Calculus BC Bible
(3rd most important book in the world)
(To be used in conjunction with the Calculus AB Bible)
PG.
Topic
2
3− 4
4−5
5−6
6−7
8−9
9
10 −12
13
14
15
15
16
16
16
16 −17
18
18
19
19
20
21
22
23-24
Rectangular and Parametric Equations (Arclength, Speed)
Polar (Polar Area)
Integration by Parts
Trigonometric Integrals
Trigonometric Substitutions
Integration Techniques (Long Division, Partial Fractions)
Improper Integrals
Convergence/Divergence Tests
Taylor/MacLaurin Series
Radius and Interval of Convergence
Writing Taylor series using known series
Separating Variables
L'Hopital's Rule
Work
Hooke's Law
Logistical/Exponential Growth
Euler's Method
Slope Fields
Lagrange Error Bound
Area as a Limit
Summary of Tests for Convergence
Power Series for Elementary Functions
Special Integration Techniques
After Calculus BC (Vector Calculus and Calculus III)
1
APPLICATIONS OF THE INTEGRAL
Rectangular Equations ( y = f (x))
Length of a Curve (Arclength)
b
L=
∫
Surface Area
b
1 + ( f ′ ( x )) dx
2
S.A. =
a
2 π f ( x ) 1 + ( f ' ( x )) dx
∫
2
a
Parametric Equations
Parametric equations are set of equations, both functions of t such that:
x = f (t )
y = g (t )
Length of a Curve
b
L=
∫
Surface Area
( f ' (t )) + ( g ' (t )) dt or
2
2
a
b
∫
a
2
2
b
⎛ dx ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dt
dt
dt
S.A. =
∫ 2π g (t ) ( f ' (t )) + ( g ' (t ))
2
2
dt
a
Speed Equation :
2
⎛ dy ⎞
⎛ dx ⎞
speed = ⎜ ⎟ + ⎜ ⎟
⎝ dt ⎠
⎝ dt ⎠
2
EX#1 :
If a smooth curve c is given by the equation,
x = f (t) and y = g(t), then the slope
of c at (x, y) is
dy
dx
dy dt
=
,
≠0
dx
dt
dx
dt
d ⎛ dy ⎞
d 2 y d ⎛ dy ⎞ dt ⎜⎝ dx ⎟⎠
= ⎜ ⎟=
dx
dx 2 dx ⎝ dx ⎠
dt
d ⎛ d2y⎞
⎜ 2⎟
d 3 y d ⎛ d 2 y ⎞ dt ⎝ dx ⎠
=
=
⎜
⎟
dx
dx 3 dx ⎝ dx 2 ⎠
dt
x= t
y=
dx 1
= t
dt 2
−1
2
(
1 2
t −4
4
)
t≥0
dy 1
= t
dt 2
3 12
d y 2t
=
= 3t
1
dx 2
2
2t
dy
t
dy dt
2
=
=
1
dx dx
dt 2 t
d 3y
dx 3
1
1 2
(t − 4)
4
t=4
3=
2
t=4
dy
dx
=8
(2,3)
d2y
dx 2
2
= 12 Concave up
t=4
2
1
3
= 6t 2
1
2t
Find slope and concavity at (2, 3)
Must find t
2=t
=
1
1
2
Polar Coordinates
Polar coordinates are defined as such:
x = r cosθ
b
y = r sin θ
r 2 = x 2 + y2
θ = tan −1
Area Under a Curve
A=
y
x
∫
a
Area Between Two Curves
b
1
( f (θ ))2 dθ
2
A=
1⎡
2
2
f (θ )) − ( g (θ )) ⎤ dθ
(
⎦
2⎣
∫
a
or
or
b
A=
∫
a
b
1 2
r dθ
2
A=
∫
a
1
⎡( R )2 − ( r )2 ⎤ dθ
⎦
2⎣
EX#1 : Find the Area of one leaf of the three-leaf rose r = 3cos 3θ . (One leaf is formed from
π
A=
1 6
(3cos 3θ )2 dθ =
2 −∫π 6
π
π
9 6
2 −∫π 6
⎛ 3π
⎞ ⎛ −3π
⎞
= ⎜
− 0⎟ − ⎜
− 0⎟ =
⎝ 8
⎠ ⎝ 8
⎠
cos 2 3θ dθ =
1 − cos 6θ
dθ =
2
9 6
2 −∫π 6
5π
6
∫
π
(1 − 2 sin θ ) dθ =
2
6
π
=
1
2
5π
1
at
2
= π −2 3+
π
∫ (1 − 4 sinθ + 4 sin θ ) dθ =
6
2
π
6
3
sin 2θ
( 3 − 4 sin θ − 2 cos 2θ ) dθ = θ + 2 cosθ −
2
2
1 6
2 −∫π 6
5π
π
6
6
and
π
1 6
2 −∫π 6
5π
6
=
3π
)
6
−π 6
1 − cos 2θ ⎞
⎛
⎜⎝ 1 − 4 sin θ + 4 ⋅
⎟⎠ dθ
2
6
∫ (1 − 4 sinθ + 4 sin θ ) dθ = ∫
2
2
2
5π
5π
6
3π
2
6
= 2π + 2 3 −
6
⎛ 5π
3⎞ ⎛ π
3⎞
= ⎜
− 3+
−⎜ + 3−
⎟
4 ⎠ ⎝4
4 ⎟⎠
⎝ 4
3π
∫ ( 3 − 4 sinθ − 2 cos 2θ ) dθ = 3θ + 4 cosθ − sin 2θ
5π
π
3
= 0.543516
2
1 2
A = 2 ⋅ ∫ (1 − 2 sin θ )2 dθ =
2 5π 6
3π
to
to form the inner loop)
EX#3 : Find the Total Area of r = 1 − 2 sin θ . (Half the figure is between
3π
6
3π
4
EX#2 : Find the Inner Loop of r = 1 − 2 sin θ . (sin θ =
1
A=
2
9
3sin 6θ
θ−
8
4
π
-π
5π
2
6
5π
6
and
3π
2
, so we double that area)
1 − cos 2θ ⎞
⎛
⎟⎠ dθ
⎜⎝ 1 − 4 sin θ + 4 ⋅
2
3⎞
⎛ 9π
⎞ ⎛ 5π
= ⎜
− 0 + 0⎟ − ⎜
−2 3+
⎝ 2
⎠ ⎝ 2
2 ⎟⎠
6
3
= 8.8812614
2
EX#4 : Find the Area of the Outer Loop of r = 1 − 2 sin θ . (Whole figure − Inner Loop)
A = 8.881261 − 0.543516 = 8.337745
3
EX#5 : Find the Area of the enclosed region between r1 = 2 − 2 cos θ and r2 = 2 cos θ .
π
1
5π
2 − 2 cos θ = 2 cos θ when cos θ = which is θ = and θ =
.
2
3
3
π
3
in the first enclosure. You have to figure out which equation outlines your encosure. r1 outlines the enclosure
π
π
π
from θ = 0 to θ = and r2 outlines the enclosure from θ = to θ = .
