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Notes on
Calculus II
Integral Calculus
Miguel A
...
Integrals
1
...
Areas and Distances
...
2
...
3
...
4
...
5
...
6
...
7
...
8
...
9
...
10
...
Applications of Integration
2
...
More about Areas
2
...
Volumes
2
...
Arc Length, Parametric Curves
2
...
Average Value of a Function (Mean Value Theorem)
2
...
Applications to Physics and Engineering
2
...
Probability

50
50
52
57
61
63
69

Chapter 3
...
1
...
2
...
3
...
Infinite Sequences and Series
4
...
Sequences
4
...
Series
4
...
The Integral and Comparison Tests
4
...
Other Convergence Tests
4
...
Power Series
4
...
Representation of Functions as Power Series
4
...
Taylor and MacLaurin Series
3

83
83
88
92
96
98
100
103

CONTENTS

4
...
Applications of Taylor Polynomials

4

109

Appendix A
...
1
...
Various Formulas
B
...
Summation Formulas

118
118

Appendix C
...
I may keep working on this
document as the course goes on, so these notes will not be completely
finished until the end of the quarter
...
), Brooks/Cole
...

If you find any typos or errors, or you have any suggestions, please,
do not hesitate to let me know
...
math
...
edu/~mlerma/courses/math214-2-02f/

Miguel A
...
northwestern
...
1
...
The Definite Integral
1
...
1
...
The Definite Integral
...

Y

y=f(x)

S

O

a

b

X

In order to estimate that area we begin by dividing the interval
[a, b] into n subintervals [x0 , x1 ], [x1 , x2 ], [x2 , x3 ],
...

Y

O

a

b
6

X

1
...
AREAS AND DISTANCES
...
So the total area under the
curve is approximately the sum
n
X
f (x∗i ) ∆x = f (x∗1 ) ∆x + f (x∗2 ) ∆x + · · · + f (x∗n ) ∆x
...

The estimation is better the thiner the strips are, and we can identify the exact area under the graph of f with the limit:
n
X
A = lim
f (x∗i ) ∆x
n→∞

i=1

As long as f is continuous the value of the limit is independent of the
sample points x∗i used
...
The symbol dx was interpreted as the length of
an “infinitesimal” interval, sort of what ∆x becomes for infinite n
...

Remark : Note that in intervals where f (x) is negative the graph of
y = f (x) lies below the x-axis and the definite integral takes a negative
value
...
e
...

1
...
2
...
We will soon study simple and efficient methods to evaluate integrals, but here we will look at how to
evaluate integrals directly from the definition
...


1
...
AREAS AND DISTANCES
...
Next we must choose some point x∗i in each subinterval
[xi−1 , xi ]
...

Hence the Riemann sum associated to this partition is:
n
n µ ¶2
X
2 + 3/n + 2/n2
i
1 X 2
1 2n3 + 3n2 + n
=

...

n→∞
6
3

x2 dx = lim

0

In order to check that the result does not depend on the sample
points used, let’s redo the computation using now the left endpoint of
each subinterval:
¶2
n µ
n
X
2 − 3/n + 2/n2
i−1
1 X
1 2n3 − 3n2 + n
=

...

n→∞
6
3

x2 dx = lim

0

1
...
3
...
The Midpoint Rule consists of computing Riemann sums using xi = (xi−1 + xi )/2 = midpoint of each
interval as sample point
...

a

i=1

Example: Use the Midpoint Rule with n = 5 to approximate

R1
0

x2 x
...
2], [0
...
4], [0
...
6], [0
...
8],
[0
...
1, 0
...
5, 0
...
9, and ∆x = 1/5, so
Z 1
¤
1 £ 2
x2 dx ≈
0
...
32 + 0
...
72 + 0
...
65/5 = 0
...


1
...
AREAS AND DISTANCES
...
1
...
The Distance Problem
...
In particular
we study the problem of finding the distance traveled by an object with
variable velocity during a certain period of time
...
Otherwise we can approximate the total distance
traveled by dividing the total time interval into small intervals so that
in each of them the velocity varies very little and can can be considered
approximately constant
...
e
...
We divide the time interval into n small
intervals [ti−1 , ti ] of length ∆t = (tend − tstart )/n, choose some instant t∗i
between ti−1 and ti , and take v = f (t∗i ) as the approximate velocity of
the body between ti−1 and ti
...
1
...
Properties of the Definite Integral
...


(2) Linearity:
Z b
Z b
Z b
[f (x) + g(x)] dx =
f (x) dx +
g(x) dx
...


a

1
...
AREAS AND DISTANCES
...


a

Z

f (x) dx = −

b

f (x) dx =

10

b

f (x) dx,

a

f (x) dx = 0
...

Z
a

b

Z
f (x) dx ≥

(c) m ≤ f (x) ≤ M ⇒ m (b − a) ≤

b

g(x) dx
...


1
...
THE EVALUATION THEOREM

11

1
...
The Evaluation Theorem
1
...
1
...
If f is a continuous function
and F is an antiderivative of f , i
...
, F 0 (x) = f (x), then
Z b
f (x) dx = F (b) − F (a)
...


Answer : An antiderivative of x2 is x3 /3, hence:
· 3 ¸1
Z 1
1
13 03
x
2
−
=
...
2
...
Indefinite Integrals
...
e
...
The family of all antiderivatives of f is called
indefinite integral of f and represented:
Z
f (x) dx = F (x) + C
...

3

1
...
3
...
We can make an integral
table just by reversing a table of derivatives
...

(1)
xn dx =
n+1
Z
1
dx = ln |x| + C
...

Z
ax
+ C
...

Z
(6)
cos x dx = sin x + C
...


1
...
THE EVALUATION THEOREM

Z
(8)
(9)
(10)

Z
Z
Z

12

csc2 x dx = − cot x + C
...

csc x cot x dx = − csc x + C
...

+
1
Z
dx
√
(12)
= sin−1 x + C
...

2
x x −1
(11)

x2

1
...
4
...
The integral of a rate of change
is the total change:
Z b
F 0 (x) dx = F (b) − F (a)
...

As an example of application we find the net distance or displacement, and the total distance traveled by an object that moves along a
straight line with position function s(t)
...
The net distance or displacement is the difference between
the final and the initial positions of the object, and can be found with
the following integral
Z t2
v(t) dt = s(t2 ) − s(t1 )
...
If we want to find
the total distance traveled we need to add all distances with a positive
sign, and this is accomplished by integrating the absolute value of the
velocity:
Z t2
|v(t)| dt = total distance traveled
...


1
...
THE EVALUATION THEOREM

13

Answer : The displacement is
· 3
¸4
Z 4
t2
t
2
(t − t − 6) dx =
− − 6t
3
2
1
1
¶ µ 3
¶
µ 3
42
12
1
4
−
−6·4 −
−
−6
=
3
2
3
2
µ
¶
37
32
9
= −
=− − −
3
6
2
In order to find the total distance traveled we need to separate the
intervals in which the velocity takes values of different signs
...
e
...
Since we are interested only in what happens in [1, 4]
we only need to look at the intervals [1, 3] and [3, 4]
...
Hence:
Z 3
Z 4
Z 4
|v(t)| dt =
[−v(t)] dt +
v(t) dt
1

Z

1

2

=
1

·

3

3

(t − t − 6) dt +

t3 t2
= − + + 6t
3
2

¸3

=

Z

4
3

·

(t2 − t − 6) dt
¸4

t3 t2
− − 6t
+
3
2
1

22 17
61
+
=

...
3
...
3
...
3
...
The Fundamental Theorem of Calculus
...
It says the following:
Suppose f is continuous on [a, b]
...
e
...

(2) (Evaluation Theorem) If F is an antiderivative of f , i
...
F 0 (x) =
f (x), then
Z b
f (x) dx = F (b) − F (a)
...

(1)
dx a
Z b
(2)
F 0 (x) dx = F (b) − F (a)
...
e
...

Z x2
d
t3 dt
...
First directly:
· 4 ¸ x2
Z x2
(x2 )4
x8
t
3
t dt =
=
=
,
g(x) =
4 0
4
4
0
hence
g 0 (x) =

8x7
= 2x7
...
3
...


2

Now we have g(x) = h(x ), hence (using the chain rule):
g 0 (x) = h0 (x2 ) · 2x = (x2 )3 · 2x = 2x7
...
4
...
4
...
4
...
The Substitution Rule
...
It is based on the following identity between
differentials (where u is a function of x):
du = u0 dx
...

Example: Find

Z √

f (u) du

1 + x2 2x dx
...


Most of the time the only problem in using this method of integration is finding the right substitution
...

R
Answer : We want to write the integral as cos u du, so cos u =
cos 2x ⇒ u = 2x, u0 = 2
...
4
...

2

In general we need to look at the integrand as a function of some
expression (which we will later identify with u) multiplied by the derivative of that expression
...

Answer : We see that x is “almost”, the derivative of −x2 , so we use
the substitution u = −x2 , u0 = −2x, hence in order to get u0 inside the
integral we do the following:
Z

−x2

e

1
x dx = −
2
1
=−
2

Z

Z

2

e−x (−2x) dx
|{z} | {z }
eu

du

2

eu du = − 12 eu + C = − 12 e−x + C
...

Z
Example: Find tan x dx
...
4
...

In general we need to identify inside the integral some expression of
the form f (u) u0 , where f is some function with a known antiderivative
...

Example: Find
e2x + 1
Answer : Let’s write
Z

ex
dx = k
e2x + 1

Z

f (u) u0 dx

(where k is some constant to be determined later) and try to identify
the function f , the argument u and its derivative u0
...

There is no much more that can be said in general, the way to learn
more is just to practice
...
4
...
Other Changes of Variable
...

