The Calculus BC Bible
the third most important book in the world
to be used in conjunction with The Calculus AB Bible
Applications of the Integral
Length of a Curve
b
L= 
1  ( f '( x)) 2 dx
a
Surface Area
b
A=  2
f ( x) 1  ( f '( x)) 2 dx
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= 
( f '(t )) 2  ( g'(t )) 2 dt
a
Surface Area
b
A =  2
g(t ) ( f '(t )) 2  ( g'(t )) 2 dt
a
Polar Coordinates
Polar coordinates are defined as such:
x = r cos 
y = r sin 
r2 = x2 + y2
tan  =y/x
Area Under a Curve
b
A =  1 f '( )) 2 d
2
a
Area Between Two Curves
b
A =  1 ( f '( )) 2  ( g'( ")) 2 d
2
a


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   vdu
Refer to the Calculus AB Bible for the general technique.
If the derivative of one product will eventually reach zero, use this
technique:
List the term that will reach zero on the left and keep on 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 cos x dx
x
1
0
 x cos x dx
cos x
sin x
-cos x
= x sin x + cos x
Multiply the terms as shown. The first one the left multiplies with the
second on the right and the sign is positive, the second on the left matches
with the third on the right and the sign is negative, and continue on until
the last term on the right has been matched up, alternating signs as you go.
Trigonometric Integrals
 sin
m
x cosn x dx
if n is odd, subsititute u = sin x
Useful Identity: cos2 x = 1 - sin2 x
if m is odd, substitute u = cos x
Useful identity: sin2 x = 1 - cos2 x
if m and n are both even, then reduce so that either m or n are odd,
using trigonometric (double angle) identities.
Example on the next page
EX.
 sin
2
x cos4 x dx
 (sin x cos x)
2
cos2 x dx
2
 sin(2 x) 2   1  cos(2 x) 
  2    2  dx
1
1
sin(2 x ) 2 dx   sin(2 x ) 2 cos(2 x) dx

8
8
 tan
m
x sec n x dx
If n is even, substitute u = tan x
Useful identity: sec2 x = 1 + tan2 x
If m is odd, substitute u = sec x
Useful identity: tan2 x = sec2 x - 1
If m is even and n is odd, reduce powers of sec x alone
Useful identity: tan2 x = sec2 x - 1
Trigonometric Substitutions
If the integrand contains
x = a sin u
a 2  x 2 then substitute
dx = a cos u du
The radicand becomes (a2 - a2 sin2 u) = a2(1-sin2 u) = a2 cos u,
the square root of which is (a cos u)
If the integrand contains
x = a tan u
a 2  x 2 then substitute
dx = a sec2 u du
The radicand becomes (a2 + a2 tan2 u) = a2(1+tan2 u) = a2 sec2 u
the square root of which is (a sec u)
If the integrand containes
x = a sec u
x 2  a 2 then substitute
dx = a sec u tan u du
The radicand becomes (a2 sec2 u - a2) = a2(sec2 u - 1) = a2 tan2 u
the square root of which is (a tan u)
Partial Fractions
Step 1 - If the degree of the numerator is greater than that of the
denominator, divide the numerator by the denominator
2x 3
6x
EX 1. 2
 2x  2
x 3
x 3
Separate the integrand into separate integrals for each term
Step 2 - Factor the numerator and the denominator of remaining fractions
2x  3
2x  3
EX 2. 3

2
x  2 x  x x ( x  1) 2
Step 3 - Separate into partial fractions
Rewrite as fractional terms - arbitrary constants divided by
eachfactor of the denominator. If a factor is repeated more
than once, write that many terms for that factor (Since the
factor (x + 1) is multiplied twice, make two terms for that
factor, the second term with (x + 1)2 being the
denominator)
2x  3
A
B
C
 

2
x x  1 ( x  1) 2
x ( x  1)
Multiply by the common denominator.
2 x  3  A( x  1) 2  Bx( x  1)  Cx
Plug in the roots of the denominator and solve for the constants
3 = A(1) + B(0) + C(0)
x=0
1 = A(0) + B(0) + C(-1)
x = -1
A=3
B = -3
C = -1
Therefore,
2x  3
3
3
1
 

2
x x  1 ( x  1) 2
x ( x  1)
The integral has been reduced down to something much more
reasonable.
Improper Integrals

b
 f ( x) dx  lim  f ( x) dx
b
a


c
f ( x) dx 



a

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.
e
x
dx
1
b
lim  e  x dx
b
1
lim(e  x |1b )
b
lim(e b  e 1 )
b
0  e 1  e 1
Series and Sequences
Taylor's Theorem and Related Series
Taylor's theorem for approximating f(x) to the nth term:
f ''(a )
f n (a )
2
f ( x )  f (a )  f '(a )( x  a ) 
( x  a ) 
( x  a) n
n!
n!
Taylor series of f(x) - this is a power series

f n (a )
( x  a) n

n!
n0
Series for ex:

xn
x2 x3

1

x

 

2! 3!
n  0 n!
Power Series

c
A power series is defined as
a
x n  c0  c1 x  c2 x 2  c3 x 3 
n0
General Series and Sequences

For the sequence a n  n  m , if lim a n exists, then the sequence converges
n
to L.
If lim a n does not exists, then the sequence diverges to  or 
n
Boundness refers to whether or not a sequence has a finite range


If a n  n  m converges, then a n  n  m is bounded
If a n  n  m diverges, then a n  n  m is unbounded


The jth partial sum (sj) refers to the sum of the first j terms of a sequence.