3
3
2
π
π
3
2
⎡1
⎤
1
π
2
A = 2 ⋅ ⎢ ∫ (2 − 2 cos θ )2 dθ +
2
cos
θ
d
θ
⎥ = 0.402180
(
)
∫
2 π3
3
⎢⎣ 2 0
⎥⎦
They form two identical enclosures. Find the Area of one enclosure and double it. They meet at θ =
5π
3
INTEGRATION TECHNIQUES
Integration by Parts
Integration by Parts is done when taking the integral of a product
in which the terms have nothing to do with each other.
∫ u dv =
uv − ∫ v du
Refer to the Calculus AB Bible for the general technique.
If the derivative of one product will eventually reach zero, use the tabular method.
Tabular method :
List the term that will reach zero on the left and keep taking the derivative of that term until it reaches zero.
List the other term on the right and take the integral for as many times as it takes for the left side to reach zero.
EX :
x
1
∫ x cos x dx
+
−
cos x
sin x
− cos x
0
Multiply the terms as shown. The first one on the left multiplies with the second one on the right and the
sign is positive, the second term on the left matches with the third one on the right and is negative, and
continue on until the last term on the right is matched up, alternating signs as you go.
∴
∫ x cos x dx = x sin x + cos x + C
4
Special cases (Integration by Parts)
( ln x must be u = when doing integration by parts )
∫ ln x dx
x ln x − ∫ dx
=
=
x ln x − x + C
u = ln x dv = dx
1
du =
v=x
x
(When neither function goes away, you may pick either function to be u = , but you must pick the same kind
of function throughout) ( i.e. If you pick the trig. function to be u = then you must pick the trig. function
to be u = the next time also. You will perform integration by parts twice.)
∫e
1st time
u = sin x
x
sin x dx = e x sin x − ∫ e x cos x dx
= e x sin x − ⎡⎣ e x cos x − ∫ −e x sin x dx ⎤⎦
dv = e x
= e x sin x − e x cos x − ∫ e x sin x dx
du = cos x dx v = e x
2 ∫ e x sin x dx = e x sin x − e x cos x
2nd time
u = cos x
dv = e
du = − sin x dx
(Add to other side)
e x sin x − e x cos x
+C
∫ e sin x dx =
2
x
x
v = ex
Trigonometric Integrals
∫ sin
m
x cos n x dx
if n is odd, substitute u = sin x
Useful Identity: cos2 x = 1 − sin 2 x
if m is odd, substitute u = cos x
Useful identity: sin 2 x = 1 − cos 2 x
If m and n are both even, then reduce so that either m or n are odd,
using trigonometric (double angle) identities.
Double angle identities :
Half - Angle Identities :
sin 2 x = 2 sin x cos x cos 2 x = cos 2 x − sin 2 x
1 − cos 2 x
1 + cos 2 x
cos 2 x =
sin 2 x =
2
∫ sin x cos x dx
= ∫ sin x (1 − cos 2 x ) cos 4 x dx
= ∫ sin x ( cos 4 x − cos 6 x ) dx
= − ∫ ( u 4 − u 6 ) du
EX#1 :
3
4
(u = cos x du = -sin x dx)
−u
u
=
+
+C
5
7
5
7
∫ sin x cos x dx
2
= ∫ sin 2 x (1 − sin 2 x ) cos x dx (Each
= ∫ ( sin 2 x − 2 sin 4 x + sin 6 x ) cos x dx
= ∫ ( u 2 − 2u 4 + u 6 ) du
2
EX#2 :
5
(u = sin x du = cos x dx)
u 3 2u 5 u 7
=
−
+
+C
3
5
7
− ( cos x )
( cos x ) + C
=
+
5
7
5
2
7
=
(sin x )3 − 2 (sin x )5 + (sin x )7 + C
3
5
5
7
cos 2 x = 1 − sin 2 x)
∫ sin x cos x dx
2
= ∫ ( sin x cos x ) cos 2 x dx
2
EX#3 :
4
2
⎛ sin 2x ⎞ ⎛ 1 + cos 2x ⎞
i
= ∫⎜
⎟⎠ dx
⎝ 2 ⎟⎠ ⎜⎝
2
1 + cos 2x ⎞
⎛
2
⎜⎝ sin 2x = 2 sin x cos x and cos x =
⎟⎠
2
1
1
sin 2 2x dx + ∫ sin 2 2x cos 2x dx
∫
8
8
1 1 − cos 4x
1 1
= ⋅∫
dx + ⋅ ∫ u 2 du ( u = sin 2x
8
2
8 2
=
x sin 4x ( sin 2x )
=
−
+
+C
16
64
48
3
sin 4x ⎞ 1 u 3
1⎛
x
−
+C
=
⎜
⎟+
4 ⎠ 16 3
16 ⎝
∫ tan
m
1 − cos 4x ⎞
⎛
du = 2 cos 2x ) ⎜ sin 2 2x =
⎟⎠
⎝
2
x sec n x dx
If n is even, substitute u = tan x
Useful identity: tan 2 x = sec 2 x − 1
If m is odd, substitute u = sec x
Useful identity: tan 2 x = sec 2 x − 1
Useful identity: tan 2 x = sec 2 x − 1
If m is even and n is odd, reduce powers of sec x alone
∫ sec x tan x dx
= ∫ sec 2 x (1 + tan 2 x ) tan 2 x dx
4
EX#1 :
(sec
2
x = 1 + tan 2 x
2
(
)
∫ sec x tan x dx
= ∫ sec x tan x ( sec 4 x tan 2 x ) dx
= ∫ sec x tan x ( sec 4 x (1 − sec 2 x )) dx
)
= ∫ sec 2 x tan 2 x + tan 4 x dx
=
∫ (u
2
+u
4
) du
( tan
=
=
x = 1 − sec 2 x
(
)
)
(u = sec x du = sec x tan x dx)
5
u
u
+
+C
3
5
=
∫ (u
4
)
− u 6 du
u5 u7
=
−
+C
5
7
( tan x )3 + ( tan x )5 + C
3
2
3
= ∫ sec x tan x sec 4 x − sec 6 x dx
(u = tan x du = sec 2 x dx)
3
5
EX#2 :
5
=
(sec x )5 − (sec x )7 + C
5
Trigonometric Substitutions
If the integrand contains a 2 − x 2 then substitute
The radicand becomes
(a
2
)
(
If the integrand contains
The radicand becomes
(a
2
2
)
)
x = a tan u
(
dx = a sec 2 u du
)
+ a 2 tan 2 u = a 2 1 + tan 2 u = a 2 sec 2 u = a secu
x 2 − a 2 then substitute
(a
dx = a cosu du
− a 2 sin 2 u = a 2 1 − sin 2 u = a 2 cos 2 u = a cosu