Z √
Example: Find
1 − x2 dx
...

| {z} | {z }
2

x

dx

1
...
THE SUBSTITUTION RULE

19

Since we do not know yet how to integrate cos2 t we leave it like this
and will be back to it later (after we study integrals of trigonometric
functions)
...
4
...
The Substitution Rule for Definite Integrals
...

a

(2) Use the substitution rule for definite integrals:
Z b
Z u(b)
0
f (u) u dx =
f (u) du
...

Z 4
√
Example: Find
2x + 1 dx
...

3
Then we use it for computing the definite integral:
·
¸4
Z 4
√
1
1
27 1
26
1
3/2
2x + 1 dx = (2x + 1)
= 93/2 − 13/2 =
− =

...
4
...

− =
3
3
3

1
...
INTEGRATION BY PARTS

21

1
...
Integration by Parts
The method of integration by parts is based on the product rule for
differentiation:
[f (x)g(x)]0 = f 0 (x)g(x) + f (x)g 0 (x) ,
which we can write like this:
f (x)g 0 (x) = [f (x)g(x)]0 − f 0 (x)g(x)
...
e
...


Writing u = f (x), v = g(x), we have du = f 0 (x) dx, dv = g 0 (x) dx,
hence:
Z
Z
u dv = uv − v du
...


Answer : In integration by parts the key thing is to choose u and
dv correctly
...
We see that the choice is right because the new
integral that we obtain after applying the formula of integration by
parts is simpler than the original one:
Z
Z
x
x
x e dx = x e −
ex dx = xex − ex + C
...

A couple of additional typical examples:
Z
Example: x sin x dx = · · ·

1
...
INTEGRATION BY PARTS

22

u = x, dv = sin x dx, so du = dx, v = − cos x:
Z
Z
··· =
x sin x dx = x (− cos x) − (− cos x) dx
|{z} | {z } |{z} | {z }
| {z } |{z}
u

u

dv

v

v

du

= −x cos x + sin x + C
...

Sometimes we need to use the formula more than once
...

u = x2 , dv = ex dx, so du = 2x dx, v = ex :
Z
Z
2
x
2 x
··· =
x e dx = x e − ex 2x dx =
...

In the following example the formula of integration by parts does
not yield a final answer, but an equation verified by the integral from
which its value can be derived
...


1
...
INTEGRATION BY PARTS

23

u = sin x, dv = ex dx, so du = cos x dx, v = ex :
Z
Z
x
x
· · · = sin x e dx = sin x · e − ex cos x dx =
...
= sin x · e − cos x ex dx
| {z } | {z }
u
dv
Z
x
x
= sin x · e − cos x · e − ex sin x dx
Hence the integral I =

R

sin x ex dx verifies

I = sin x · ex − cos x · ex − I ,
i
...
,

2I = sin x · ex − cos x · ex ,

hence

I = 12 ex (sin x − cos x) + C
...
5
...
Integration by parts for Definite Integrals
...

a

a

Z

1

Example:
0

tan−1 x dx = · · ·

1
u = tan−1 x, dv = dx, so du =
dx, v = x:
1 + x2
Z 1
Z 1
1
−1
−1
1
··· =
tan x dx = [ tan x · x ]0 −
x
dx
1 + x2
0 | {z } |{z}
0 |{z} |
| {z } |{z}
{z }
u

dv

u

v

v

£
¤
= tan−1 1 · 1 − tan−1 0 · 0 −
Z 1
π
x
= −
dx
2
4
0 1+x

Z
0

du

1

x
dx
1 + x2

1
...
INTEGRATION BY PARTS

24

The last integral can be computed with the substitution t = 1 + x2 ,
dt = 2x dx:
Z

1
0

1
x
dx
=
1 + x2
2

Z

2
1

ln 2
1
1
dt = [ln t]21 =

...

4
2

1
...
2
...
Assume that we want to find the
following integral for a given value of n > 0:
Z

xn ex dx
...


On the right hand side we get an integral similar to the original one
but with x raised to n−1 instead of n
...
Using this same formula several
taking
R xtimes, and
x
into account that for n = 0 the integral becomes e dx = e + C, we
can evaluate the original integral for any n
...


1
...
INTEGRATION BY PARTS

25

Another example:
Z
Z
n
sin x dx =
sinn−1 x sin x dx
| {z } | {z }
u

n−1

= − sin

dv

Z

x cos x + (n − 1)

= − sinn−1 x cos x + (n − 1)
− (n − 1)

Z
Z

cos2 x sinn−2 dx
| {z }

1−sin2 x

sinn−2 dx
sinn x dx

Adding the last term to both sides and dividing by n we get the following reduction formula:
Z
Z
sinn−1 x cos x n − 1
n
sin x dx = −
+
sinn−2 x dx
...
6
...
SUBSTITUTIONS

26

1
...
Trigonometric Integrals and Trigonometric
Substitutions
1
...
1
...
Here we discuss integrals of powers of trigonometric functions
...

2
Remember also the identities:
sin2 x + cos2 x = 1 ,
sec2 x = 1 + tan2 x
...
6
...
1
...
We will study
now integrals of the form
Z
sinm x cosn x dx ,
including cases in which m = 0 or n = 0, i
...
:
Z
Z
n
sinm x dx
...

Z
Example: sin4 x cos x dx = · · ·
(u = sin x, du = cos x dx)
Z
··· =

u5
sin5 x
u du =
+C =
+C
...


1
...
TRIGONOMETRIC INTEGRALS AND TRIG
...

3
5

=

If all the exponents are even then we use the half-angle identities
...

8
32

1
...
1
...
Integrals of Secants and Tangents
...
Analogously
Z
Z
Z
cos x
du
cot x dx =
dx =
= ln |u| + C = ln | sin x| + C ,
sin x
u
where u = sin x
...
6
...
SUBSTITUTIONS

28

The integral of sec x is a little tricky:
Z

Z
sec x dx =

Z
sec x tan x + sec2 x
sec x (tan x + sec x)
dx =
dx =
sec x + tan x
sec x + tan x
Z
du
= ln |u| + C = ln | sec x + tan x| + C ,
u

where u = sec x + tan x, du = (sec x tan x + sec2 x) dx
...

More generally an integral of the form
Z
tanm x secn x dx
can be computed in the following way:
(1) If m is odd, use u = sec x, du = sec x tan x dx
...

Z
Example:

tan3 x sec2 x dx = · · ·

Since in this case m is odd and n is even it does not matter which
method we use, so let’s use the first one:
(u = sec x, du = sec x tan x dx)
Z
Z
2
··· =
tan x sec x tan x sec x dx = (u2 − 1)u du
| {z } | {z } |
{z
}
u
du
u2 −1
Z
= (u3 − u) du
=
=

u4 u2
−
+C
4
2
1
4

sec4 x − 12 sec2 x + C
...
6
...
SUBSTITUTIONS

(u = tan x, du = sec2 x dx)
Z
Z
u4
3
2
+C =
tan x sec x dx = u3 du =
4
| {z } | {z }

1
4

29

tan4 x + C
...

{z
}
|
previous answer

1
...
2
...
Here we study substitutions of the form x = some trigonometric function
...

Answer : We make x = sin t, dx = cos t dt, hence
p
√
√
1 − x2 = 1 − sin2 t = cos2 t = cos t ,
and
Z √

Z
1−

x2

dx =

cos t cos t dt
Z

=
Z
=

cos2 t dt
1
(1
2

+ cos 2t) dt

(half-angle identity)

t sin 2t
+
+C
2
4
t 2 sin t cos t
= +
+C
(double-angle identity)
2
4
p
t sin t 1 − sin2 t
= +
+C
2
2
√
sin−1 x x 1 − x2
=
+
+C
...
6
...
SUBSTITUTIONS

expression substitution

30

identity

a2 − u2

u = a sin t

1 − sin2 t = cos2 t

a2 + u2

u = a tan t

1 + tan2 t = sec2 t

u2 − a2

u = a sec t

sec2 t − 1 = tan2 t

So for instance, if an integral contains the expression a2 −u2 , we may
try the substitution u = a sin t and use the identity 1 − sin2 t = cos2 t
in order to transform the original expression in this way:

a2 − u2 = a2 (1 − sin2 t) = a2 cos2 t
...


= 27

where x = 3 sin t, dx = 3 cos t dt
...
6
...
SUBSTITUTIONS

Example:
Z √
Z q
9
2
9 + 4x dx = 2
+ x2 dx
(x = 32 tan t)
4
Z p
3
3
=2
1 + tan2 t sec2 t dt
2
2
Z
9
=
sec3 t dt
2
9
= (sec t tan t + ln | sec t + tan t|) + C1
4µ
¯¶
¯
q
q
¯
¯
9 2
4 2
¯ 2 x + 1 + 4 x2 ¯ + C1
=
x
1
+
x
+
ln
9
9
¯
¯3
4 3
√
√
x 9 + 4x2 9
=
+ ln |2x + 9 + 4x2 | + C
...

where x = sec t, dx = sec t tan t dt
...
7
...
7
...
7
...
Rational Functions and Partial Fractions
...

Q(x)

Here we discuss how to integrate rational functions
...
This can be done in the following way:
(1) Use long division of polynomials to get a quotient p(x) and a
remainder r(x)
...

(2) Factor the denominator Q(x) = q1 (x)q2 (x)
...

(3) Decompose r(x)/Q(x) into partial fractions of the form:
r(x)
= F1 (x) + F2 (x) + F3 (x) + · · ·
Q(x)
where each fraction is of the form
A
Fi (x) =
(ax + b)k
or
Ax + B
Fi (x) =
,
2
(ax + bx + c)k
where 1 ≤ k ≤ n (n is the exponent of ax + b or ax2 + bx + c
in the factorization of Q(x)
...
7
...