A geometric series
 cr
n
converges iff r  1 .
nm

If a geometric series converges, then
cr m
1 r
 cr n =
nm

If
a


and
n
n1
b
n
both converge, then
 an
n1
n
 bn does also and
n 1
n1

a


n1
n 1
+  bn =  a n  bn
Convergence Tests
Integral Test for Convergence (for nonnegative sequences)

 a n converges iff
n1

 f ( x) dx converges
1
Comparison Test for Convergence and Divergence (for nonnegative
sequences)

If
If

 bn converges and 0  an  bn then
a
n1
n1

 bn diverges and 0  bn  an then
n1
n
converges

a
n1
n
diverges
Limit Comparison Test (for nonnegative sequences)
a
Assume lim n > 0
n  b
n

If
If

 bn converges, then
a
n1

n1
n
converges

 bn diverges, then
a
n1
n1
n
diverges
Ratio Test (for nonnegative sequences)

The sequence
a
n
converges iff each element of the sequence
n1
a n  n1
is smaller then the preceding one, that is,
lim(a n 1 )
n
an
1
Root Test for Convergence (for nonnegative sequences)

The sequence
a
n
converges if lim n a n  1
n
diverges if lim n a n  1
n1

The sequence
a
n
n
n1
If lim n a n  1 , then this test is inconclusive
n
Alternating Series Test for Convergence (for alternating sequences)

The series
  1
n 1
2
a n converges if a n  n1 is a positive

decreasing sequence and lim an  0
n
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.
 is the angle betweem two vectors.
Length (norm) of a vector: a  a1 2  a 2 2
Unit vectors: i = (1,0)
j = (0,1)
a + b = (a1 + b1, a2 + b2)
a - b = (a1 - b1, a2 - b2)
ca = (ca1, ca2)
|| ca || = | c | || a ||
ab = || a || || b || cos 
If ab = 0, then a and b are perpendicular
If b = ca, then a and b are parallel
a = a1i + a2j
ab = a1b1 + a2b2
a x a = || a ||2
cos  =
ab_____
||a|| x ||b||
Projection (of vector b onto a)
a b
pra b =
(||a||) 2
b = prab + pra'b
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) = cos ti + 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  lim f 1 (t ))i  lim f 2 (t )) j
t c
t c
t c
A vector valued function is continuous at t0 if its component functions are
continuous at t0
F'(t) = f1'(t) + f2'(t)
 F(t )dt   f 1 (t )dt   f 2 (t )dt
Motion (defined as a vector valued function of time)
Position:
r(t) = x(t)i + y(t)j
Velocity:
v(t) = x'(t)i + y'(t)j
Speed:
||v(t)||
Acceleration: 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) ||
Differential Equations
Separable Differential Equations
dy
If P( x)  Q( y )
 0 then
dx
 Q( y) dy    P( x) dx
Linear First Order Differential Equations
A linear first order differential equation has this form:
dy
 P( x ) y  Q( x )
dx
Solve using this equation:
where S ( x)   P( x) dx
y  e  S ( x )  e S ( x ) Q( x) dx
Second Order Linear Differential Equations
A second order differential equation has the 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 s2 
s1 
2
2
There are three cases depending on the solution to b2 - 4c
If b2 - 4c > 0
y  C1e s1x  C2 e s2 x
where C1 and C2 are constants
If b2 - 4c = 0
y  C1e s1x  C2 xe s1x
where C1 and C2 are constants
If b2 - 4c < 0
y  C1e ux sin vx  C2 e ux cos vx where C1 and C2 are
constants
b
1
and
u
v
4c  b 2
2
2
Nonhomogeneous Equations
A second order differential equation is nonhomogeneous if g(x)
does not equal zero
To solve:
d2y
dy
b
 cy  0
2
dx
dx
See "Homogeneous Equations"
1. Find the solution for
2. Find yp
yp = u1y1 + u2y2
where:
( y2 g )
y1 y 2 '  y1 ' y 2
y1 g
u2 ' 
y1 y 2 '  y1 ' y 2
and y1 and y2 are solutions in Step 1
u1 ' 
3. Express the solution as
y = yp + the solution given by Step 1
Happy studying and good luck!
compiled by kris chaisanguanthum, class of '97
edited and recompiled by Michael Lee, class of '98