If the integrand contains a 2 + x 2 then substitute
The radicand becomes
x = a sin u
)
x = a secu
(
)
dx = a secu tan u du
sec 2 u − a 2 = a 2 sec 2 u − 1 = a 2 tan 2 u = a tan u
6
7
∫
EX#1 :
9 − x 2 dx =
∫
9 − ( 3sin θ ) 3cosθ dθ
2
∫ 9 (1 − sin θ ) 3cosθ dθ
=
2
3
1 + cos 2θ
dθ
∫
2
2
9θ 9 sin 2θ
9
⎛ x⎞ 9 ⎛ x⎞ ⎛ 9 − x ⎞
=
+
+ C = arcsin ⎜ ⎟ + ⎜ ⎟ ⎜
⎟ +C
⎝ 3⎠ 2 ⎝ 3⎠ ⎝
2
2
4
3 ⎠
x = 3sin θ
9 cos 2 θ 3cosθ dθ
=
dx = 3cosθ dθ
x
= ∫ 9 cos 2 θ dx = 9 ∫
θ
9 − x2
x = 3sin θ
x
= sin θ
3
2
9
⎛ x⎞ x 9 − x
+C
= arcsin ⎜ ⎟ +
⎝ 3⎠
2
2
∫
EX#2 :
x
16 + x
dx
2
(Substitute x = 4 tanθ
=
∫
=
∫
4 tan θ
16 + 16 tan θ
4 tan θ
2
(
16 1 + tan θ
2
4 tan θ
=
∫
=
∫ 4 tanθ secθ dθ
16 sec θ
2
)
dx = 4 sec 2 θ dθ
⋅ 4 sec 2 θ dθ
⋅ 4 sec 2 θ dθ
⋅ 4 sec θ dθ
2
= 4 sec θ + C
=
=
=
16 + x
+C
4
∫
∫
dx = 5 sec θ tan θ dθ )
25 sec 2 θ − 25
⋅ 5 sec θ tan θ dθ
5 sec θ
(
) ⋅ 5 secθ tanθ dθ
25 sec 2 θ − 1
5 sec θ
25 tan 2 θ
⋅ 5 sec θ tan θ dθ
5 sec θ
=
∫
=
∫ 5 tan θ dθ
2
∫ ( 5 sec θ − 5 ) dθ
=
2
= 4
)
x 2 − 25
EX#3 : ∫
dx
x
(Substitute x = 5 secθ
2
=
∫ 5 (sec
)
θ − 1 dθ
= 5 tan θ − 5θ + C
⎛ x 2 − 25 ⎞
⎛ x⎞
= 5⎜
⎟ − 5arcsec ⎜ ⎟ + C
⎝ 5⎠
5
⎝
⎠
16 + x 2 + C
=
⎛ x⎞
x 2 − 25 − 5arcsec ⎜ ⎟ + C
⎝ 5⎠
x
16 + x 2
x = 4 tan θ
x
= tan θ
4
2
x = 5 sec θ
x
= sec θ
5
x
θ
x 2 − 25
θ
5
4
7
Integration Techniques
Long Division (Must use long division if the top function has the same or greater power as the bottom function)
Step 1 - Divide the numerator by the denominator. Separate the integrand into separate integrals for each term
2x 3
∫ x 2 + 3 dx
EX.
∫
=
2x −
6x
dx
x +3
2
(
)
= x 2 − 3ln x 2 + 3 + C
2x
x 2 + 3 2x 3
2x 3 + 6x
− 6x
Add and subtract terms in the numerator
∫x
=
2
2x
dx =
+ 6x + 13
∫x
2
∫x
2
2x + 6
6
− 2
dx =
+ 6x + 13 x + 6x + 13
∫x
2
2x + 6
6
− 2
dx
+ 6x + 13 x + 6x + 9 + 13 − 9
6
2x + 6
⎛ x + 3⎞
+C
−
dx = ln x 2 + 6x + 13 + 3arctan ⎜
2
⎝ 2 ⎟⎠
+ 6x + 13 ( x + 3) + 4
Partial Fractions ( Used when the denominator is factorable )
Before performing partial fractions, you must use long division if the degree of the numerator is
greater or equal to the denominator.
Step 1 - Factor the denominator of the function
Step 2 - Separate into partial fractions
Rewrite as fractional terms - arbitrary constants divided by each factor of the denominator.
If a factor is repeated more than once, write that many terms for that factor
4x − 1
4x − 1
A
B
EX#1 : ∫ 2
dx = ∫
dx = ∫
+
dx
x−4 x+3
x − x − 12
( x − 4 ) ( x + 3)
Multiply by the common denominator.
4x − 1 = A ( x + 3) + B ( x − 4 )
Plug in the roots of the denominator and solve for the constants
15
Let x = 4 ⇒ 15 = 7A
⇒ A=
7
13
Let x = − 3 ⇒ −13 = − 7B ⇒ B =
7
The integral has been reduced down to something much more reasonable and now we are able to integrate.
Therefore,
4x − 1
∫ x 2 − x − 12 dx =
15
13
15
13
7 +
∫ x − 4 x +73 dx = 7 ln x − 4 + 7 ln x + 3 + C
8
2x + 3
2x + 3
A
B
C
∫ x 3 + 2x 2 + x dx = ∫ x ( x + 1)2 dx = ∫ x + x + 1 + ( x + 1)2 dx
EX#2 :
Multiply by the common denominator.
2x + 3 = A ( x + 1) + Bx ( x + 1) + Cx
2
Plug in the roots of the denominator and solve for the constants
Let x = 0 ⇒ 3 = A
Let x = −1 ⇒ 1 = − C ⇒ C = −1
Let x = 1 ⇒ 5 = 4A + 2B + C ⇒ B = − 3
The integral has been reduced down to something much more reasonable and now we are able to integrate.
Therefore,
2x + 3
∫ x ( x + 1)2 dx
3
3
1
∫ x − x + 1 − ( x + 1)2 dx
=
= 3ln x − 3ln x + 1 +
1
+C
x +1
Simple partial fractions ( Other method )
8
∫ ( x + 3)( x − 5 ) dx
EX#3 :
8 = A ( x − 5 ) + B ( x + 3)
A+ B = 0
=
A
B
∫ x + 3 + x − 5 dx
⇒
−1
1
∫ x + 3 + x − 5 dx
8 = Ax − 5A + Bx + 3B
= − ln x + 3 + ln x − 5 + C
= ln
x-5
+C
x+3
3A + 3B = 0
−5A + 3B = 8
− 5A + 3B = 8
8A = −8
A = −1
B=1
Improper Integrals
∞
∫
f ( x ) dx = lim
b
∫
b→∞ a
a
∞
f ( x ) dx
f ( x ) dx =
∫
−∞
c
∫
−∞
∞
f ( x ) dx + ∫ f ( x ) dx
c
In the first case:
If f ( x ) converges then the integral exists
If f ( x ) diverges then the integral is ∞ or − ∞
∞
EX #1 :
∫
1
∞
EX #2 :
∫
1
∞
EX #3 :
b→∞
1
dx =
x
∫
1
dx =
x2
1
1
dx =
x2
1
EX #4 :
b
∫
0
(
)
(
)
e− x dx = lim ∫ e− x dx = lim − e− x 1 = lim − e−b + e−1 = 0 + e−1 =
b=∞
∫
1
b=∞
1
b→∞
b
1
b=∞
dx = ln x 1 = lim ( ln b ) − ln1 = ∞
b→∞
x
b=∞
∫
−1
x1
1
1
dx =
x2
−1
=
x a=0
∫
a=0
1
∞
So
∫
1
−1
⎛ −1 ⎞
− lim− ⎜ ⎟ = −1 + ∞
1 a→0 ⎝ a ⎠
9
∞
∫
e− x dx converges
1
1
dx diverges.
x
∞
⎛ −1 ⎞ −1
= lim ⎜ ⎟ −
= 0 +1= 1
b→∞ ⎝ b ⎠
1
1
dx =
x2
1
b→∞
1
So
e
So
∫
1
1
So
∫
0
1
dx converges.
x2
1
dx diverges.
x2
General Series and Sequences
For the sequence {an }n=m , if lim an exists, then the sequence converge to L.