A
B
x+3
=
+

...


(∗)

Now there are two ways of finding A and B:
Method 1
...
In (*) give x two different values (as many as the number
of coefficients to determine), say x = 1 and x = −1
...

(x + 1)(x − 1)
x+1 x−1
Finally:
x3 + x2 + 2
1
2
=x+1−
+

...
7
...
Factoring a Polynomial
...

Then for each real root r write a factor of the form (x − r)k where k is
the multiplicity of the root
...
7
...
Finally multiply by the leading coefficient
of Q(x)
...
In that
case we get as many roots as we can, and divide Q(x) by the factors
found
...
So pose the algebraic equation
q(x) = 0
and try to solve it for this new (and simpler) polynomial
...

Answer : The roots of Q(x) are 1 (simple), −2 (triple) and 3 (double), hence:
Q(x) = (x − 1)(x + 2)3 (x − 3)2
...

Answer : Q(x) has a simple real root x = −1
...

1
...
3
...
Assume that Q(x)
has already been factored and degree of r(x) is less than degree of Q(x)
...
Ak are coefficients to be determined
...
Bk and C1
...


1
...
PARTIAL FRACTIONS

35

(3) Multiply by Q(x) and simplify
...

Finally determine the coefficients Ai , Bi , Ci
...
Another way is to write a
system of equations with unknowns Ai , Bi , Ci by giving x
various values
...

(x − 1)2 (x2 + x + 1)2
Answer : The denominator is already factored, so we proceed with
the next step:
4x5 − 2x4 + 2x3 − 8x2 − 2x − 3
=
(x − 1)2 (x2 + x + 1)2
B
Cx + D
Ex + F
A
+
+ 2
+ 2

...

Identifying coefficients

A
+



A + B −



A + 2B
−A + 3B −





 −A + 2B +
−A + B

on both sides we get:
C
C +
−
C
C −
+

=
D
=
D + E
=
− 2E + F =
D + E − 2F =
D
+ F =

4
−2
2
−8
−2
−3

1
...
PARTIAL FRACTIONS

36

The solution to this system of equations is A = 2, B = −1, C = 2, D =
−1, E = 1, F = 1, hence:
4x5 − 2x4 + 2x3 − 8x2 − 2x − 3
=
(x − 1)2 (x2 + x + 1)2
1
x+1
2
2x − 1
−
+ 2
+ 2
2
x − 1 (x − 1)
x + x + 1 (x + x + 1)2
1
...
4
...
After decomposing
the rational function into partial fractions all we need to do is to integrate expressions of the form A/(x − r)k and (Bx + C)/(ax2 + bx + c)k
...

x2 − 4x + 4

Answer : First we decompose the integrand into partial fractions:
(1)

x3 − x2 − 7x + 8
x−4
=x+3+ 2
2
x − 4x + 4
x − 4x + 4

(2) x2 − 4x + 4 = (x − 2)2
...
7
...

2
x − 4x + 4
x − 2 (x − 2)2
Z

Finally we integrate:
x3 − x2 − 7x + 8
dx =
(x − 2)2
=

Z

Z
(x + 3) dx +

1
dx −
x−2

Z

2
dx
(x − 2)2

2
x2
+ 3x + ln |x − 2| +
+C
...
7
...
Completing the Square
...
e
...
g
...

In general:

µ
2

ax + bx + c =

√

b
x a+ √
2 a

¶2
−

b2 − 4ac

...

+ c−
2
4
This result is of the form u2 ± A2 , where u = x + b/2
...


(u = x + 3)

1
...
PARTIAL FRACTIONS

Example:
Z √
Z
2
x − 4x + 5 dx =
Z
=
Z
=
Z
=

p
√

(x − 2)2 + 1 dx

u2 + 1 du

p

tan2 t + 1 · sec2 t dt

sec3 t dt

sec t tan t 1
+ ln | sec t + tan t| + C
2
√2
2
√
u u +1 1
+ ln |u + u2 + 1| + C
=
2 √
2
2
(x − 2) x − 4x + 5
=
2
¯
√
1 ¯¯
¯
+ ln ¯(x − 2) + x2 − 4x + 5¯ + C
...
8
...
8
...
However we often need to modify slightly the
original integral and perhaps complete or simplify the answer
...

x
Answer : In the tables we find the following formula No
...

x
|x|
Example: Find the integral:
Z
x2
√
dx = · · ·
9 + 4x2
Answer : In the tables the formula that resembles this integral most
is No
...

=
9 + 4x2 −
8
16
Example: Find the same integral using Maple
...
8
...
On the other
hand, it involves an inverse hyperbolic function:
³
´
√
arcsinh x = ln x + 1 + x2 ,
hence the answer provided by Maple is:
Ã
!
r
2
2x
9
4x
x√
ln
+ 1+
=
9 + 4x2 −
8
16
3
9
´
√
9 ³
9
x√
9 + 4x2 −
ln 2x + 9 + 4x2 +
ln(3) ,
8
16
32
9
so it differs from the answer found using the tables in a constant 32
ln(3)
which can be absorbed into the constant of integration
...
9
...
9
...
e
...
However, in those cases we
still can find an approximate value for the integral of a function on an
interval
...
9
...
Trapezoidal Approximation
...

i=1

Where ∆x = xi − xi−1 = (b − a)/n and x∗i is some point in the interval
[xi−1 , xi ]
...

i=1

The trapezoidal approximation is the average of Ln and Rn :
1
∆x
{f (x0 )+2f (x1 )+2f (x2 )+· · ·+2f (xn−1 )+f (xn )}
...


R1
0

x2 dx with trapezoidal approximation us-

Solution: We have ∆x = 1/4 = 0
...
The values for xi and f (xi ) =
x2i can be tabulated in the following way:
i
0
1
2
3
4

xi
0
0
...
5
0
...
0625
0
...
5625
1

1
...
NUMERICAL INTEGRATION

42

Hence:
L4 = 0
...
0625 + 0
...
5625) = 0
...
25 · (0
...
25 + 0
...
468750
...
218750 + 0
...
34375
...
3333
...
9
...
Midpoint Approximation
...
Then the midpoint approximation of a f (x) dx is
Mn = (∆x){f (x1 ) + f (x2 ) + · · · + f (xn )}
...


R1
0

x2 dx with midpoint approximation using

Solution: We have:
i
1
2
3
4

xi
0
...
375
0
...
875

f (xi )
0
...
140625
0
...
765625

Hence:
M4 = 0
...
015625 + 0
...
390625 + 0
...
328125
...
9
...
Simpson’s Approximation
...
, [xn−2 , xn ] (of length

1
...
NUMERICAL INTEGRATION

43

2∆x each):
1
S2n = (2Mn + Tn )
3·
1
2(2∆x){f (x1 ) + f (x3 ) + · · · + f (x2n−1 )}
=
3
¸
2∆x
{f (x0 ) + 2f (x2 ) + 2f (x4 ) + · · · + 2f (xn−2 ) + f (xn )}
+
2
∆x
{f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + 2f (x4 ) + · · ·
=
3
+ 2f (x2n−2) + 4f (x2n−1 ) + f (x2n )}
...


R1
0

x2 dx with Simpson’s approximation us-

Solution: We use the previous results and get:
1
1
S8 = (2M4 + T4 ) = (2 · 0
...
34375) = 1/3
...

1
...
4
...
Here we give a way to estimate the error
or difference E between the actual value of an integral and the value
obtained using a numerical approximation
...
9
...
1
...
Suppose
|f 00 (x)| ≤ K for a ≤ x ≤ b
...

12n2

1
...
4
...
Error Bound for the Midpoint Approximation
...
Then the error EM in the trapezoidal approximation
verifies:
K(b − a)3
|EM | ≤

...
9
...
3
...
Suppose |f (4) (x)| ≤
K for a ≤ x ≤ b
...

180n4

1
...
NUMERICAL INTEGRATION

44

Example: Approximate the value of π using the trapezoidal, midpoint and Simpson’s approximations of
Z 1
4
dx
2
0 1+x
for n = 4
...

Answer : First note that:
Z 1
£ −1 ¤1
π
1
dx
=
4
tan x 0 = 4 = π ,
4
2
4
0 1+x
so by approximating the given integral we are in fact finding approximated values for π
...
131176470
...
1666
...
146800518
...
08333
...
9
...
141568627
For the error estimate we now need the fourth derivative:
f (4) (x) = 96(5x4 − 10x2 + 1)/(1 + x2 )5 ,
so

96(5 + 10 + 1)
= 1536
1
for 0 ≤ x ≤ 1
...
0333
...
10
...
10
...
10
...
Improper Integrals
...
Improper integrals are integrals in which one or both of
these conditions are not met, i
...
,
(1) The interval of integration is not bounded:
[a, +∞) ,
e
...
:

(−∞, a] ,
Z

(−∞, +∞) ,

∞

1
dx
...

x→c
e
...
:
Z 1
1
√ dx
...
10
...
Infinite Limits of Integration
...
In case one of the limits of integration is infinite, we define:
Z ∞
Z t
f (x) dx = lim
f (x) dx ,
or

Z

t→∞

a
a

a

Z
f (x) dx = lim

t→−∞

−∞

a

f (x) dx
...

