<
Advanced
Calculus
F
o
r
Data
Science
Mik
e
Ca
rr
Contents
Intro
duction
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2
1
Review
of
Algeb
ra
and
Calculus
5
1.1
Graphs
of
F
unctions
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6
1.2
Limits
and
Derivatives
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16
1.3
Applications
of
Derivatives
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35
1.4
Definite
Integrals
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44
2
Advanced
Integration
and
Applications
59
2.1
Area
Bet
w
een
Curves
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60
2.2
V
olumes
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75
2.3
Integration
b
y
P
a
rts
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91
2.4
App
ro
ximate
Integration
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101
2.5
Imp
rop
er
Integrals
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120
2.6
Probabilit
y
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138
2.7
F
unctions
of
Random
V
a
riables
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158
3
Series
169
3.1
T
a
ylo
r
P
olynomials
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170
3.2
Sequences
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188
3.3
Series
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198
3.4
P
o
w
er
Series
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218
3.5
T
a
ylo
r
Series
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227
4
Multiva
riable
Functions
241
4.1
Three-Dimensional
Co
o
rdinate
Systems
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242
4.2
F
unctions
of
Several
V
a
riables
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259
4.3
Limits
and
Continuit
y
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275
4.4
P
a
rtial
Derivatives
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281
4.5
Linea
r
App
ro
ximations
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295
5
V
ectors
in
Calculus
305
5.1
V
ecto
rs
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306
5.2
The
Dot
Pro
duct
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321
5.3
No
rmal
Equations
of
Planes
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330
5.4
The
Gradient
V
ecto
r
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342
5.5
The
Chain
Rule
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359
5.6
Maximum
and
Minimum
V
alues
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375
5.7
Lagrange
Multipliers
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393
6
Multiva
riable
Integration
409
6.1
Double
Integrals
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410
6.2
Double
Integrals
over
General
Regions
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424
6.3
Joint
Probabilit
y
Distributions
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437
6.4
T
riple
Integrals
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463
1
Intro
duction
So
far
in
calculus
y
ou
have
developed
the
to
ols
to
answer
the
follo
wing
questions
about
a
function
of
one
va
riable:
1
How
quickly
do
es
the
value
of
the
function
change
as
the
input
changes?
2
How
do
we
estimate
the
value
of
the
function
near
a
p
oint?
3
What
are
the
maximum
and
minimum
values
of
the
function?
4
What
is
the
area
under
the
graph
of
the
function?
What
do
es
it
mean?
These
are
all
useful
to
ols,
but
they
don’t
necessarily
apply
to
the
types
of
data
that
w
e
encounter
in
the
w
o
rld.
Data
generally
takes
the
fo
rm
of
a
set
of
observations,
rather
than
an
algebraic
function.
Ho
w
do
w
e
p
erform
calculus
with
such
a
set?
W
e
cannot
integrate
it
without
an
antiderivative.
In
some
cases,
the
b
est
functions
to
mo
del
our
data
are
difficult
to
w
ork
with.
W
e
tak
e
fo
r
granted
that
sin
x
is
a
2
useful
function,
but
ho
w
do
w
e
even
evaluate
a
quantit
y
like
sin(7
.
52)
?
In
all
these
circumstances,
the
b
est
we
can
do
is
appro
ximate.
W
e
will
develop
metho
ds
to
appro
ximate
integrals
and
to
appro
ximate
functions.
Click to Load Applet
Figure:
App
ro
ximations
of
an
integral
and
of
a
function
Many
measurable
quantities
can
b
e
found
to
dep
end
on
the
value
of
multiple
inputs.
These
are
multiva
riable
functions
lik
e
z
=
F
(
x,
y
)
,
where
z
is
a
function
of
t
wo
indep
endent
variables.
Examples
app
ea
r
in
all
the
sciences
1
Chemistry:
V
=
nr
t
P
2
Physics:
F
=
GM
m
r
2
3
Economics:
P
=
P
0
e
rt
Figure:
The
graph
of
a
tw
o-variable
function
W
e
want
to
understand
how
to
measure
rates
of
change
of
these
functions,
and
what
these
mea-
surements
can
do
fo
r
us.