∞
n→∞
If
lim an
n→∞
does not exist, then the sequence diverges to ∞ or − ∞.
Boundness refers to whether or not a sequence has a finite range
{an }∞n=m converges, then {an }∞n=m is bounded
∞
∞
If {an }n=m diverges, then {an }n=m is unbounded
nth partial sum ( Sn ) refers to the sum of the first n terms of a sequence.
If
The
∞
∞
∞
n=1
n=1
n=1
If ∑ an and ∑ bn both converge, then ∑ ( an + bn ) does also and
∞
∞
∞
n=1
n=1
n=1
∑ an + ∑ bn = ∑ ( an + bn )
Convergence Tests ( PARTING CC )
nth term test (Used to show immediate divergence)
If top power is the same or greater than the bottom power then the series is divergent.
If bottom power is greater we must proceed and use a different test, because this test cannot prove convergence.
∞
EX#1 :
∑
n=2
n3 + 2
and
n3 − 5
∞
∑
n =1
( −1)n n 5
3n 3 + 2
These are divergent by the nth term test.
P - Series
∞
1
∑n
∞
EX#1 :
∞
1
∑ n 2 and
n =1
∞
EX#2 :
1
∑n
∞
1
∑n
n =1
1.1
n =1
1
∑n
and
2
n =1
converges if p > 1 and diverges if p ≤ 1
p
n =1
3
Converge by p-series
Diverge by p-series
Integral Test for Convergence (for nonnegative sequences) (Used if you know the integral of the series)
∞
∑ an converges iff
n =1
∞
EX#1 :
∑n
n=0
∞
EX#2 :
2
1
⇒
+1
1
∑n ⇒
n =1
∞
1
0
2
∞
∑ an diverges iff
n =1
π
π
1
b=∞
dx = arctan x 0 = lim ( arctan b ) − arctan 0 = − 0 =
b→∞
+1
2
2
∫ x dx = ln x
1
∫
f ( x ) dx converges
1
∞
∫x
∞
b=∞
1
= lim ( ln b ) − ln1 = ∞ − 0 = ∞
b→∞
10
so series Diverges
∞
∫ f ( x ) dx diverges
1
so series Converges
Alternating Series Test for Convergence (for alternating sequences)
∞
The series ∑ ( −1) an converges if {an }n =1 is a positive decreasing sequence and lim an = 0.
∞
n
n→∞
n =1
∞
EX#1 :
( −1)n
∑ n +1
Converges because the denominator is larger in the alternating series.
n =1
Alternating series can not be used to prove divergence.
Direct Comparison Test for Convergence and Divergence (for positive sequences)
∞
∞
If ∑ bn converges and 0 ≤ an ≤ bn then ∑ an converges
n =1
n =1
∞
∞
n =1
n =1
If ∑ bn diverges and 0 ≤ bn ≤ an then ∑ an diverges
∞
EX#1 :
∑
n=0
∞
EX#2 :
∑
n=0
1
1
1
1
⇒ Diverges since
> and we know diverges by p-series.
n −1 n
n
n −1
n
n
2n
2n
⎛ 2⎞
⎛ 2⎞
⇒ Converges since n
< ⎜ ⎟ and we know ⎜ ⎟ converges by geometric series.
n
⎝ 7⎠
7 +2
7 + 2 ⎝ 7⎠
Limit Comparison Test (for nonnegative sequences) (Used on messy algebraic series)
an
>0
n→∞ b
n
∞
∞
n =1
n =1
If ∑ bn converges, then ∑ an converges.
Assume lim
1
1
1
n2
1
n
−
1
⇒ lim
= lim
⋅
=1
EX#1 : ∑
n→∞
n→∞
1
n −1 1
n −1
n=0
1
n2
Diverges since we compared with a divergent series.
∞
∞
n =1
n =1
If ∑ bn diverges, then ∑ an diverges.
∞
7
3
7
n3
7
⇒ lim n + 5 = lim 3
⋅
=7
EX#2 : ∑ 3
n→∞
n→∞ n + 5
1
1
n=0 n + 5
n3
Converges since we compared with a convergent series.
Limit is finite and positive.
∞
Limit is finite and positive.
Geometric Series Test
∞
A geometric series
∑ a ⋅ r n converges iff r < 1.
∞
A geometric series
n= m
∞
If a geometric series converges, then
∑ a ⋅ rn =
n= m
⎛ 3⎞
n
⎛ 7⎞
EX#2 : ∑ 6 ⋅ ⎜ ⎟
⎝ 2⎠
n=0
n
∞
EX#1 :
∑ 2 ⋅ ⎜⎝ 5 ⎟⎠
Converges because r =
n=0
∞
Diverges because r =
3
.
5
7
.
2
ar m
1− r
∑a⋅r
diverges iff r ≥ 1.
n= m
Sum : S =
a
1− r
The sum of the series is S =
2
=5
1 − 35
No sum since series diverges.
11
n
Ratio Test (for nonnegative series) (Used for really ugly problems)
∞
The series ∑ an converges if lim
n→∞
n =1
∞
The series ∑ an diverges if lim
n→∞
n =1
If lim
n→∞
∞
EX#1 :
∑
( −1)n n!
e
n =1
n
an +1
>1
an
an +1
= 1, then this test is inconclusive.
an
( n + 1)!
en +1
n!
en
lim
n→∞
( n + 1)5
= lim
n→∞
( n + 1)! ⋅ en
e
n +1
n!
= lim
n→∞
( n + 1)
e
= ∞ so series Diverges
n +1
n + 1) 2 n
(
( n + 1) = 1 so series Converges
2
lim
= lim
⋅ 5 = lim
EX#2 : ∑
5
n
n +1
n→∞
n→∞
n→∞
n
2
2
2
n
2n 5
n =1
n
2
Root Test for Convergence (for nonnegative series)
∞
( −1)
an +1
<1
an
n
n
5
5
5
∞
The sequence∑ an converges if lim n an < 1
n =1
n→∞
∞
The sequence ∑ an diverges if lim n an > 1
n =1
n→∞
If lim n an = 1, then this test is inconclusive.
n→∞
∞
EX#1 :
∑
( −1)
EX#2 :
∑
2n
en
n =1
∞
n
( −1)n 7 n
n =1
5n
lim
n
2n
2 2
= lim =
so series Converges
n
n→∞ e
e
e
lim
n
7n
7 7
= lim =
so series Diverges
n
n→∞
5 5
5
n→∞
n→∞
Telescoping Series ( Limit = 0 and the terms get smaller as we approach ∞ )
If a series is a telescoping series then it is convergent.
∞
EX#1 :
∑
n=1
∞
EX#1 :
∑
n=1
1
1⎞ ⎛ 1 1⎞ ⎛ 1 1 ⎞
1
⎛
−
is convergent. The sum = ⎜ 1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ..... = 1
⎝
2 ⎠ ⎝ 2 3⎠ ⎝ 3 4 ⎠
n n +1
3
3
⎛3
−
is convergent. The sum = ( 3 − 1) + ⎜ −
⎝2
n n+2
12
3⎞ ⎛ 3 3⎞
3
3⎞ ⎛
⎟⎠ + ⎜⎝ 1 − ⎟⎠ + ⎜⎝ − ⎟⎠ + ..... = 3 + =
5
4 6
2
4
9
2
SERIES AND SEQUENCES
∞
Power Series : A power series is defined as
∑c x
n=0
n
a
= c0 + c1 x + c2 x 2 + c3 x 3 + ...