−∞

−∞

c

If the limits defining the integral exist the integral is called convergent, otherwise it is called divergent
...

t→∞

1
...
IMPROPER INTEGRALS

Analogously:

47

[F (x)]a−∞ = lim [F (x)]at ,
t→−∞

and
c
∞
c
t
[F (x)]∞
−∞ = [F (x)]−∞ + [F (x)]c = lim [F (x)]t + lim [F (x)]c
...

t→∞
x2
x 1
t
1
Example: For what values of p is the following integral convergent?:
Z ∞
1
dx
...
Now suppose p 6= 1:
½
· −p+1 ¸t
¾
Z t
1
1
1
x
dx =
=
−1
p
−p + 1 1 1 − p tp−1
1 x
If p > 1 then p − 1 > 0 and
½
¾
Z ∞
1
1
1
dx = lim
− 1 = 0,
t→∞ 1 − p
xp
tp−1
1
hence the integral is convergent
...
So:
Z ∞
1
dx is convergent if p > 1 and divergent if p ≤ 1
...
10
...
10
...
Infinite Integrands
...

Assume f is defined in [a, b) but
lim f (x) = ±∞
...


a

Analogously, if f is defined in (a, b] but
lim f (x) = ±∞
...


t

Finally, if f (x) has an infinite discontinuity at c inside [a, b], then
the definition is
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
...

Remark : If the interval of integration is [a, b) sometimes we write
[F (x)]ba as an abbreviation for lim− [F (x)]ta —and analogously for intert→b

vals of the form (a, b]
...

t→0−
x
0
Z
Example: Evaluate

1

ln x dx
...
10
...
Hence:
x→0
Z 1
Z 1
ln x dx = lim+
ln x dx
t→0

0

t

= lim+ [x ln x − x]1t
t→0

= lim+ {(1 ln 1 − 1) − (t ln t − t)}
t→0

= lim+ {t − 1 − t ln t}

( lim+ t ln t = 0)

t→0

t→0

= −1
...
10
...
Comparison Test for Improper Integrals
...

R∞
R∞
(1) If Ra f (x) dx if convergent thenR a g(x) dx is convergent
...

A similar statement holds for type 2 integrals
...

0

Answer : We have:
Z
Z ∞
−x2
e
dx =
0

1

−x2

e
0

Z
dx +

∞

2

e−x dx
...
In fact,
2
for x ≥ 1 we have x2 ≥ x, so e−x ≤ e−x
...
Hence, by the comparison theorem
is convergent, QED
...
1
...
1
...
Area Between Two Curves
...

A=
a

a

a

Calling yT = f (x), yB = g(x), we have:
Z b
A=
(yT − yB ) dx
a

Example: Find the area between y = ex and y = x bounded on the
sides by x = 0 and x = 1
...
So:
¸1 µ
¶ µ
¶
·
Z 1
x2
12
02
x
x
1
0
− e −
A=
(e − x) dx = e −
= e −
2 0
2
2
0
= e − 12 − 1 = e −

3
2


...
First find the
intersection points a and b by solving the equation f (x) = g(x)
...

a

a

a

If the result is negative that means that we have subtracted wrong
...

50

2
...
MORE ABOUT AREAS

51

Example: Find the area between y = x2 and y = 2 − x
...

We get x = −2 and x = 1
...

−2

−2

Hence the area is 9/2
...
If xL (y) ≤ xR (y) for p ≤ y ≤ q, then the
area between the graphs of x = xL (y) and x = xR (y) and the horizontal
lines y = p and y = q is:
Z q
A=
(xR − xL ) dy
p

Example: Find the area between the line y = x−1 and the parabola
y 2 = 2x + 6
...
In this case it is easier to write
xL = 12 y 2 − 3 ,

xR = y + 1 ,

and integrate from y = −2 to y = 4:
Z 4
Z 4
©
ª
A=
(xR − xL ) dx =
(y + 1) − ( 12 y 2 − 3) dx
−2

−2
4

Z
=
·

−2

¡ 1 2
¢
− 2 y + y + 4 dx

y3 y2
+ 4y
= − +
6
2
= 18

¸4
−2

2
...
VOLUMES

52

2
...
Volumes
2
...
1
...
First we study how to find the volume
of some solids by the method of cross sections (or “slices”)
...

The volume of the solid is the sum of the volumes of its slices
...
2
...
Volume of Cylinders
...
The volume of a cylinder is the product of its height and the area of its base:
V = Ah
...
2
...
Volume by Cross Sections
...
For each x in [a, b] we denote A(x)
the area of the cross section of the solid by a plane perpendicular to
the x-axis at x
...
The planes that are perpendicular to the
x-axis at the points x0 , x1 , x2 ,
...
If
the cross section of R changes little along a subinterval [xi−1 , xi ], the
slab positioned alongside that subinterval can be considered a cylinder
of height ∆x and whose base equals the cross section A(x∗i ) at some
point x∗i in [xi−1 , xi ]
...

The total volume of the solid is
n
n
X
X
V =
∆Vi ≈
A(x∗i ) ∆x
...
In the limit as
n → ∞ we get the so called Cavalieri’s principle:
Z b
V =
A(x) dx
...
For instance
if a solid lies alongside some interval [a, b] on the y axis, the formula
becomes
Z
b

V =

A(y) dy
...


2
...
VOLUMES

53

Answer : Assume that the cone is placed with its vertex in the origin
of coordinates and its axis on the x-axis
...
The cross section of the cone at each point x is a
circular disk of radius xr/h, hence its area is A(x) = π(xr/h)2 =
πr2 x2 /h2
...

2
2
2
h
h
3 0
h 3
3
0
2
...
4
...
Consider the plane region between
the graph of the function y = f (x) and the x-axis along the interval
[a, b]
...
Now each cross section is a circular disk of radius y, so its
area is A(x) = πy 2 = π[f (x)]2
...

a

a

Example: Find the volume of a cone of radius r and height h
...
This cone can be obtained by
revolving the area under the line y = rx/h between x = 0 and x = h
around the x-axis
...

2
2
2
h
h
h
3 0
h 3
3
0
0
If the revolution is performed around the y-axis, the roles of x and
y are interchanged, so in that case the formula is
Z b
πx2 dy ,
V =
a

where x must be written as a function of y
...
)
The area of the annular ring is A(x) = π(f (x)2 − g(x)2 ), hence the
volume of the solid will be:
Z b
Z b
£
¤
£
¤
2
2
V =
π (yT ) − (yB ) dx =
π f (x)2 − g(x)2 dx
...
2
...

a

Example: Find the volume√of the solid obtained by revolving the
area between y = x2 and y = x around the x-axis
...

10

2
...
5
...
Next we study how to find the volume
of some solids by the method of shells
...

2
...
6
...
A cylindrical shell is the
region between two concentric circular cylinders of the same height h
...

2
...
7
...
Consider the solid generated by revolving around the y-axis the region under the graph of
y = f (x) between x = a and x = b
...
The volume V of
the solid is the sum of the volumes ∆Vi of the shells determined by the
partition
...
Its height

2
...
VOLUMES

55

is f (x∗i ), where x∗i is the midpoint of [xi−1 , xi ]
...
Its
average radius is x∗i
...

V =
i=1

i=1

As n → ∞ the right Riemann sum converges to the following integral:
Z b
Z b
V =
2πxf (x) dx =
2πxy dx
...

Answer : The graph intersects the positive x-axis at x = 1, so the
interval is [0, 1]
...

2

2
...
8
...
Here we
find the volume of the solid obtained by revolving around the y-axis
the area between two curves y = f (x) and y = g(x) over an interval
[a, b]
...

a

a

Example: Find the volume of the solid obtained by revolving the
plane region limited by the curves y = x and y = x2 over the interval
[0, 1]
...

12
6

2
...
VOLUMES

56

If the region is revolved around the x-axis then the variables x and
y reverse their roles:
Z b
V =
2πy (xR − xL ) dy
...
2
...
Revolving Around an Arbitrary Line
...

V =
a

a

Similarly, if the region is revolved around the horizontal line x = c,
the formula becomes:
Z b
Z b
V =
2π(y − c)(f (y) − g(y)) dy =
2π(y − c)(xR − xL ) dy ,
a

a

where y − c must be replaced by c − y if c > y
...
3
...
3
...
3
...
Parametric Curves
...
For each value of t we get a point of the curve
...


The equations x = f (t), y = g(t) are called parametric equations
...
For instance
t can be eliminated from x = cos t, y = sin t by using the trigonometric
relation cos2 t + sin2 t = 1, which yields the (non-parametric) equation
for a circle of radius 1 and center (0, 0):
x2 + y 2 = 1
...

Answer : We eliminate t by isolating it from the second equation:
t = (y − 1) ,
and plugging it in the first equation:
x = (y − 1)2 − 2(y − 1)
...
e
...


2
...
2
...
Here we describe how to find the length of a
smooth arc
...

If the arc is just a straight line between two points of coordinates
(x1 , y1 ), (x2 , y2 ), its length can be found by the Pythagorean theorem:
p
L = (∆x)2 + (∆y)2 ,
where ∆x = x2 − x1 and ∆y = y2 − y1
...
3
...
Assume that the arc is given by the parametric
functions x = f (x), y = g(x), a ≤ t ≤ b
...
The
corresponding points in the arc have coordinates (f (ti ), g(ti )), so two
consecutive points are separated by a distance equal to
p
Li = [f (ti ) − f (ti−1 )]2 + [g(ti ) − g(ti−1 )]2
...
On the other hand, by the mean
value theorem
f (ti ) − f (ti−1 ) = f 0 (t∗i ) ∆t
g(ti ) − f (ti−1 ) = g 0 (t∗i ) ∆t
for some t∗i in [ti−1 , ti ]
...

The total length of the arc is
L≈

n
X

si =

i=1

n
X
p

[f 0 (t∗i )]2 + [g 0 (t∗i )]2 ∆t ,

i=1

which converges to the following integral as n → ∞:
Z bp
L=
[f 0 (t)]2 + [g 0 (t)]2 dt
...

ds = (dx) + (dy) = dt
dt
dt
Example: Find the arc length of the curve x = t2 , y = t3 between
(1, 1) and (4, 8)
...
3
...