F
urthermo
re,
real
wo
rld
data
do
es
not
come
prepack
aged
with
a
differentiable
function
to
describ
e
it.
One
approach
is
to
find
a
line
of
b
est
fit.
Doing
so
requires
optimizing
tw
o
variables
at
once
(slop
e
and
intercept)
to
find
the
b
est
fit.
3
Intro
duction
Click to Load Applet
Figure:
Fitting
a
line
to
a
set
of
data
p
oints
The
values
of
y
ma
y
not
b
e
a
function
of
x
at
all.
Another
view
p
oint
is
to
see
(
x,
y
)
as
a
randomly
chosen
p
oint
in
the
plane.
T
o
mo
del
such
random
choices,
we
use
a
tw
o-variable
density
function.
V
olumes
under
its
graph
(computed
b
y
integrals)
tell
us
where
these
random
p
oints
are
likely
to
lie.
Click to Load Applet
Figure:
A
function
that
mo
dels
the
outcomes
of
a
random
process
These
app
roaches
will
requires
us
to
use
derivatives
and
integrals
of
multivariable
functions.
4
Chapter
1
Review
of
Algeb
ra
and
Calculus
This
chapter
reviews
the
most
imp
ortant
information
about
functions,
limits,
derivatives,
and
integrals.
It
is
not
meant
to
teach
this
material
to
a
first-time
learner,
but
can
serve
as
a
reference
or
reminder.
Contents
1.1
Graphs
of
F
unctions
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6
1.2
Limits
and
Derivatives
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16
1.3
Applications
of
Derivatives
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35
1.4
Definite
Integrals
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44
Section
1.1
Graphs
of
F
unctions
Goals:
1
Graph
algebraic
and
trigonometric
functions.
2
Solve
equations
using
inverse
functions.
3
Solve
equations
containing
quotients.
4
Graph
transformations
of
functions.
Definition
The
graph
of
an
equation
is
the
set
of
o
rdered
pairs
(
x,
y
)
that
satisfy
the
equation.
These
are
the
p
oints
that,
when
their
co
ordinates
are
plugged
in
for
x
and
y
,
the
t
w
o
sides
of
the
equation
are
equal.
Linea
r
F
unctions
Linea
r
functions
can
b
e
written
in
slop
e-intercept
form
:
f
(
x
)
=
mx
+
b.
The
graph
y
=
mx
+
b
of
a
linear
function
is
a
line.
m
is
the
slop
e,
which
is
the
change
in
y
over
the
change
in
x
b
etw
een
any
tw
o
p
oints
on
the
line.
(0
,
b
)
is
the
y
intercept.
If
we
have
the
slop
e
and
a
known
p
oint
(
x
0
,
y
0
)
on
a
line.
We
can
write
its
equation
in
p
oint-slop
e
fo
rm.
y
−
y
0
=
m
(
x
−
x
0
)
If
w
e
have
b
oth
the
x
-
and
y
-intercepts
of
the
line,
it
is
convenient
to
write
it
in
normal
form
ax
+
by
+
c
=
0
6
Monomials
A
monomial
is
a
function
of
the
fo
rm:
f
(
x
)
=
x
n
where
n
is
an
integer
greater
than
0
.
F
o
r
n
≥
2
the
graph
y
=
x
n
curves
up
w
a
rd
over
the
p
ositive
values
of
x
.
Higher
values
of
n
have
lo
w
er
values
when
0
<
x
<
1
but
higher
values
after
x
>
1
.
F
o
r
even
values
of
n
the
graph
is
symmetric
across
the
y
-axis,
curving
up
when
x
is
negative.
F
o
r
o
dd
values
of
n
the
graph
curves
down
when
x
is
negative.
It
is
anti-symmetric
across
the
x
=
0
.
Figure:
Graphs
of
monomials
of
o
dd
and
even
p
ow
ers
Monomials
of
Negative
P
o
w
er
Monomials
of
negative
p
o
w
er
have
the
fo
rm
f
(
x
)
=
x
−
n
.
They
are
also
commonly
written
f
(
x
)
=
1
x
n
.
The
graph
y
=
1
x
n
has
a
vertical
asymptote
at
x
=
0
.
