three possibilities:  
k
• The series converges only if x = a. (R = 0) for n ≥ 1, and = 1.
0
• The series converges for all x. (R = ∞)
• The series converges for |x − a| < R and di- Vectors
verges for |x − a| > R. For |x − a| = R other
means must point out whether convergence Notation:
or divergence occurs. A vector a is often written as:
The number R is called the radius of convergence,
and can often be found using the ratio test. a = ax i + ay j + az k (32)

Differentiation and integration: Where i, j and k are unit vectors.

Differentiation and integration of power functions
is possible in the interval (a − R, a + R), where the Vector length:
function does not diverge. It goes as follows: q
!0 |a| = a2x + a2y + a2z (33)
X∞ X∞
n
cn (x − a) = ncn (x − a)n−1 (26)
n=0 n=1 Vector addition and subtraction:
∞ ∞ n+1
(x − a)
Z X X
cn (x − a)n dx = cn (27) a + b = (ax + bx )i + (ay + by )j + (az + bz )k (34)
n=0 n=0
n+1
a − b = (ax − bx )i + (ay − by )j + (az − bz )k (35)
Representation of functions as power series:
The first way to represent functions as power series Dot product:
is simple, but doesn’t always work. To find the
representation of f (x), first find a function g(x) a · b = ax bx + ay by + az bz (36)
1
such that f (x) = axb 1−g(x) , where a and b are
constants. The power series is then equal to: a · a = |a|2 (37)
∞ a·b
X cos θ = (38)
f (x) = a · g(x)n+b (28) |a||b|
n=0

3
Cross product: Expressing acceleration in unit vectors:
 
i
 j k  a(t) = |v(t)|0 T(t) + κ|v(t)|2 N(t) (53)
a × b = ax ay az  (39)
 bx by bz  r0 (t) · r00 (t) |r0 (t) × r00 (t)|
a(t) = T(t) + N(t)
|r0 (t)| |r0 (t)|
a × b = (ay bz − az by )i+ (54)
(40)
+(az bx − ax bz )j + (ax by − ay bx )k
Calculus - Period 3
Vector Functions:
Functions of Multiple Variables
Notation:
Definitions:
r(t) = f (t)i + g(t)j + h(t)k (41) The domain D is the set (x, y) for which f (x, y)
exists. The range is the set of values z for which
there are x, y such that z = f (x, y). The level
Differentiation and integration: curves are the curves with equations f (x, y) = k
where k is a constant.
r0 (t) = f 0 (t)i + g 0 (t)j + h0 (t)k (42)
Checking for Limits:
R(t) = F (t)i + G(t)j + H(t)k + D (43)
If f (x, y) → L1 as (x, y) → (a, b) along a path
C1 and f (x, y) → L2 as (x, y) → (a, b) along a
Function dependant unit vectors: path C2 , where L1 6= L2 then lim(x,y)→(a,b) f (x, y)
does not exist. Also f is continuous at (a, b) if
r0 (t) lim(x,y)→(a,b) f (x, y) = f (a, b)
T(t) = (44)
|r0 (t)|
Partial Derivatives:
T0 (t)
N(t) = (45) The partial derivative of f with respect to x at
|T0 (t)| (a, b) is:
B(t) = T(t) × N(t) (46)
fx (a, b) = g 0 (a) where g(x) = f (x, b) (55)

Trajectory length: In words, to find fx , regard y as constant and dif-

p ferentiate f (x, y) with respect to x. fy is defined
ds(t) = (dx)2 + (dy)2 + (dz)2 = |r0 (t)|dt (47) similarly. If fxy and fyx are both continuous on D,
Z t then fxy = fyx .
s(t) = |r0 (t)|dt (48)
a Tangent Planes:
For points close to z0 = f (x0 , y0 ) the curve of
Trajectory velocity and acceleration: f (x, y) can be approximated by:

ds(t) z−z0 = fx (x0 , y0 )(x−x0 )+fy (x0 , y0 )(y−y0 ) (56)

|v(t)| = = |r0 (t)| (49)
dt
The plane described by this equation is the plane
a(t) = v0 (t) = r00 (t) (50) tangent to the curve of f (x, y) at (x0 , y0 ).