=
27
2

=

(u = 4 + 9t2 )

In cases when the arc is given by an equation of the form y = f (x)
or x = f (x) the formula becomes:
Z

b

L=

s

µ

1+

a

dy
dx

¶2
dx

or
Z
L=
a

b

sµ

dx
dy

¶2
+ 1 dy

Example: Find the length of the arc defined by the curve y = x3/2
between the points (0, 0) and (1, 1)
...
3
...

=
27
27
0
µ

60

2
...
AVERAGE VALUE OF A FUNCTION (MEAN VALUE THEOREM)

61

2
...
Average Value of a Function (Mean Value Theorem)
2
...
1
...
The average value of finitely
many numbers y1 , y2 ,
...

n

The average value has the property that if each of the numbers y1 , y2 ,
...

a

a

Hence:
fave

1
=
b−a

Z

b

f (x) dx
...
4
...
The Mean Value Theorem for Integrals
...
e
...


Example: Assume that in a certain city the temperature (in ◦ F) t
hours after 9 A
...
is represented by the function
T (t) = 50 + 14 sin

πt

...
M
...
M
...
4
...
9
...
5
...
5
...
5
...
Work
...
If the force is constant, the work done
is the product
W = F · d
...
In the American system a unit of work is the foot-pound
...
224809 lb and 1 m = 3
...
737561 ft lb
...
Let F (x) be the force function
...
In order to
find the total work done by the force we divide the interval [a, b] into
small subintervals [xi−1 , xi ] so that the change of F (x) is small along
each subinterval
...
So,
the total work is
n
n
X
X
W =
∆Wi ≈
F (x∗i ) ∆x
...

a

2
...
2
...
Consider a spring on the x-axis so that its
right end is at x = 0 when the spring is at its rest position
...
e
...

The energy needed to stretch the spring from 0 to a is then the
integral
Z a
a2
W =
kx dx = k
...
5
...
5
...
Work Done Against Gravity
...

The energy needed to lift a body from a point at distance R1 from
the center of the Earth to another point at distance R2 is given by the
following integral
·
¸R
µ
¶
Z R2
k 2
1
k
1
W =
dr = −
=k
−

...
The Earth radius is 6378 Km
...
Since the weight of the body for r = 6378 Km is 1 N we
have k/63782 = 1, so k = 63782
...
462 N Km
...
462 N Km = 864
...

2
...
4
...
Consider a tank whose bottom is at some height y = a and its top is at y = b
...
We fill the tank by lifting from
the ground (y = 0) tiny layers of thickness dy each
...
The work needed to lift each layer is
dW = dF · y = ρyA(y) dy
...

a

2
...
APPLICATIONS TO PHYSICS AND ENGINEERING

65

2
...
5
...
Consider a tank like the one in the
previous paragraph
...
The analysis of the problem is similar to the previous paragraph, but now the work done to pump a tiny layer of thickness dy
is
dW = dF · (h − y) = ρ (h − y)A(y) dy
...

a

2
...
6
...

The pressure p of an homogeneous liquid of density ρ at depth h is
p = ρh
...

However the pressure against a vertical wall is not constant because it
depends on the depth
...
The force
exerted at y (so at depth h = c − y) against a small horizontal strip of
height dy and width w(t) (area = w(t) dy) is
dF = ρ (c − y)w(y) dy
...


Example: A cylindrical tank of radius 1 m and full of water (ρ =
9800 N/m3 ) is lying on its side
...
On the other
hand we obtain geometrically w(y) = 2 1 − y 2
...
6 N
...
5
...
5
...
Moments, Center of Mass
...
That point is called center of mass
...
If two
mases m1 and m2 are attached to a rod of negligible mass on opposite
sides of a fulcrum at distance d1 , d2 from the fulcrum, the rod will
balance if
m1 d1 = m2 d2
...

m1 + m2
The numbers m1 x1 and m2 x2 are called moments of the masses m1
and m2
...
More generally, for a solid made up of n particles of
masses mi placed on the x-axis at points xi :
n
P
mi xi
M
i=1
,
x= P
=
n
m
mi
i=1

where M =

n
X

mi xi is the total moment of the solid, and m =

i=1

n
X

mi

i=1

is its total mass
...

i=1

My measures the tendency of the solid to rotate about the y-axis, and
Mx measures the tendency of the solid to rotate about the x-axis
...
5
...

m

Now consider a flat plate of uniform density ρ occupying the plane
region under a curve y = f (x) between x = a and x = b
...
The sum of their
moments in the limit as ∆x → 0 is the integral:
Z
My = ρ

b

xf (x) dx
...
Adding and
taking the limit we get the integral
Z
Mx = ρ

b

a

1
[f (x)]2
2

dx
...


a

Hence, the coordinates of its center of mass are:
Rb
Rb
ρ a xf (x) dx
xf (x) dx
My
x=
= Ra b
= Rb
m
ρ a f (x) dx
f (x) dx
a
Rb
Rb 1
ρ a 21 [f (x)]2 dx
[f (x)]2 dx
Mx
y=
=
= aR 2b

...

Answer : We use coordinates
so that the plate occupies the region
√
under the graph of y = r2 − x2 , −r ≤ x ≤ r
...
By symmetry x = 0, so we only need to

2
...
APPLICATIONS TO PHYSICS AND ENGINEERING

find y:

Z
Z r √
1 r 1
1
2
1
y=
{f (x)} dx = 2 · 2
( r2 − x2 )2 dx
A −r 2
πr /2
−r
¸r
·
Z r
2
2
x3
2
2
2
= 2
(r − x ) dx = 2 r x −
dx
πr 0
πr
3 0
µ
¶
2
r3
2 2r3
4r
3
= 2 r −
= 2
=

...
6
...
6
...
6
...
Continuous Random Variables
...
g
...
In this example the random
variable can take only a discrete set of values
...
g
...

Given a random variable X, its probability distribution function
is the function F (x) = P (X ≤ x) = probability that the random
variable X takes a value less than or equal to x
...
7) = P (X ≤ 4
...
7, i
...
, the number of points
is 1, 2, 3 or 4, so the probability is 4/6 = 2/3, and F (4
...

If the random variable is continuous then we can also define a probability density function f (x) equal to the limit as ∆x → 0 of the probability that the random variable takes a value in a small interval of
length ∆x around x divided by the length of the interval
...
The probability that the random
variable takes a value in some interval [a, b] is
Z
P (a ≤ X ≤ b) = F (b) − F (a) =

b

f (x) dx
...

Z ∞
(2)
f (x) dx = 1 (the probability of a sure event is 1)
...
Find the probability density
of the uniform distribution on the interval [2, 5]
...
On the other

2
...
PROBABILITY

70

hand f (x) = 0 for x outside [2, 5], hence:
Z ∞
Z 5
1=
f (x) dx =
c dx = c(5 − 2) = 3c ,
−∞

2

so c = 1/3
...


2
...
2
...
The mean or average of a discrete random variable
that takes values x1 , x2 ,
...
, pn respectively is
n
X
x = x1 p1 + x2 p2 + · · · + xn pn =
xi pi
...
5
...
5 points per roll
...

−∞

2
...
3
...
The time that we must wait for some
event to occur (such as receiving a telephone call) can be modeled with
a random variable of density
(
0
if t < 0,
f (t) =
−ct
ce
if t ≥ 0,
were c is a positive constant
...

−∞

0

u→∞

2
...
PROBABILITY

71

The mean waiting time can be computed like this:
Z ∞
µ=
tf (t) dt
−∞
Z ∞
=
tce−ct dx
0
Z ∞
£
¤
−ct ∞
= −te
+
e−ct dx (I
...
So we can rewrite the density function like this:
(
0
if t < 0,
f (x) = 1 −t/µ
e
if t ≥ 0,
µ
Example: Assume that the average waiting time for a catastrophic
meteorite to strike the Earth is 100 million years
...
Find the probability that no such catastrophic event will
happen in the next 5 billion years
...


So the answer to the first question is
Z 100
h
i100
−8
−8
−8
−6
10−8 e−10 t dt = 1 − e−10 t
= 1−e−10 ·100 = 1−e−10 ≈ 10−6 ,
0

0

i
...
, about 1 in a million
...


2
...
PROBABILITY

72

2
...
4
...
Many important phenomena follow a so called Normal Distribution, whose density function is:
1
2
2
f (x) = √ e−(x−µ) /2σ
...
The positive constant σ is called standard deviation; it
measures how spread out the values of the random variable are
...
What proportion
of the population has an IQ between 70 and 130?
Answer : The integral cannot be evaluated in terms of elementary
functions, but it can be approximated with numerical methods:
Z
P (70 ≤ X ≤ 130) =

130

70

1
2
2
√ e−(x−100) /2·15 dx ≈ 0
...

15 2π

Another approach for solving these kinds of problems is to use the
error function, defined in the following way:
2
φ(x) = √
π

Z

x

2

e−t dt
...
0
0
...
0
0
...
0
0
...
1
0
...
1
0
...
1
0
...
2
0
...
2
0
...
2
0
...
3
0
...
3
0
...
3
0
...
4
0
...
4
0
...
4
0
...
5
0
...
5
0
...
5
1
...
6
0
...
6
0
...
6
1
...
7
0
...
7
0
...
7
1
...
8
0
...
8
0
...
8
1
...