The
graph
app
roaches
the
x
-axis,
y
=
0
as
x
gets
large.
F
o
r
even
values
of
n
,
the
graph
is
ab
ove
the
x
-axis.
F
o
r
o
dd
values
of
n
,
the
graph
is
ab
ove
the
x
-axis
fo
r
p
ositive
x
and
b
elo
w
it
fo
r
negative
x
.
A
la
rger
choice
of
n
mak
es
the
function
app
roach
the
x
-axis
mo
re
quickly
.
7
Section
1.1
Graphs
of
Functions
Figure:
Graphs
of
monomials
of
negative
o
dd
and
even
p
ow
ers
Ro
ots
A
ro
ot
functiom
is
a
function
of
the
form:
f
(
x
)
=
n
√
x
where
n
is
an
integer
greater
than
0
.
The
domain
of
n
√
x
is
[0
,
∞
)
if
n
is
even
and
all
real
numb
ers
if
n
is
o
dd.
The
x
and
y
intercept
of
y
=
n
√
x
is
at
(0
,
0)
.
Ro
ot
functions
a
re
increasing.
At
x
=
0
,
they
travel
straight
up.
Figure:
The
graphs
of
y
=
√
x
and
y
=
3
√
x
8
Exp
onential
F
unctions
An
exp
onential
function
has
the
form:
f
(
x
)
=
a
x
where
a
is
a
numb
er
greater
than
0
.
a
is
called
the
base
of
the
exp
onential
function.
The
graph
y
=
a
x
passes
through
(0
,
1)
.
If
a
>
1
then
f
(
x
)
increases
quickly
as
x
takes
on
p
ositive
values.
Higher
values
of
a
give
a
steep
er
increase.
f
(
x
)
approaches
0
as
x
go
es
to
−∞
.
Higher
values
of
a
give
a
faster
approach.
The
graph
do
es
not
touch
o
r
cross
the
x
-axis.
If
a
<
1
,
then
the
ab
ove
is
reversed.
e
is
a
commonly
used
base.
e
is
appro
ximately
2
.
718
.
Figure:
The
graphs
of
exp
onential
functions
Loga
rithms
A
loga
rithmic
function
have
the
form:
f
(
x
)
=
log
x
a
where
a
is
a
numb
er
greater
than
1
.
log
a
x
is
the
numb
er
b
such
that
a
b
=
x
.
a
b
can
never
b
e
0
o
r
less.
The
domain
of
f
(
x
)
=
log
x
a
is
(0
,
∞
)
.
As
x
go
es
to
0
,
log
a
x
go
es
to
−∞
.
y
=
log
a
x
has
an
x
intercept
at
(1
,
0)
.
9
Section
1.1
Graphs
of
Functions
Figure:
The
graphs
of
logarithm
functions
Loga
rithms
and
exp
onents
are
inverse
functions.
W
e
solve
exp
onential
equations
by
applying
a
loga
rithm
to
b
oth
sides.
We
solve
logarithm
equations
by
exp
onentiating
b
oth
sides.
a
x
=
c
x
=
log
a
c
log
a
x
=
c
x
=
a
c
T
rigonometric
F
unctions
f
(
x
)
=
sin
x
and
f
(
x
)
=
cos
x
are
p
erio
dic
functions.
sin
x
and
cos
x
have
a
range
of
[
−
1
,
1]
.
These
functions
a
re
p
erio
dic.
This
means
that
for
all
x
,
f
(
x
+
2
π
)
=
f
(
x
)
.
Figure:
The
graphs
of
y
=
sin
x
and
y
=
cos
x
10
The
other
trigonometric
functions
can
b
e
written
in
terms
of
sine
and
cosine.
tan
x
=
sin
x
cos
x
cot
x
=
cos
x
sin
x
sec
x
=
1
cos
x
csc
x
=
1
sin
x
Since
trigonometric
functions
obtain
the
same
values
infinitely
many
times,
the
do
not
technically
have
inverse
functions.
How
ever,
we
define
inverse
trigonometric
functions
on
a
restricted
range.
−
π
2
≤
sin
−
1
x
≤
π
2
0
≤
cos
−
1
x
≤
π
−
π
2
≤
tan
−
1
x
≤
π
2
These
functions
provide
one
solution
to
a
trigonometric
equation.