Taylor’s Theorem and Related Series (If the series is centered at c = 0 then it is a MacLaurin Series)
Taylor’s theorem for approximating f ( x ) to the nth term:
f ′ (c)
f ′′ ( c )
f ′′′ ( c )
f n (c)
(x − c)n
( x − c) +
( x − c )2 +
( x − c )3 + ...... +
1!
2!
3!
n!
∞
f n( c )
n
Taylor series of f ( x ) → this is a power series
∑ n! ( x − c )
n=0
f ( x ) = f (c) +
EX#1 : Find the fourth degree MacLaurin Series ( centered at c = 0 ) for f ( x ) = cos x.
f ( x ) = cos x
f (0) = 1
f ′ ( x ) = − sin x
f ′ (0) = 0
f ′′ ( x ) = − cos x
f ′′ ( 0 ) = −1
f ′′′ ( x ) = sin x
f ′′′ ( 0 ) = 0
f 4 ( x ) = cos x
f ( x)
1
2
3
4
x − 0)
x − 0)
x − 0)
x − 0)
(
(
(
(
= 1+ 0
−1
+0
+1
1!
2!
3!
2
4
( x − 0) + 1 ( x − 0)
f ( x) = 1 − 1
2!
4!
n
∞
−1) x 2n
x2 x4
(
f ( x) = 1 −
+
= ∑
2! 4!
( 2n )!
n=0
4!
f 4 (0) = 1
EX#2 : Find the fourth degree Taylor Series for f ( x ) = ln x centered at c = 1.
f ( x ) = ln x
f ′( x) =
1
x
−1
x2
2
f ′′′ ( x ) = 3
x
−6
f 4 ( x) =
x4
f ′′ ( x ) =
f (1) = 0
f ( x)
f ′ (1) = 1
1
2
3
4
x − 1)
x − 1)
x − 1)
x − 1)
(
(
(
(
= 0 +1
−1
+2
−6
1!
f ( x ) = ( x − 1) − 1
( x − 1)
2!
2
+2
( x − 1)
3!
3
−6
( x − 1)
2!
3!
4!
2
3
( x − 1) + ( x − 1) − ( x − 1)4 =
f ( x ) = ( x − 1) −
2
3
4
f ′′ (1) = −1
f ′′′ (1) = 2
EX #3 : f ( 2 ) = 5
f 4 (1) = − 6
f ′ (2) = 4
f ′′ ( 2 ) = 2
f ′′′ ( 2 ) = 17
3rd Degree Taylor Polynomial : P3 ( x ) = 5 + 4 ( x − 2 )
2
3
x − 2)
x − 2)
(
(
+2
+ 17
2!
Using 3rd Degree Taylor Polynomial to approximate x = 2.1 :
P3 ( 2.1) = 5 + 4 ( 2.1 − 2 )
4!
4
2
3
2.1 − 2 )
2.1 − 2 )
(
(
+2
+ 17
= 5.41283
2!
3!
13
3!
∞
∑
n=0
( −1)n ( x − 1)n +1
n +1
Radius of convergence / Interval of convergence
EX#2 : Find the radius and interval of convergence
EX#1 : Find the radius and interval of convergence
(−1)n (x − 4)n
n ⋅ 3n
n=0
∞
n
∞ ⎛ x + 1⎞
(x + 1)n
= ∑⎜
∑
⎟ Geometric Series
n
2
n=0
n=0 ⎝ 2 ⎠
x +1
<1
Converges iff r < 1
2
x +1
<1
−1 <
2
−2 < x + 1 < 2
− 3< x <1
∞
∑
lim
n→∞
(x − 4) n
⋅
n→∞
3
n +1
x−4
−1 <
<1
3
−3 < x − 4 < 3
1< x < 7
Check endpoints
endpoints of geometric series never work.
I'll check anyway.
Check endpoints
2n
n=0
2n
Check − 3, ∑
Check 1, ∑
n=0
2
∞
(−1)2n 3n
n=0
n ⋅ 3n
Check 1, ∑
Diverges
∞
(−1)n 2 n
∞
(x − 4)n+1
n ⋅ 3n
⋅
( n + 1) ⋅ 3n+1 (x − 4)n
lim
Don't have to check endpoints because
∞
n
Use Ratio Test
Check 7, ∑
Diverges
(−1)n 3n
Converges
n ⋅ 3n
Since 7 works include 7 in interval.
n=0
Diverges
Center of convergence: − 1
Center of convergence: 4
Radius of convergence: 2
Radius of convergence: 3
Interval of Convergence: − 3 < x < 1
Interval of Convergence: 1 < x ≤ 7
When finding the interval of convergence using ratio test :
If lim
=0
then the series converges from − ∞ to ∞.
If lim
=∞
then the series converges at the center c only.
n→∞
n→∞
(x − 4)n
n
n=0 3 ⋅ n!
∞
EX#1 : ∑
lim
n→∞
(x − 4)n+1
3n+1 ⋅ ( n + 1)!
(x − 4)n
3n ⋅ n!
= lim
n→∞
(x − 4)n+1
(x − 4)
3n ⋅ n!
⋅
= lim
=0
n+1
n
n→∞ 3 ( n + 1)
3 ⋅ ( n + 1)! (x − 4)
so the series converges from − ∞ to ∞.
(x + 1)n n!
EX#2 : ∑
2n
n=0
∞
(x + 1)n+1 ( n + 1)!
2 n+1
lim
n→∞
(x + 1)n n!
2n
so the series converges at the center − 1 only.
= lim
(x + 1)n+1 ( n + 1)!
n→∞
14
2 n+1
⋅
(x + 1) ( n + 1)
2n
= lim
=∞
n
n→∞
2
(x + 1) n!
Memorize these known series
= 1+ x +
Series for e x :
Series for sin x : =
∞
x2 x3
xn
+
+ ... = ∑
2! 3!
n=0 n!
3
Series for
( −1) x
=∑
n=0 ( 2n + 1)!
∞
5
x
x
x−
+
− ...
3! 5!
n
2n+1
∞
1
: = 1 + x + x 2 + x 3 ...
1− x
2
=∑ xn
n=0
( −1)n x 2n
=∑
( 2n )!
n=0
∞
4
x
x
Series for cos x : = 1 −
+
− ...
2! 4!
EX#1 : Use the series from above to write each of the following series
f ( x ) = cos x 2
⇒
f ( x ) = x cos x ⇒
f ( x ) = sin x ⇒
f ( x ) = e2 x
⇒
f ( x) = 1 −
∞
( −1) x 4n
x4 x8
+
− ... = ∑
2! 4!
( 2n )!
n=0
n
∞
( −1) x 2n+1
x3 x5
f ( x) = x −
+
− ... = ∑
2! 4!
( 2n )!
n=0
n
f ( x) = x −
f ( x ) = 1 + 2x +
( 2x )2
f ( x ) = ∫ e dx ⇒ f ( x ) = x + x +
2x
2
2!