Answer : We need to transform our integral into another expression
containing the error function:

0
...
797
1
...
993
2
...
000

2
...
PROBABILITY

Z
P (70 ≤ X ≤ 130) =

130

70

1
=√
π

Z

73

1
2
√ e−(x−100) /450 dx
15 2π
√

2

√
− 2
√
Z 2

2

e−u du

2
2
=√
e−u du
π 0
√
= φ( 2) ≈ φ(1
...
952

√
[u = (x − 100)/15 2]
(by symmetry)

CHAPTER 3

Differential Equations
3
...
Differential Equations and Separable Equations
3
...
1
...
The growth of a population is usually modeled with an equation of the form
dP
= kP ,
dt
where P represents the number of individuals an a given time t
...

A solution to this equation is the exponential function:
P (t) = Cekt
...

A more realistic model takes into account that any environment has
a limited carrying capacity K, so if P reaches K the population stops
growing
...

= kP 1 −
dt
K
This is called the logistic differential equation
...
1
...
Motion of a Spring
...
According to Hook’s law the restoring force of
a spring stretched (or compressed) a distance x from its natural length
is
F = −kx ,
where k is a positive constant (the spring constant) and the negative
sign expresses that the sense of the force is opposite to the sense of
the stretching
...
1
...

dt2
m
This is an example of a second order differential equation because
it involves second order derivatives
...
1
...
General Differential Equations
...
The order of the differential equation is the
order of the highest derivative that occurs in the equation
...
1
...
First-order Differential Equations
...
A solution of the differential
equation is a function y(x) such that y 0 (x) = F (x, y(x)) for all x in
some appropriate interval
...

dx
x
A possible solution for that equation is, for instance, y = x2 , because
dy
= y 0 (x) = (x2 )0 = 2x ,
dx
and
2
hence y 0 (x) =

y
x2
=2
= 2x ,
x
x

2 y(x)
for all x 6= 0
...
1
...
1
...
Separable Differential Equations
...
In particular this is true if the equation is of the
form
dy
= g(x) φ(y) ,
dx
where the right hand side is a product of a function of x and a function
of y
...

φ(y)
Given the equation
f (y) dy = g(x) dx ,
we can solve it by integrating both sides
...

Next we will try to solve this equation algebraically in order to either
write y as a function of x, or x as a function of y
...

dx
The right hand side is the product of a function of x and a function of
y, so it is separable:
1
dy = x dx
...

y=−

x2

3
...
DIFFERENTIAL EQUATIONS AND SEPARABLE EQUATIONS

77

3
...
6
...
A differential equation together
with an initial condition

 dy = F (x, y)
dx

y(x0 ) = y0
is called an initial value problem
...

Example: Solve the following initial value problem:

 dy = y 2 x
dx

y(0) = 1
Solution: We already found the general solution to the differential
equation:
2

...

C
So the solution is
2
2
y=− 2

...
2
...
2
...

3
...
1
...
Consider a differential equation of the form
dy
= F (x, y)
...
So the right hand side of the equation can
be interpreted as a slope field in the xy-plane
...
Each solution curve is
a particular solution of the slope field
...

3
...
2
...
This method consists of interpreting the
differential equation as a slope field and sketch solutions just by following the field
...
2
...
2
...
Euler’s Method
...
So
assume that we want to find approximate values of a solution for an
initial-value problem
½ 0
y = F (x, y)
y(x0 ) = y0
at equally spaced points x0 , x1 = x0 + h, x2 = x1 + h,
...
We take (x0 , y0 ) as the initial point of the solution
...
e
...
Proceeding
in the same way we get in general:
y1 = y0 + hF (x0 , y0 )
y2 = y1 + hF (x1 , y1 )
···
yn = yn−1 + hF (xn−1 , yn−1 )
···
Example: Use Euler’s method with step size 0
...
1) ≈ y1 = y0 + hF (x0 , y0 ) = 1 + 0
...
1
y(0
...
1 + 0
...
1 + 1
...
22
y(0
...
22 + 0
...
2 + 1
...
362
···

3
...
EXPONENTIAL GROWTH AND DECAY

80

3
...
Exponential Growth and Decay
3
...
1
...
Consider population with P (t) individuals at time t and with constant birth rate β (births per unit of time)
and death rate δ (deaths per unit of time)
...
Since P in fact varies, we need to use smaller
intervals of time [t, t + ∆t] in which P can be considered almost constant
...
So the change in the population will
be
∆P = P (t + ∆t) − P (t) ≈ βP ∆t − δP ∆t
...
e
...
With x(t) in place of P (t) we get the Natural Growth
Equation:
dx
= kx
...
Putting t = 0 we see that A = x0 = x(0), hence:
x(t) = x0 ekt
...
021 and
δ = 0
...
Assuming the birth and death rates remain
constant, find the population of the Earth in the year 2100
...
So we have

3
...
EXPONENTIAL GROWTH AND DECAY

81

x0 = 6 billion, and k = β − δ = 0
...
009 = 0
...
012t
in billions
...
012×100 = 19
...

3
...
2
...
Consider a given sample of radioactive material with N (t) atoms at time t
...
The solution to this equation is
N (t) = N0 e−kt ,
where No is the number of atoms at time t = 0
...
e
...

k
3
...
3
...
The air in the atmosphere contains
two carbon isotopes: 12 C, which is stable, and 14 C, which is radioactive
with a half-life of about 5700 years—so k = ln 2/τ = ln 2/5700 =
0
...

While an organism is alive, it absorbs both carbon isotopes by
breathing air, so the proportion of those isotopes in living matter is
the same as in air
...
So by measuring the proportion of
14
C in an organism we can estimate for how long it has been dead
...
When did that individual die?
Answer : We have:
0
...
3
...
80
ln 0
...

k
0
...
3
...
Continuously Compounded Interest
...
Let A(t) be the number of dollars in the account at time t
...
The
interest produced is rA(t)∆t, so
A(t + ∆t) = A(t) + rA(t)∆t ,
i
...


∆A
= rA(t)
...
In
that case we get:
dA
= rA(t)
...


CHAPTER 4

Infinite Sequences and Series
4
...
Sequences
A sequence is an infinite ordered list of numbers, for example the
sequence of odd positive integers:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
...
, an ,
...

Notation: the sequence a1 , a2 , a3 ,
...

Some sequences can be defined with a formula, for instance the
sequence 1, 3, 5, 7,
...

A recursive definition consists of defining the next term of a sequence as a function of previous terms
...

4
...
1
...
The limit of a sequence is the value to which its
terms approach indefinitely as n becomes large
...


n→∞

For instance

1
= 0,
n→∞ n
lim

83

4
...
SEQUENCES

84

n+1
= 1,
n→∞
n
lim

etc
...

If the sequence becomes arbitrarily large then we write
lim an = ∞
...


n→∞

4
...
2
...
Let f be a function defined in [1, ∞]
...
e
...
)
ln n

...
But now we can use
L’Hˆopital’s Rule:
Answer : According to the theorem that limit equals lim

ln x
(ln x)0
1/x
= lim
= 0,
= lim
0
x→∞ x
x→∞ (x)
x→∞ 1
lim

hence
ln n
=0
...

n→∞

Answer : This limit is the same as that of the exponential function
r , hence


0 if 0 < r < 1
n
lim r = 1 if r = 1
n→∞

∞ if r > 1
x

4
...
SEQUENCES

85

4
...
3
...
If an → a and bn → b then:
(an + bn ) → a + b
...

can → ca for any constant c
...

a
an
→ if b 6= 0
...

n2 + n + 1

...

3
n→∞ 2n2 + 3
n→∞
2+0
2
2 + n2
4
...
4
...
If an ≤ bn ≤ cn for every n ≥ n0 and
lim an = lim cn = L, then lim bn = L
...

n→∞

n→∞

cos n

...

n→∞
n
4
...
5
...

4
...
5
...
Increasing, Decreasing, Monotonic
...
It is decreasing if an+1 < an for every n
...

Example: Probe that the sequence an =

n+1
is decreasing
...
1
...


Answer : an+1 − an =
an+1

86

< 0, hence

4
...
5
...
Bounded
...
It is bounded below if there is a number
m such that m ≤ an for all n
...

Example: Probe that the sequence an =

n+1
is bounded
...
To prove that it is bounded above note that
n+1
1
= 1 + ≤ 2
...

4
...
6
...
Every bounded monotonic sequence is convergent
...


Next example shows that sometimes in order to find a limit you
may need to make sure that the limits exists first
...
Find it:
r
q
q
√
√
√
2, 2 + 2, 2 + 2 + 2,
...
First we will prove by induction that
0 < an < 2, so the sequence is bounded
...

Next the induction step
...
From here we must prove
that the same is true for the next value of n, i
...
that 0 < a√
n+1 < 2
...
e
...
So by the induction principle all terms of the sequence verify
that 0 < an < 2
...
1
...

Finally, since the given sequence is bounded and increasing, by the
monotonic sequence theorem it has a limit L
...

Since an → L and an+1 → L we have:
√
L = 2 + L ⇒ L2 = 2 + L

⇒ L2 − L − 2 = 0
...

Note that the trick works only when we know for sure that the limit
exists
...
(f1 = 1, f2 = 1, fn = fn−1 + fn−2 ),
calling L the “limit” we get from the recursive relation that L = L + L,
hence L = 0, so we “deduce” limn→∞ fn = 0
...


4
...
SERIES

88

4
...
Series
A series is an infinite sum:
∞
X
an = a1 + a2 + a3 + · · · + an + · · ·
n=1

In order to define the value of these sum we start be defining its
sequence of partial sums
n
X
ai = a1 + a2 + · · · + an
...

n→∞

n=1

Otherwise the series is called divergent
...

+ 2 + 3 + ··· + n + ··· =
2 2
2
2
2n
n=1
∞

Its partial sums are:
n
X
1
1
1
1
1
sn =
= + 2 + ··· + n = 1 − n
...