Trajectory curvature: Differentials:

  0
∂z ∂z
 == |T (t)|
 dT(t)  dz = fx (x, y)dx+fy (x, y)dy = dx+ dy (57)
κ(t) =  0
(51) ∂x ∂y
ds(t)  |r (t)|
If z = f (x, y), x = g(s, t) and y = h(s, t) then:
|r0 (t) × r00 (t)|
κ(t) = (52)
|r0 (t)|3 dz ∂z dx ∂z dy
= + (58)
ds ∂x ds ∂y ds

4
Directional Derivatives: If D2 is the region such that D2 = {(x, y)|a ≤ y ≤
The directional derivative of f at (x0 , y0 ) in the b, h1 (y) ≤ x ≤ h2 (y)}, then:
direction of a unit vector (meaning, |u| = 1) u = Z bZ
ZZ h2 (y)
ha, bi is:
f (x, y) dA = f (x, y) dx dy (65)
D2 a h1 (y)
Du f (x0 , y0 ) = fx (x, y)a + fy (x, y)b = ∇f · u (59)

grad f = ∇f = hfx (x, y), fy (x, y)i (60) Integrating over Polar Coordinates
The maximum value of Du f (x, y) is |∇f (x, y)| and y
occurs when the vector u = ha, bi has the same r2 = x2 + y 2 tan θ = (66)
x
direction as ∇f (x, y).
x = r cos θ y = r sin θ (67)
Local Maxima and Minima: If R is the polar rectangle such that R = {(r, θ)|0 ≤
If f has a local maximum or minimum at (a, b), a ≤ r ≤ b, α ≤ θ ≤ β} where 0 ≤ β − α ≤ 2π, then:
then fx (a, b) = 0 and fy (a, b) = 0. If fx (a, b) = 0
and fy (a, b) = 0 then (a, b) is a critical point. If ZZ Z βZ b
(a, b) is a critical point, then let D be defined as: f (x, y) dA = f (r cos θ, r sin θ) r dr dθ
R α a
2 (68)
D = D(a, b) = fxx (a, b)fyy (a, b) − (fxy (a, b)) If D is the polar rectangle such that D = {(r, θ)|0 ≤
(61) h1 (θ) ≤ r ≤ h2 (θ), α ≤ θ ≤ β} where 0 ≤ β − α ≤
• If D > 0 then: 2π, then:
– If fxx (a, b) > 0, then f (a, b) is a minimum. Z βZ
ZZ h2 (θ)
– If fxx (a, b) < 0, then f (a, b) is a maximum.
f (x, y) dA = f (r cos θ, r sin θ) r dr dθ
• If D < 0, then f (a, b) is a saddle point. R α h1 (θ)
(69)
Absolute Maxima and Minima:
To find the absolute maximum and minimum val- Applications:
ues of a continuous function f on a closed bounded If m is the mass, and ρ(x, y) the density, then:
set D, first find the values of f at the critical points ZZ
of f in D. Then find the extreme values of f on the m= ρ(x, y) dA (70)
boundary of D. The largest of these values is the D
absolute maximum. The lowest is the minimum.
The x-coordinate of the center of mass is:
RR
x ρ(x, y) dA
Multiple Integrals x = RRD (71)
D
ρ(x, y) dA
Integrals over Rectangles: The moment of inertia about the x-axis is:
If R is the rectangle such that R = {(x, y)|a ≤ x ≤ ZZ
b, c ≤ y ≤ d}, then: Ix = y 2 ρ(x, y) dA (72)
D
ZZ Z bZ d
f (x, y) dA = f (x, y) dy dx (62) The moment of inertia about the origin is:
R a c
ZZ
ZZ Z dZ b I0 = (x2 + y 2 )ρ(x, y) dA = Ix + Iy (73)
f (x, y) dA = f (x, y) dx dy (63) D
R c a