We
can
obtain
the
others
by
using
the
p
erio
dic
b
ehavio
r
of
trognometric
funtions.
sin
x
=
c
x
=
sin
−
1
c
+
2
π
n
or
π
−
sin
−
1
c
+
2
π
n
cos
x
=
c
x
=
cos
−
1
c
+
2
π
n
or
−
cos
−
1
c
+
2
π
n
tan
x
=
c
x
=
tan
−
1
c
+
π
n
Where
n
can
b
e
any
integer.
Question
1.1.1
Ho
w
Do
T
ransfo
rmations
Affect
the
Graph
of
a
F
unction?
T
ransfo
rmations
Supp
ose
w
e
w
ould
lik
e
to
transform
the
graph
y
=
f
(
x
)
.
Here
are
four
wa
ys
we
can.
The
graph
of
y
=
af
(
x
)
is
stretched
by
a
factor
of
a
in
the
y
direction.
The
graph
of
y
=
f
(
x
)
+
b
is
shifted
by
b
in
the
p
ositive
y
direction.
The
graph
of
y
=
f
(
cx
)
is
compressed
by
a
factor
of
c
in
the
x
direction.
The
graph
of
y
=
f
(
x
+
d
)
is
shifted
by
d
in
the
negative
x
direction.
11
Question
1.1.1
Ho
w
Do
T
ransfo
rmations
Affect
the
Graph
of
a
F
unction?
Click to Load Applet
Figure:
The
graphs
of
y
=
f
(
x
)
and
it
transformation
Example
1.1.2
A
Equation
with
Quotients
An
equation
of
the
fo
rm
f
(
x
)
g
(
x
)
=
0
is
satisfied
whenever
f
(
x
)
=
0
but
g
(
x
)
=
0
.
Example
Solve
2
x
2
−
3
x
−
5
x
2
+
3
x
+
2
=
0
12
Solution
2
x
2
−
3
x
−
5
=
0
set
numerato
r
=
0
(2
x
−
5)(
x
+
1)
=
0
facto
r
x
=
5
2
o
r
x
=
−
1
Then
w
e
must
check
that
neither
of
these
causes
the
denominator
to
b
e
0
.
5
2
2
+
3
5
2
+
2
=
63
4
(
−
1)
2
+
3(
−
1)
+
2
=
0
So
x
=
5
2
is
the
only
solution.
If
there
are
terms
b
esides
the
quotient,
move
them
all
to
the
same
side
of
the
equation
and
use
a
common
denominato
r
to
combine
them.
Example
Solve
2
+
x
+
3
x
+
1
=
4
x
Solution
2
+
x
+
3
x
+
1
−
4
x
=
0
move
to
one
side
2
x
2
+
2
x
x
2
+
x
+
x
2
+
3
x
x
2
+
x
−
4
x
+
4
x
2
+
x
=
0
common
denominato
r
3
x
2
+
x
−
4
x
2
+
x
=
0
combine
set
3
x
2
+
x
−
4
=
0
(3
x
+
4)(
x
−
1)
=
0
facto
r
x
=
−
4
3
o
r
x
=
1
Then
w
e
must
check
that
neither
of
these
causes
the
denominator
to
b
e
0
.
−
4
3
2
+
−
4
3
=
4
9
1
2
+
1
=
2
Both
solutions
a
re
valid.
x
=
−
4
3
o
r
x
=
1
.
13
Section
1.1
Exercises
1.1
Q1
Simplify
5
2
5
4
3
Q2
Simplify
e
5
(
e
4
)
3
Q3
Comp
ress
2
log
5
x
+
log
5
y
−
3
log
5
z
into
a
single
logarithm.
Q4
Comp
ress
3
ln(
x
+
y
)
−
ln(
x
2
+
2
xy
+
y
2
)
into
a
single
loga
rithm.
Q5
Solve
2
e
x
−
7
=
22
Q6
Solve
4
cos(2
x
)
=
1
Q7
Solve
2
sin
2
x
−
1
=
0
Q8
Solve
2
ln(
x
−
5)
=
16
Q9
Solve
4
3
x
−
2
=
15
Q10
Solve
log
7
(
x
2
+
5)
−
3
=
11
1.1.1
Q11
Graph
y
=
3
sin(2
x
)
.