2n n→∞ i=1 2i n→∞
2
n=1
4
...
1
...
A series verifying an+1 = ran , where r
is a constant, is called geometric series
...

n=0

The partial sums are now:
sn =

n
X
i=0

ari
...
2
...

1−r
If |r| < 1 we can rewrite the result like this:
a
a n+1
sn =
−
r
,
1−r 1−r
and then get the limit as n → ∞:
a
a
a
−
lim rn+1 =
s = lim sn =
n→∞
1 − r 1 − r |n→∞{z } 1 − r
sn =

↓

0

So for |r| < 1 the series is convergent and
∞
X

arn =

n=0

a

...

4
...
2
...
A telescopic series is a series whose
terms can be rewritten so that most of them cancel out
...

n(n + 1)

1
1
1
= −

...

n+1
=

4
...
SERIES

Hence, the sum of the series is
s = lim sn = lim
n→∞

n→∞

µ

90

1
1−
n+1

4
...
3
...
If the series
limn→∞ an = 0
...


an is convergent then

Proof
P : If the series is convergent then the sequence of partial sums
sn = ni=1 ai have a limit s
...

The converse is not true in general
...

4
...
4
...
The following series is called harmonic series:
∞
X
1
1 1 1
= 1 + + + + ···
n
2 3 4
n=1
The main fact about it is that it is divergent
...
, so in general s2n > 1+ n2 , hence the sequence of partial sums grows
without limit and the series diverges
...
2
...
Test for Divergence
...

n=1

Example: Show that

∞
X
n=1

n
diverges
...
Since the nth term of the series
n→∞ n + 1
does not tend to 0, the series diverges
...
2
...


n=1

Answer : All we need to show is that sin n does not tend to 0
...
84 · · · 6= 0
So if a term sin n is close to zero, the next term sin (n + 1) will be far
from zero, so it is impossible for sin n to get permanently closer and
closer to 0
...
2
...
Operations with Series
...
3
...
3
...
3
...
The Integral Test
...
Then the convergence
or
R ∞divergence of the series n=1 an is the same as that of the integral
f (x) dx, i
...
:
1
Z

∞

(1) If

f (x) dx is convergent then

1

Z

∞

(2) If

∞
X

f (x) dx is divergent then

1

an is convergent
...


n=1

The best way to see why the integral test works is to compare the
area under the graph of y = f (x) between 1 and ∞ to the sum of the
areas of rectangles of height f (n) placed along intervals [n, n + 1]
...
2

1
...
8

0
...
6

y 0
...
4

0
...
2

0
...
3
...

1

i=1

1

The first inequality shows that if the integral diverges so does the series
...

P∞Example: Use the integral test to prove that the harmonic series
n=1 1/n diverges
...
3
...

4
...
2
...
The following series is called p-series:
∞
X
1

...
For p = 1
we have seen that it diverges
...

t→∞ 1 xp
t→∞ 1 − p
t→∞ 1 − p
xp
1−p
1
1
For 0 < p < 1 the limit is infinite, and for p > 1 it is zero so:
∞
X
1
The p-series
is
np
n=1

(

convergent if p > 1
divergent
if p ≤ 1

P
P
4
...
3
...
Suppose that an and bn are series
with positive terms and suppose that an ≤ bn for all n
...

P
P
(2) If
an is divergent then
bn is divergent
...


n2

converges or di-

Answer : We have
0<

cos2 n
1
≤
n2
n2

for all n ≥ 1

∞
X
1
and we know that the series p-series
converges
...
5736380465
...


4
...
THE INTEGRAL AND COMPARISON TESTS

P 4
...
4
...
Suppose that
bn are series with positive terms
...

Example: Determine whether the series

∞
X
n=1

diverges
...
We have
series
n
n=1
1/n
lim √
= lim
n→∞ 1/ 1 + 4n2
n→∞

√

1 + 4n2
n
r
1 + 4n2
= lim
n→∞
n2
r
√
1
= lim
+
4
=
4 = 2,
n→∞
n2

so the given series has the same behavior as the harmonic series
...

4
...
5
...
The difference between the sum
s
=
n=1 an of a convergent series and its
P
nth partial sum sn = i=1 ai is the remainder :
Rn = s − sn =

∞
X

ai
...
Namely: If an converges by the Integral
Test and Rn = s − sn , then
Z ∞
Z ∞
f (x) dx ≤ Rn ≤
f (x) dx
n+1

n

4
...
THE INTEGRAL AND COMPARISON TESTS

Equivalently (adding sn ):
Z ∞
Z
sn +
f (x) dx ≤ s ≤ sn +
n+1

∞

95

f (x) dx

n

∞
X
1
Example: Estimate
to the third decimal place
...
0005, i
...
, we
need to find some n such that
Z ∞
1
dx < 0
...

x4
n
We have
·
¸∞
Z ∞
1
1
1
dx
=
−
=
,
x4
3x3 n
3n3
n
hence
r
1
3
3
= 18
...
,
<
0
...
0005
so we can take n = 19
...
082278338
...
08227
...
0005 = 1
...
and 1
...
+0
...
08277
...
082
...
0823232337
...
4
...
4
...
4
...
Alternating Series
...
, for instance
X (−1)n+1
1 1 1
1 − + − + ··· =

...
4
...
1
...
If the sequence of positive
terms bn verifies
(1) bn is decreasing
...

Example: The alternating harmonic series
X (−1)n+1
1 1 1
1 − + − + ··· =
2 3 4
n
n=1
∞

converges because 1/n → 0
...
6931471806
...
4
...
2
...
If s = ∞
n=1 (−1) bn
is the sum of and alternating series verifying that bn is decreasing and
bn → 0, then the remainder of the series verifies:
|Rn | = |s − sn | ≤ bn+1
...
4
...
Absolute Convergence
...

P
Absolute convergence implies convergence, i
...
, if a series
an is
absolutely convergent, then it is convergent
...
For instance, the alternatP
(−1)n+1
is convergent but it is not absolutely
ing harmonic series ∞
n=1
n
convergent
...
4
...

P∞ | cos n|
Answer : We see that the series of absolute values
is
n=1 n2
P∞ 1
convergent by comparison with n=1 n2 , hence the given series is absolutely convergent, therefore it is convergent (its sum turns out to be
1/4 − π/2 + π 2 /6 = 0
...
, but the proof of this is beyond
the scope of this notes)
...
4
...
The Ratio Test
...

¯
¯
∞
X
¯ an+1 ¯
¯ = L > 1 (including L = ∞) then the series
(2) If lim ¯¯
an
n→∞
an ¯
n=1

is divergent
...

Example: Test the series
∞
X
n=1

(−1)n

n!
nn

for absolute convergence
...


4
...
POWER SERIES

98

4
...
Power Series
A power series is a series of the form
∞
X

c 0 xn = c 0 + c 1 x + c 2 x 2 + · · · + c n x n + · · ·

n=0

where x is a variable of indeterminate
...
The cn ’s are the coefficients of the series
...

xn = 1 + x + x2 + · · · + xn + · · · =
1−x
n=0
More generally given a fix number a, a power series in (x − a), or
centered in a, or about a, is a series of the form
∞
X

c0 (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + · · · + cn (x − a)n + · · ·

n=0

∞
X

4
...
1
...
For a given power series
cn (x − a)n there are only three possibilities:

n=1

(1) The series converges only for x = a
...

(3) There is a number R, called radius of convergence, such that
the series converges if |x − a| < R and diverges if |x − a| > R
...

Example: Find the radius of convergence and interval of convergence
of the series
∞
X
(x − 3)n

...
5
...
Consequently, the radius of convergence is R = 1
...
For x = 4 the series
becomes
∞
X
1
,
n
n=0
i
...
, the harmonic series, which we know diverges
...
e
...
So the interval
of convergence is [2, 4)
...
6
...
6
...
A slightly different matter is that sometimes a given function
can be written as a power series
...
e
...

4
...
1
...
Since
the sum of a power series is a function we can differentiate it and integrate it
...
The main related result is that the derivative or integral of a power series can be computed by term-by-term
differentiation and integration:
4
...
1
...

Differentiation and Integration
...

Example: Find a power series representation for the function
1
f (x) =

...
6
...

Example: Find a power series representation for tan−1 x
...

3
5
7

Since tan−1 0 = 0 then C = 0, hence
−1

tan

x=

∞
X
n=0

(−1)n

x 3 x5 x 7
x2n+1
=x−
+
−
+
...

Example: Find a power series representation for ln (1 + x)
...
6
...


102

4
...
TAYLOR AND MACLAURIN SERIES

103

4
...
Taylor and MacLaurin Series
4
...
1
...
Assume that we have a
function f for which we can easily compute its value f (a) at some
point a, but we do not know how to find f (x) at other points x close
to a
...
1? One
way to deal with the problem is to find an approximate value of f (x)
...


y

y=f(a)+f’(x)(x-a)
y=f(x)

a

x

Figure 4
...
1
...

The equation of the tangent line to y = f (x) at x = a is
y = f (a) + f 0 (a)(x − a) ,
hence

f (x) ≈ f (a) + f 0 (a)(x − a) ,
for x close to a
...

For a = 0 we get:
sin(x) ≈ sin 0 + cos 0 · (x − 0) = x ,
so sin(0
...
1
...
1) = 0
...
, which is close to
0
...

The tangent line is the graph of the first degree polynomial
T1 (x) = f (a) + f 0 (a)(x − a)
...
7
...
For instance, we may
try a second degree polynomial of the from:
T2 (x) = c0 + c1 (x − a) + c2 (x − a)2 ,
with the following conditions:
T2 (a) = f (a)
T20 (a) = f 0 (a)
T200 (a) = f 00 (a)
i
...
:



 c0 = f (a)
c1 = f 0 (a)


2c2 = f 00 (a)

After solving the system of equations obtained we get:
c0 = f (a)
c1 = f 0 (a)
c2 =

f 00 (a)
2

hence:
T2 (x) = f (a) + f 0 (a)x +

f 00 (a) 2
x
...