Triple Integrals
Integrals over Regions:
If E is the volume such that E = {(x, y, z)|a ≤
If D1 is the region such that D1 = {(x, y)|a ≤ x ≤
x ≤ b, g1 (x) ≤ y ≤ g2 (x), h1 (x, y) ≤ z ≤ h2 (x, y)},
b, g1 (x) ≤ y ≤ g2 (x)}, then:
then:
ZZ Z bZ g2 (x) ZZZ Z bZ g2 (x)Z h2 (x,y)
f (x, y) dA = f (x, y) dy dx (64) f (x, y, z)dV = f (x, y, z)dzdydx
D1 a g1 (x)
E a g1 (x) h1 (x,y)
(74)

5
Calculus - Period 4 • A piecewise-smooth curve - A union of a fi-
nite number of smooth curves.
• A closed curve - A curve of which its terminal
Three-Dimensional Integrals point coincides with its initial point.
• A simple curve - A curve that doesn’t inter-
Cylindrical Coordinates: sect itself anywhere between its endpoints.
• An open region - A region which doesn’t con-
x = r cos θ y = r sin θ z=z (75) tain any of its boundary points.
• A connected region - A region D for which
y
r2 = x2 + y 2 tan θ = z=z (76) any two points in D can be connected by a
x path that lies in D.
• A simply-connected region - A region D such
Integrating Over Cylindrical Coordinates: that every simple closed curve in D encloses
RRR R β R h2 (θ) only points that are in D. It contains no
f (x, y, z)dV = ... holes and consists of only one piece.
R u (r cosEθ,r sin θ) α h1 (θ)
. . . u12(r cos θ,r sin θ) rf (r cos θ, r sin θ, z)dzdrdθ • Positive orientation - The positive orienta-
(77) tion of a simple closed curve C refers to a
single counterclockwise traversal of C.
Spherical Coordinates:

x = ρ cos θ sin φ y = ρ sin θ sin φ z = ρ cos φ Vector Field:

(78) A vector field on Rn is a function F that assigns
ρ2 = x2 + y 2 + z 2 (79) to each point (x, y) in an n-dimensional set an n-
dimensional vector F(x, y). The gradient ∇f is
defined by:
Integrating Over Spherical Coordinates:
If E is the spherical wedge given by E = {(ρ, θ, φ)|a ≤ ∇f (x, y, . . .) = fx i + fy j + . . . (83)
ρ ≤ b, α ≤ θ ≤ β, c ≤ φ ≤ d}, then:
RbRβRd and is called the gradient vector field. A vector
2
RRR
E
f (x, y, z)dV = a α c ρ sin φ . . . field F is called a conservative vector field if it is
(80) the gradient of some scalar function.
. . . f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)dρdθdφ

Line Integrals:
Change of Variables: The line integral of f along C is:
The Jacobian of the transformation T given by x = s 
Z b 2  2
g(u, v) and y = h(u, v) is:
Z
dx dy
f (x, y)ds = f (x(t), y(t)) + dt
  C a dt dt
∂x ∂x 
∂(x, y)  ∂u ∂v  = ∂x ∂y − ∂x ∂y (84)
=  ∂y ∂y (81)
∂(u, v)  ∂u  ∂u ∂v ∂v ∂u

∂v
The line integral of f along C with respect to x is:
Z Z b
If the Jacobian is nonzero and the transformation dx
f (x, y)dx = f (x(t), y(t)) dt (85)
is one-to-one, then: C a dt
ZZ ZZ  
 ∂(x, y) 
f (x, y)dA = f (x(u, v), y(u, v)) 
  dudv The line integral of a vector field F along C is:
R S ∂(u, v) 
Z Z b Z
(82) F·dr = F(r(t))·r0 (t) dt = F·T ds (86)
This method is similar to the one for triple inte- C a C
grals, for which the Jacobian has a bigger matrix 0
and the change-of-variable equation has some more Where T = |rr0 | is the unit tangent vector.
terms.
Conservative Vector Fields:
If C is the curve given by r(t) (a ≤ t ≤ b), then:
Basic Vector Field Theorems Z
∇f · dr = f (r(b)) − f (r(a)) (87)
Definitions C

6
 R
The integral RC F · dr is independent of path in D For a surface graph of g(x, y), the normal vector is
if and only if C F · dr = 0 for every closed path C given by:
in D.
∂g ∂g
If F(x, y) = P (x, y)i + Q(x, y)j is a conservative − ∂x i − ∂y j+k
n= r (94)
vector field, then:  2  2
∂g ∂g
1 + ∂x + ∂y
∂P ∂Q
= (88)
∂y ∂x
Also, if D is an open simply-connected region, and Flux:
∂Q If F is a vector field on a surface S with unit normal
if ∂P
∂y = ∂x , then F is conservative in D.
vector n, then the surface integral of F over S is:
ZZ ZZ
Surfaces F · dS = F · n dS (95)
S S

Parametric Surfaces: This integral is also called the flux of F across S.