2 ⋅ 3!
3
(Replace x with 2x in the known e series)
n!
n=0
4
+
2 1⎞
⎛
1
+... 0 = ⎜ 1 + 1 + + ⎟ − ( 0 ) 3 (Take the definite integral of e2 x )
⎝
2 ⋅ 4!
3 3⎠
= ∑ ( −1) x n
n
(Replace x with − x in the known series
n=0
x − x + x − x ...
2
=∑
+...
3!
∞
1
: = 1 − x + x 2 − x 3 ...
1+ x
x
: =
f ( x) =
1+ x
( −1)n x 2n+1
(Take the integral of the known cosine series)
n=0 ( 2n + 1)!
∞
( 2x )3
( 2x )n
x
=∑
( 2x )3 ( 2x )4
0
2
+
(Multiply the known cosine series by x)
∞
x3 x5
+
− ...
3! 5!
1
f ( x) =
( Replace x with x 2 in the known cosine series)
6
∞
= ∑ ( −1) x n+2
n
(Multiply the found series
n=0
DIFFERENTIAL EQUATIONS
1
)
1− x
1
by x 2 )
1+ x
(Separating Variables )
Separable Differential Equations
dy
If P ( x ) + Q ( y ) = 0 then Q ( y ) dy = − P ( x ) dx then ∫ Q ( y ) dy = − ∫ P ( x ) dx
dx
dy x 2
EX#1 : Find the general solution given
=
dx y
dy x 2
=
dx y
⇒ y dy = x 2 dx
⇒
2
∫ y dy = ∫ x dx
⇒
y2 x 3
=
+C
2
3
⇒
y2 =
2x 3
+ C1
3
EX#2 : Find the particular solution y = f (x) for EX#1 given ( 3, − 5 )
2x 3
y =
+ C1
3
2
⇒ 25 = 18 + C1
⇒ 7 = C1
2x 3
⇒ y =
+7
3
EX#3 : Find the particular solution y = f (x) given
dy
= 6x dx ⇒
y
∫
2
2x 3
⇒ y=−
+7
3
dy
= 6xy and ( 0, 5 )
dx
2
2
2
dy
= ∫ 6x dx ⇒ ln y = 3x 2 + C ⇒ y = e 3x +C ⇒ y = C1e3x ⇒ 5 = C1 (1) ⇒ y = 5e3x
y
15
L’ Hôpital’s Rule : If lim
x→a
f (x) 0
f (x)
f '(x)
∞
= or , then lim
= lim
x→a g(x)
x→a g '(x)
g(x) 0
∞
Reminder: In limits
1
1
= 0 and = ∞
∞
0
∞
∞
∞
ex
ex
ex
ex
=
so
lim
=
=
lim
=
=
lim
= ∞
x→∞ x 3
x→∞ 3x 2
x→∞ 6x
x→∞ 6
∞
∞
∞
x 2 + 5x − 14 0
2x + 5
EX #2 : lim
=
so lim
=9
x→2
x→2
x−2
1
0
0
∞⎞
Indeterminate forms : 0 ⋅ ∞, ∞ − ∞, 1∞ , ∞ 0 , 0 0 ⎛⎜ Indeterminate forms must be changed to
or
⎟
⎝
0
∞⎠
1
ln x
0
x = lim − x = 0
EX #3 : lim+ x ⋅ ln x = 0 ⋅ ∞ so we must convert to lim+
=
so lim+
1
−1 2
x→0
x→0
x→0
x→0 +
0
x
x
EX #1 : lim
1⎞
⎛
EX #4 : lim ⎜ 1 + ⎟
x→∞ ⎝
x⎠
1⎞
⎛
ln y = lim ln ⎜ 1 + ⎟
x→∞
⎝
x⎠
ln y = lim
x→∞
1
1
1+
x
x
x
∞
=1
1⎞
⎛
so we must bring down the x so let y = lim ⎜ 1 + ⎟
x→∞ ⎝
x⎠
1⎞
⎛
ln ⎜ 1 + ⎟
⎝
x⎠ 0
=
ln y = lim
x→∞
1
0
x
1⎞
⎛
ln y = lim x ln ⎜ 1 + ⎟ = 0 ⋅ ∞
x→∞
⎝
x⎠
ln y = 1
x
and take ln of both sides.
⎛
⎞
1
⎜
⎟ −1
⋅
⎜
1 ⎟ x2
⎜⎝ 1 + ⎟⎠
x
ln y = lim
x→∞
−1
x2
y=e
Work
W = F ⋅ d cosθ
or the dot product of the Force vector and the distance vector.
W = F ⋅d
b
W = ∫ F ( x ) dx ; where F ( x ) is a continuously varying force.
a
Hooke’s Law : The force F required to compress or stretch a spring (within its elastic limits) is
proportional to the distance d that the spring is compressed or stretched from its original length. That is,
F = kd
b
W = ∫ k x dx
a
where the constant of proportionality k (the spring constant) depends on the specific nature of the spring.
EX : Compressing a Spring
A force of 750 pounds compresses a spring 3 inches from its natural length of 15 inches.
Find the work done in compressing the spring an additional 3 inches.
750 = k ( 3) k = 250
F ( x ) = k x so F ( x ) = 250x
6
W=
∫
250x dx = 3375 inch − pounds
3
16
Logistical Growth
L
dy
y⎞
⎛
= ky ⎜ 1 − ⎟ ; y =
⎝
1 + be− k t
dt
L⎠
k = constant of proportionality
L = carrying capacity
Exponential Growth
; b=
L − Y0
Y0
dy
= ky ; y = Cek t
dt
k = growth constant
C = initial amount
Y0 = initial amount
dy
y
= ky(1 − )
dt
L
where L is the carrying capacity (upper bound) and k is the constant of proportionality.
If you solve the differential equation using very tricky separation of variables, you get the general solution:
L − Y0
L
y=
;
b=
− kt
1 + be
Y0
Logistic Growth
EX#1 :
A highly contagious “pinkeye” (scientific name: Conjunctivitus itchlikecrazius) is ravaging the local
elementary school. The population of the school is 900 (including students and staff), and the rate of
infection is proportional both to the number infected and the number of students whose eyes are pus-free.
If seventy-five people were infected on December 15 and 250 have contracted pinkeye by December 20,
how many people will have gotten the gift of crusty eyes by Christmas Day?
Solution
Because of the proportionality statements in the problem, logistic growth is the approach we should take.
The upper limit for the disease will be L = 900; it is impossible for more than 900 people to be infected
since the school only contains 900 people. This gives us the equation
900 − 75
900
L = 900 b =
= 11
⇒
y=
75
1 + 11e− kt
Five days later, 250 people have contracted pinkeye, so plug that information to find k:
900
⇒ 250(1 + 11e− 5 k ) = 900
250 =
− k⋅5
1 + 11e
13
13
⇒ e− 5 k =
11e− 5 k =
5
55
13
⎛ ⎞
ln e− 5 k = ln ⎜ ⎟
⇒ − 5k = −1.442383838
⎝ 55 ⎠
k = 0.2884767656
900
. We want to find the number of infections on
Finally, we have the equation y =
1 + 11e− 0.2884767656t
December 25, so t = 10.