Q12
Graph
y
=
−
ln
x
+
5
.
Q13
Graph
y
=
e
x
−
4
.
Q14
Graph
y
=
3
√
x
+
3
.
Q15
Graph
y
=
1
(
x
−
2)
2
.
14
Q16
Graph
y
=
−
2
√
x
+
1
+
4
.
1.1.2
Q17
Solve
fo
r
x
:
x
2
+
5
x
−
6
x
−
1
=
0
Q18
Solve
fo
r
x
:
e
x
−
2
x
2
+
2
x
−
3
=
0
Q19
Solve
fo
r
x
:
3
x
2
−
5
2
e
x
−
7
=
0
Q20
Solve
fo
r
t
:
ln
t
−
4
3
−
t
=
0
Q21
Solve
fo
r
x
:
ln
x
−
4
3
−
x
=
0
Q22
Solve
fo
r
x
:
3
x
+
2
=
7
x
+
4
Q23
Solve
fo
r
u
:
5
(
u
+
1)
2
=
u
u
+
1
15
Section
1.2
Limits
and
Derivatives
Goals:
1
Compute
limits
of
functions.
2
Verify
that
a
function
is
continuous.
3
Compute
derivatives.
4
Use
derivatives
to
understand
graphs
and
vice
versa.
Question
1.2.1
What
Is
a
Limit?
The
Limit
of
a
F
unction
If
we
can
make
f
(
x
)
a
rbitrarily
close
to
some
numb
er
L
by
considering
only
x
in
a
small
interval
(
a,
a
+
δ
)
then
we
say
the
limit
of
f
as
x
approaches
a
from
the
right
is
L
.
We
write:
lim
x
→
a
+
f
(
x
)
=
L
If
f
(
x
)
cannot
b
e
made
a
rbitra
rily
close
to
any
numb
er,
then
this
limit
do
es
not
exist
.
Simila
rly
,
if
w
e
can
mak
e
f
(
x
)
arbitra
rily
close
to
some
numb
er
L
b
y
considering
only
x
in
a
small
interval
(
a
−
δ,
a
)
then
we
say
the
limit
of
f
as
x
approaches
a
from
the
left
is
L
.
We
write:
lim
x
→
a
−
f
(
x
)
=
L
If
f
(
x
)
cannot
b
e
made
a
rbitra
rily
close
to
any
numb
er,
then
this
limit
do
es
not
exist
.
If
b
oth
lim
x
→
a
+
f
(
x
)
=
L
and
lim
x
→
a
−
f
(
x
)
=
L
,
we
say
the
tw
o-sided
limit
o
r
just
limit
of
f
as
x
app
roaches
a
is
L
.
We
write
lim
x
→
a
f
(
x
)
=
L
If
the
either
the
limit
from
the
left
or
the
limit
from
the
right
does
not
exist,
or
if
they
do
exist
but
a
re
not
equal
to
each
other,
then
the
t
w
o
sided
limit
do
es
not
exist.
16
Click to Load Applet
Figure:
An
interval
of
x
values
that
produce
values
in
a
small
neighb
orhoo
d
of
L
when
plugged
into
f
(
x
)
.
Infinite
Limits
If
f
(
x
)
can
b
e
made
arbitra
rily
large
b
y
considering
only
x
in
a
small
interval
(
a,
a
+
δ
)
then
w
e
say
the
limit
of
f
as
x
approaches
a
from
the
right
is
∞
.
lim
x
→
a
+
f
(
x
)
=
∞
This
is
a
w
ay
of
representing
growth
without
b
ound.
Infinite
limits
from
the
left
are
defined
anal-
ogously
.
Also
analogous
is
our
treatment
of
a
function
then
decreases
without
bound.
We
say
these
functions
limit
to
−∞
.
If
either
one-sided
limit
at
x
=
a
is
infinite,
then
the
line
x
=
a
is
a
vertical
asymptote
of
y
=
f
(
x
)
.