Tn(n) (a) = f (n) (a)

4
...
TAYLOR AND MACLAURIN SERIES

105

From here we get a system of n+1 equations with the following solution:
c0 = f (a)
c1 = f 0 (a)
c2 =

...


That polynomial is the so called nth-degree Taylor polynomial of
f (x) at x = a
...

T3 (x) = sin a + cos a · (x − a)2 −
2
3!
For a = 0 we have sin 0 = 0 and cos 0 = 1, hence:
T3 (x) = x −

x3

...
1 ≈ 0
...
13
= 0
...

6

The actual value of sin 0
...
1 = 0
...

4
...
2
...
The difference between the value of
a function and its Taylor approximation is called remainder :
Rn (x) = f (x) − Tn (x)
...
7
...


Example: Find the third degree Taylor approximation for sin x at
x = 0, use it to find an approximate value for sin 0
...

Answer : We already found
x3
T3 (x) = x −
,
6
and
T3 (0
...
99833333
...
1 = 0
...
0000042 = 4
...

4!
In fact the estimation is correct, the approximate value differs from the
actual value in
|R3 (0
...
1) − sin 0
...
000000083313 · · · < 8
...

4
...
3
...
If the given function has derivatives of all
orders and Rn (x) → 0 as n → ∞, then we can write
f (x) =

∞
X
f (n) (a)
n=0

n!

(x − a)n

= f (a) + f 0 (a)(x − a) +
··· +

f 00 (a)
(x − a)2 +
2!

f (n) (a)
(x − a)n + · · ·
n!

The infinite series to the right is called Taylor series of f (x) at x = a
...

Example: The Taylor series of f (x) = ex at x = 0 is:
X xn
xn
x2
+ ··· +
+ ··· =
1+x+
2
n!
n!
n=0
∞

4
...
TAYLOR AND MACLAURIN SERIES

107

For |x| < d the remainder can be estimated taking into account that
f (n) (x) = ex and |ex | < ed , hence
|Rn (x)| <

ed
|x|n+1
...
Consequently we can write:
x

e =

∞
X
xn
n=0

n!

=1+x+

x2 x3
xn
+
+ ··· +
+ ···
2!
3!
n!

For x = 1 this formula provides a way of computing number e:
e=

∞
X
1
1
1
1
= 1 + 1 + + + · · · + + · · · = 2
...

n!
2! 3!
n!
n=0

The following are Maclaurin series of some common functions:
x

e =

∞
X
xn
n=0

sin x =

n!

=1+x+

x2 x3 x4
+
+
+ ···
2!
3!
4!

∞
X
x3 x5 x7
x2n+1
=x−
+
−
+ ···
(−1)n
(2n
+
1)!
3!
5!
7!
n=0

∞
X
x2 x4 x6
x2n
cos x =
(−1)n
=1−
+
−
+ ···
(2n)!
2!
4!
6!
n=0
∞
X
xn
x2 x3 x4
ln (1 + x) = −
(−1)n
=x−
+
−
+ ···
n
2
3
4
n=1
α

(1 + x) =

X µα¶
n=0

µ ¶
µ ¶
α 2
α 3
x = 1 + αx +
x + ···
x +
n
2
3
n

µ ¶
α(α − 1)(α − 2)
...

where
=
n!
n
X
1
= (1 + x)−1 =
(−1)n xn = 1 − x + x2 − x3 + · · ·
1+x
n=0

4
...
TAYLOR AND MACLAURIN SERIES

tan−1 x =

X

(−1)n

n=0

108

x2n+1
x3 x5 x7
=x−
+
−
+ ···
(2n + 1)!
3
5
7

Remark : By letting x = 1 in the Taylor series for tan−1 x we get
the beautiful expression:
X
1 1 1
π
1
= 1 − + − + ··· =
(−1)n

...
Since the series for tan−1 x converges more quickly for
small values of x, it is more convenient to express π as a combination
of inverse tangents with small argument like the following one:
π
1
1
= 4 tan−1 − tan−1

...
Then the inverse
tangents can be computed using the Maclaurin series for tan−1 x, and
from them an approximate value for π can be found
...
7
...
Finding Limits with Taylor Series
...

x→0
x2

Example: Find lim

Answer : Replacing ex with its Taylor series:
2

3

4

(1 + x + x2 + x6 + x24 +
...

2
= lim
x→0
x2
½
¾
1 x x2
1
= lim
+ +
+
...

x→0
2 6 24
2

4
...
APPLICATIONS OF TAYLOR POLYNOMIALS

109

4
...
Applications of Taylor Polynomials
4
...
1
...
Here we illustrate an application of Taylor polynomials to physics
...

Let’s rewrite the formula in the following way:
µ
¶−1/2
v2
2
E = m0 c 1 − 2

...

2c
8c
16 c6
−1/2

If we subtract the energy at rest m0 c2 we get the kinetic energy:
µ 2
¶
1v
3 v4
5 v6
2
2
K = E − m0 c = m0 c
+
+
+ ···
...

2
2c
2
That is the expression for the usual (non relativistic or Newtonian)
kinetic energy, so this tells us at low speed the relativistic kinetic energy
is approximately equal to the non relativistic one
...
8
...
8
...
Using Series to Solve Differential Equations
...
In such cases an alternative is to represent the solution as a power series and try to determine the values of the coefficients that solve the equation
...

We start with an equation that we do know how to solve explicitly,
so we can compare the power series obtained with the explicit solution:
y0 = y
...


Next we solve it using power series
...

n=0

Its derivative is
y 0 = c1 + 2c2 x + 3c3 x2 · · · =

∞
X

(n + 1)cn+1 xn
...


n=0

In order to be equal the coefficients must be the same on both sides,
so:

c0 = c1



c
= 2c2


 1
c2 = 3c3
···


 c


 n = (n + 1)cn+1
···

4
...
APPLICATIONS OF TAYLOR POLYNOMIALS

111

This defines a sequence of coefficients in which the first one c0 is arbitrary, and the following ones verify the recursive relation
cn+1 =

cn

...
e
...

Lets look now at a more sophisticated example
...

The idea is the same as before, we replace y with a power series, find
its derivatives that appear in the equation, pose the equation with the
powers series, and find a relation among the coefficients:
y=
y0 =

∞
X
n=0
∞
X

c n xn
ncn xn−1

n=1

y 00 =

∞
X
n=2

n(n − 1)cn xn−2

4
...
APPLICATIONS OF TAYLOR POLYNOMIALS

y 00 − 2xy 0 + y =

∞
X

cn xn − x

n=0

=

∞
X

∞
X

ncn xn−1 +

n=1

cn xn −

n=0

∞
X
n=1

ncn xn +

∞
X

112

n(n − 1)cn xn−2

n=2
∞
X

n(n − 1)cn xn−2

n=2

After some reindexing and grouping we get that the equation becomes:
∞
X
{(n + 2)(n + 1)cn+2 − (2n − 1)cn } xn = 0 ,
n=0

which implies:

2n − 1
cn
...
1
...
1
...
Definitions
...
Analogously, the functions sin x, cos x, tan x,
etc
...

A
...
2
...
The hyperbolic functions verify some identities similar to those of the circular functions, except for
some occasional sign differences:
113

A
...
HYPERBOLIC FUNCTIONS

114

cosh2 x − sinh2 x = 1
1 − tanh2 x = sech2 x
coth2 x − 1 = csch2 x
sinh (x + y) = sinh x cosh y + cosh x sinh y
cosh (x + y) = cosh x cosh y + sinh x sinh y
sinh 2x = 2 sinh x cosh y
cosh 2x = cosh2 x + sinh2 x
All of them can be verified algebraically, for instance:
µ
¶2 µ
¶2
1 x
1 x
2
2
−x
−x
(e + e ) −
(e − e )
cosh x − sinh x =
2
2
1
1
= (e2x + 2 + e−2x ) − (e2x − 2 + e−2x )
4
4
µ ¶
1
1
= 1
...
1
...
Derivatives of Hyperbolic Functions
...
1
...
Integrals of Hyperbolic Functions
...
1
...
1
...
Inverse Hyperbolic Functions
...

2
Solving the equation in y we get:
³
´
√
sinh−1 x = ln x + x2 + 1
for all x
...

2
Solving the equation in y we get:
³
´
√
cosh−1 x = ln x + x2 − 1
for all x ≥ 1
...

ey + e−y

for |x| < 1
...
1
...


for 0 < x ≤ 1
...


A
...
6
...

The derivatives of the inverse hyperbolic functions can be found from
their expressions in terms of the natural logarithm, e
...
:
n ³
´o0
1 + 2√2x
√
1
2
−1
0
2
√ x +1 = √
(sinh x) = ln x + x + 1
=

...
1
...
Integrals Involving Inverse Hyperbolic Functions
...
1
...
1
...
Taylor Series of Hyperbolic Functions
...
1
...

i=

n(n + 1)

...

6
i=1
·
¸2
n
X
n(n + 1)
3
(4)
i =

...
TABLE OF INTEGRALS

120

Table of Integrals
...

Z
Z
du
du
−1
√
√
= sinh u + C
= cosh−1 u + C
2
2
u −1
u +1
Z
Z
du
du
−1
√
√
= − sech |u| + C
= − csch−1 |u| + C
2
2
u 1−u
u 1+u
Reduction Formulas
...

tan u du =
n−1
Z
Z
secn−2 u tan u n − 2
n
+
sec u du =
secn−2 u du

---