900
y=
− (0.2884767656)(10)
1 + 11e
y ≈ 557.432
So almost 558 students have contracted pinkeye in time to open presents.
17
Euler’s Method
Uses tangent lines to approximate points on the curve.
dy
dy
New y = Old y + dx ⋅
dx : change in x
= Derivative (slope) at the point.
dx
dx
EX#1 :
dy xy
Given: f (0) = 3
=
Use step of h = 0.1 to find f (0.3)
dx 2
Original (0, 3)
New y = 3 + (0.1)(0) = 3
f ( 0.1) 3
New y = 3 + (0.1)(0.15) = 3.015
f ( 0.2 ) 3.015
New y = 3.015 + (0.1)(0.3015) = 3.04515
f ( 0.3) 3.04515
dy 0 ⋅ 3
=
=0
dx
2
dy 0.1⋅ 3
At (0.1, 3)
=
= 0.15
dx
2
dy 0.2 ⋅ 3.015
At (0.2, 3.015)
=
= 0.3015
dx
2
At (0, 3)
Slope Fields
dy
Draw the slope field for
= xy
dx
dy
and graph the tangent line at the point
Plug each point into
dx
dy
dy
dy
= 1. At ( −1, 1)
= −1. At ( 0, 1)
=0
At (1, 1)
dx
dx
dx
EX : Here are the slope fields for the given differential equations
dy y
= (6 − y)
dx 8
9
•
•
•
•
•
•
•
1
−1
•
0
•
•
•
dy
= x+y
dx
3.1
-4.7
0
•
8
4.7
-3.1
The slope fields give ALL the solutions to the differential equation .
18
Lagrange Error Bound
Error = f ( x ) − Pn ( x ) ≤ Rn ( x ) where Rn ( x ) ≤
f n+1 ( z ) ( x − c )
( n + 1)!
n+1
f n+1 ( z ) = the maximum of the ( n + 1) th derivative of the function
EX #1 : cos x = 1 −
R4 ( x ) =
x2 x4
+
− ....
2! 4!
f 5 ( z )( x − c)
5!
5
cos ( 0.1) 0.99500416667
f 5 ( z ) = − sin z
We need − sin z to be as large as possible (which is 1 because −1 ≤ sin z ≤ 1)
1⋅ ( 0.1)
R4 ( 0.1) <
= 0.0000000833
5!
EX #2 : f (1) = 2
f ′ (1) = 5
f ′′ (1) = 7
5
f ′′′ (1) = 12
2nd Degree Taylor Polynomial : P2 ( x ) = 2 + 5 ( x − 1) + 7 ( x − 1)
2
P2 (1.1) = 2 + 5
Using 2nd Degree Taylor Polynomial to approximate x = 1.1:
Lagrange Error Bound : R3 ( x ) =
(1.1 − 1)1
1!
+7
(1.1 − 1)2
2!
= 2.535
f 3 ( z )( x − c)
12 (1.1 − 1)
=
= 0.002 = Error
3!
3!
3
3
Area as a limit
n
b−a
⎛
Area = lim ∑ f ⎜ a +
n→∞
⎝
n
i =1
height
⎞ ⎛ b − a⎞
i⎟ ⎜
⎠ ⎝ n ⎟⎠
i = interval
width
Summation formulas and properties
n
1)
n
∑ c = cn
∑i =
2)
i
n
4)
∑ i3 =
i
i
n 2 ( n + 1)
4
2
n
∑ ( ai ± bi )
5)
=
i
[ 0, 1]
EX#1 :
f ( x) = x3
Area =
1− 0 ⎞ ⎛ 1− 0⎞
⎛
lim ∑ f ⎜ 0 +
i⎟ ⎜
⎟
⎝
n→∞
n ⎠⎝ n ⎠
i=1
n
n ( n + 1)
2
n
3)
∑ i2 =
i
n
n
∑ ai ± ∑ bi
i
6)
i
n ( n + 1) ( 2n + 1)
6
n
n
i
i=1
∑ k ⋅ ai = k ∑ ai , k is a constant
n subdivisions Find Area under curve.
3
n
⎛ i⎞ 1
= lim ∑ ⎜ ⎟
⎝ ⎠ n
n→∞
i =1 n
= lim
n→∞
1
n
4
n
∑ i3
i=1
1 n 2 ( n + 1)
n→∞ n 4
4
= lim
2
=
1
4
1⎡ 1
2
n⎤
+
+ ......... +
⎥=
⎢
n→∞ n
n
n⎦
⎣ n
EX#2 : lim
1
Answer :
∫
x dx
(You have to recognize that the problem is asking for Area from [ 0,1] with n subdivisions)
0
19
Summary of tests for Series
Test
Series
Converges
Diverges
∞
∑a
nth-Term
lim an ≠ 0
n
n =1
n→∞
∞
Geometric Series
∑ ar
r <1
n
Comment
This test cannot be
used to show convergence.
Sum: S =
r ≥1
n=0
∞
Telescoping
∑ (b
n
n =1
∞
− bn +1 )
n =1
p >1
p
n −1
∑ ( −1) an
∞
∑ an
n =1
( an ,bn > 0 )
Limit Comparison
( an ,bn > 0 )
∫ f ( x ) dx converges
1
n
∞
∑a
n
n =1
∫ f ( x ) dx diverges
1
∞
0 < RN < ∫ f ( x ) dx
N
lim n an < 1
n→∞
lim n an > 1
n→∞
a
lim n +1 < 1
n→∞
an
∞
∑a
n
a
lim n +1 > 1
n→∞
an
0 < an ≤ bn and
∞
∑ bn converges
0 < bn ≤ an and
∞
∑ bn diverges
an
=L>0
n→∞ b
n
an
=L>0
n→∞ b
n
n=1
n =1
Remainder:
∞
Test is inconclusive if
lim n an = 1.
Test is inconclusive if
∑a
n =1
∞
n→∞
∞
Direct comparison
R N ≤ aN +1
n→∞
Integral
∞
( f is continuous, ∑ an ,
n =1
positive, and
an = f ( n ) ≥ 0
decreasing)
Ratio
Remainder:
and lim an = 0
n =1
Root
p ≤1
0 < an +1 ≤ an
∞
Alternating Series
Sum: S = b1 − L
lim bn = L
n→∞
1
∑n
p-Series
a
1− r
lim
∞
and ∑ bn converges
n =1
20
n=1
lim
∞
and ∑ bn diverges
n =1
lim
n→∞
an +1
= 1.
an
Power Series for Elementary Functions
Function
1
x
Interval of Convergence
= 1 − ( x − 1) + ( x − 1) − ( x − 1) + ( x − 1) − + ( −1) ( x − 1) + 2
3
4
n
1
1+ x
= 1 − x + x 2 − x 3 + x 4 − x 5 + + ( −1) x n + ln x
= ( x − 1) −
ex
= 1+ x +
n
−1< x <1
n
( x − 1)2 + ( x − 1)3 − ( x − 1)4 + + ( −1)n −1 ( x − 1)n + 2
3
0<x<2
4
n
x2 x3 x4 x5
xn
+ +
+
++
+
2! 3! 4! 5!
n!
0<x≤2
−∞<x<∞
−1) x 2n +1
x3 x5 x7 x9
(
sin x = x − +
−
+
−+
+
3! 5! 7! 9!