A surface described by r(u, v) is called a paramet- For a parametric surface, the flux is given by:
∂r ∂r
ric surface. ru = ∂u and rv = ∂v . For smooth ZZ ZZ
surfaces (ru × rv 6= 0 for every u and v) the tan-
F · dS = F · (ru × rv ) dA (96)
gent plane is the plane that contains the tangent S D
vectors ru and rv , and the vector ru × rv is the
normal vector to the tangent plane. For a surface graph of g(x, y), the flux is given by:
ZZ ZZ  
Surface Areas: ∂g ∂g
F · dS = −P −Q + R dA (97)
For a parametric surface, the surface area is given S D ∂x ∂y
by: ZZ
A= |ru × rv |dA (89)
D
For a surface graph of g(x, y), the surface area is
Advanced Vector Field Theorems
given by:
Curl:
If F = P i + Qj + Rk, then the curl of F, denoted
s  2  2
ZZ
∂g ∂g
A= 1+ + dA (90) by curl F or also ∇ × F, is defined by:
D ∂x ∂y
     
∂R ∂Q ∂P ∂R ∂Q ∂P
− i+ − j+ − k
Surface Integrals: ∂y ∂z ∂z ∂x ∂x ∂y
(98)
For a parametric surface, the surface integral is
If f is a function of three variables, then:
given by:
curl(∇f ) = 0 (99)
ZZ ZZ
f (x, y, z) dS = f (r(u, v))|ru ×rv |dA (91)
S D
This implies that if F is conservative, then curl F =
For a surface graph of g(x, y), the surface integral 0. The converse is only true if F is defined on all of
is given by: Rn . So if F is defined on all of Rn and if curl F = 0,
ZZ then F is a conservative vector field.
f (x, y, z) dS =
S Divergence:
ZZ
s  2  2 If F = P i + Qj + Rk, then the divergence of F,
∂g ∂g denoted by div F or also ∇ · F, is defined by:
f (x, y, g(x, y)) 1 + + dA
D ∂x ∂y
(92) ∂P ∂Q ∂R
div F = + + (100)
∂x ∂y ∂z
Normal Vectors:
For a parametric surface, the normal vector is given If F is a vector field on Rn , then div curl F = 0.
by: If div F = 0, then F is said to be incompressible.
ru × rv Note that curl F returns a vector field and div F
n= (93) returns a scalar field.
|ru × rv |

7
Green’s Theorem:
Let C be a positively oriented piecewise-smooth
simple closed curve in the plane and D be the re-
gion bounded by C. Now:
Z ZZ  
∂Q ∂P
P dx + Q dy = − dA (101)
C D ∂x ∂y

This can also be useful for calculating areas. To

calculate an area, take functions P and Q such that
∂Q ∂P
∂x − ∂y = 1 and then apply Green’s theorem.
In vector form, Green’s theorem can also be writ-
ten as:
Z ZZ
F · dr = (curl F) · k dA (102)
C D
Z ZZ
F · n ds = div F(x, y) dA (103)
C D

Stoke’s Theorem:
Let S be an oriented piecewise-smooth surface that
is bounded by a simple, closed, piecewise-smooth
boundary curve C with positive orientation. Let F
be a vector field that contains S. Then:
Z ZZ
F · dr = curl F · dS (104)
C S

The Divergence Theorem:

Let E be a simple solid region and let S be the
boundary surface of E, given with positive (out-
ward) orientation. Let F be a vector field on an
open region that contains E. Then:
ZZ ZZZ
F · dS = div F dV (105)
S E