( 2n + 1)!
n
−∞<x<∞
x2 x4 x6 x8
( −1) x 2n + cos x = 1 −
+
−
+
−+
2! 4! 6! 8!
( 2n )!
n
−∞<x<∞
x3 x5 x7 x9
( −1) x 2n +1 + arctan x = x − +
−
+
−+
3
5
7
9
2n + 1
n
arcsin x = x +
−1≤ x ≤1
x3
1⋅ 3x 5 1⋅ 3 ⋅ 5x 7
( 2n )!x 2n +1 + +
+
++
2
2⋅3 2⋅4⋅5 2⋅4⋅6⋅7
2 n n! ( 2n + 1)
(
)
k ( k − 1) x 2 k ( k − 1) ( k − 2 ) x 3 k ( k − 1) ( k − 2 ) ( k − 3) x 4
+
+
+
2!
3!
4!
*The convergence at x = ±1 depends on the value of k.
(1 + x )k = 1 + kx +
21
−1≤ x ≤1
− 1 < x < 1*
Procedures for fitting integrands to Basic Rules
Technique
Example
Expand (numerator)
∫ (1 + e )
Separate numerator
∫x
Complete the square
∫
Divide improper rational function
1 ⎞
x2
⎛
∫ x 2 + 1 dx = ∫ ⎜⎝ 1 − x 2 + 1⎟⎠ dx
Add and subtract terms in the numerator
∫x
x 2
dx =
∫ (1 + 2e
x
)
+ e2 x dx
1+ x
x ⎞
⎛ 1
dx = ∫ ⎜ 2
+ 2
dx
2
⎝ x + 1 x + 1 ⎟⎠
+1
1
2x − x
2
2
dx =
∫
1
1 − ( x − 1)
2
(
)
= arcsin ( x − 1) + C
= x − arctan x + C
2x
2
⎛ 2x + 2
⎞
dx = ∫ ⎜ 2
− 2
dx =
⎝ x + 2x + 1 x + 2x + 1 ⎟⎠
+ 2x + 1
tan 2 x = sec 2 x − 1
Multiply and Divide by Pythagorean conjugate
e2 x
+C
2
1
= arctan x + ln x 2 + 1 + C
2
dx
⎛ 2x + 2
2 ⎞
−
⎜
∫ ⎝ x 2 + 2x + 1 ( x + 1)2 ⎟⎠ dx
Use trigonometric Identities
= x + 2e x +
sin 2 x = 1 − cos 2 x
⎛
1
= ln x 2 + 2x + 1 +
1
2
+C
x +1
cot 2 x = csc 2 x − 1
⎞ ⎛ 1 − sin x ⎞
1 − sin x
dx
2
x
∫ 1 + sin x dx = ∫ ⎜⎝ 1 + sin x ⎟⎠ ⎜⎝ 1 − sin x ⎟⎠ dx = ∫ 1 − sin
1 − sin x
sin x ⎞
⎛ 1
dx = ∫ ⎜
−
dx
2
2
⎝ cos x cos 2 x ⎟⎠
cos x
=
∫
=
∫ (sec
2
)
x − sec x tan x dx
= tan x + sec x + C
Happy studying and good luck!
compiled by Kris Chaisanguanthum, class of 97
edited and recompiled by Michael Lee, class of 98
edited and recompiled by Doug Graham, class of Forever
22
VECTOR CALCULUS
A vector has two components: magnitude and direction, expressed as an ordered pair of real numbers:
( a1, a2 ) where a1 = x1 − x0 and a2 = y1 − y0 : ( x0 , y0 ) and ( x1, y1 ) represent the initial and terminal point of the
vector’s directed line segment respectively. θ is the angle between two vectors.
Length (magnitude) of a vector : a = a12 + a2 2
Unit vectors : i = (1, 0)
a + b = ( a1 + b1 , a2 + b2 )
ca = ( ca1, ca2 )
ca = c a
a •b = a
b cosθ
j = (0,1)
a - b = a1 − b1 , a2 − b2
a = a1i + a2 j
a • b = a1b1 + a2b2
2
a×a= a
(
If aib = 0, then a and b are perpendicular
a •b
Angle between two vectors : cosθ = a × b
Projection (of vector b onto a)
aib
prab =
b = prab + pra ' b
2
(a)
)
If b = ca, then a and b are parallel
where a ' is perpendicular to a
Vector - Valued Functions
A vector - valued function has a domain (a set of real numbers) and a rule (which assigns
to each number in the domain a vector)
e.g. F ( t ) = costi + sin tj
If F ( t ) = f1 ( t ) i + f2 ( t ) j
lim F ( t ) exists only if lim f1 ( t ) and f2 ( t ) exist
lim F ( t ) = lim f1 ( t )) i + lim f2 ( t )) j
t →c
t →c
t →c
A vector valued function is continuous at t 0 if its component functions are continuous at t 0
F′ ( t ) = f1′ ( t ) + f2′ ( t )
∫ F (t ) dt = ∫ f (t ) dt + ∫ f (t ) dt
1
2
Motion (defined as a vector valued function of time)
Position:
Velocity:
Speed:
Acceleration:
r (t ) = x (t ) i + y (t ) j
v ( t ) = x ′ ( t ) i + y′ ( t ) j
v (t )
a ( t ) = x ′′ ( t ) i + y′′ ( t ) j
Tangent vector:
T (t ) = r′ (t ) / r′ (t )
Curvature:
k = T′ ( t ) / r ′ ( t )
Normal vector:
N ( t ) = T′ ( t ) / T′ ( t )
23
OTHER DIFFERENTIAL EQUATIONS (Calculus III)
Linear First Order Differential Equations
dy
+ P ( x) y = Q( x)
dx
A linear first order differential equation has this form:
y = e− S ( x ) ∫ eS ( x )Q ( x ) dx
Solve using this equation:
where S ( x ) = ∫ P ( x ) dx
Second Order Linear Differential Equations
A second order differential equation has this general form:
d2y
dy
+ b + cy = g ( x )
2
dx
dx
Homogeneous Equations
A second order linear equation is homogeneous if g ( x ) = 0; thus
d2y
dy
+ b + cy = 0
2
dx
dx
To solve, make use of these equations:
−b + b 2 − 4c
−b − b 2 − 4c
and s1 =
2
2
There are three cases depending on the solution to b 2 − 4c
s1 =
If b 2 − 4c > 0
y = C1e s1 x + C2 es2 x
If b − 4c = 0
y = C1e
s1 x
If b − 4c < 0
y = C1e sin vx + C2 e cos vx where C1 and C2 are constants
2
2
u=−
b
2
+ C2 e
ux
and
v=
where C1 and C2 are constants
s2 x
where C1 and C2 are constants
ux
1
4c − b 2
2
Nonhomogeneous Equations
A second order differential equation is nonhomogeneous if g ( x ) does not equal zero
To solve:
d2y
dy
1. Find the solution for 2 + b + cy = 0
dx
dx
See “Homogeneous Equations”
2. Find y p
y p = u1 y1 + u2 y2
where:
u1′ =
u2′ =
− ( y2 g )
y1 y2′ − y1′ y2
y1g
y1 y2′ − y1′ y2
and y1 and y2 are solutions in Step 1
3. Express the solution as
y = y p + the solution given by Step 1
24