Example
Let
f
(
x
)
=
1
x
.
lim
x
→
0
+
f
(
x
)
=
∞
lim
x
→
0
−
f
(
x
)
=
−∞
17
Question
1.2.1
What
Is
a
Limit?
Figure:
The
graph
of
y
=
1
x
V
ertical
Asymptotes
There
a
re
only
t
w
o
common
algeb
raic
constructions
that
p
ro
duce
infinite
limits.
A
function
of
the
fo
rm
f
(
x
)
g
(
x
)
where
lim
x
→
a
g
(
x
)
=
0
and
lim
x
→
a
f
(
x
)
=
0
.
lim
x
→
0
+
log
a
x
=
−∞
.
Rema
rk
∞
is
not
a
numb
er,
so
if
lim
x
→
a
+
f
(
x
)
=
∞
w
e
w
ould
still
say
that
lim
x
→
a
+
f
(
x
)
do
es
not
exist.
There
a
re
several
limit
la
ws
that
allo
w
us
to
compute
limits
of
combinations
of
simpler
functions.
Theo
rem
[Limit
La
ws]
The
follo
wing
hold
limits,
p
rovided
that
lim
x
→
a
f
(
x
)
and
lim
x
→
a
g
(
x
)
exist.
lim
x
→
a
(
f
(
x
)
+
g
(
x
))
=
lim
x
→
a
f
(
x
)
+
lim
x
→
a
g
(
x
)
lim
x
→
a
(
cf
(
x
))
=
c
lim
x
→
a
f
(
x
)
lim
x
→
a
(
f
(
x
)
g
(
x
))
=
lim
x
→
a
f
(
x
)
lim
x
→
a
g
(
x
)
lim
x
→
a
f
(
x
)
g
(
x
)
=
lim
x
→
a
f
(
x
)
lim
x
→
a
g
(
x
)
!
p
rovided
that
lim
x
→
a
g
(
x
)
=
0
lim
x
→
a
f
(
g
(
x
))
=
lim
x
→
b
f
(
x
)
p
rovided
that
lim
x
→
a
g
(
x
)
=
b
W
e
can
write
simila
r
statements
fo
r
one-sided
limits,
though
w
e
need
to
b
e
careful
ab
out
directions
in
the
comp
osition
rule.
18
Question
1.2.2
What
is
Continuit
y?
Definition
A
function
f
(
x
)
is
continuous
at
a
,
if
lim
x
→
a
f
(
x
)
=
f
(
a
)
Rema
rk
This
definition
is
useful,
if
w
e
already
kno
w
w
e
are
dealing
with
a
continuous
function.
F
or
example
f
(
x
)
=
sin
x
is
continuous
so
lim
x
→
π
6
sin
x
=
sin
π
6
=
1
2
F
o
rtunately
,
many
familia
r
functions
are
continuous.
Theo
rem
The
follo
wing
functions
a
re
continuous
on
their
domains
1
Constant
functions
2
Linear
functions
3
Polynomials
4
Ro
ots
5
Exp
onential
functions
6
Logarithms
7
T
rigonometric
functions
8
f
(
x
)
=
|
x
|
Mo
re
complex
functions
made
from
continuous
functions
a
re
also
continuous.
19
Question
1.2.2
What
is
Continuit
y?
Theo
rem
If
f
(
x
)
and
g
(
x
)
are
continuous
on
their
domains,
and
c
is
a
constant,
then
the
follo
wing
are
also
continuous
on
their
domains
1
f
(
x
)
+
g
(
x
)
2
f
(
x
)
−
g
(
x
)
3
f
(
x
)
g
(
x
)
4
f
(
x
)
g
(
x
)
(note
that
any
x
where
g
(
x
)
=
0
is
not
in
the
domain)
5
f
(
x
)
g
(
x
)
as
long
as
f
(
x
)
>
0
6
f
(
g
(
x
))
Rema
rk
Putting
the
ab
ove
theorems
together,
w
e
see
that
just
ab
out
any
function
we
can
write
using
alge-
b
raic
and
trigonometric
exp
ressions
is
continuous
on
its
domain.
This
do
es
not
mean
it
is
continuous
everywhere.
f
(
x
)
=
1
x
is
not
continuous
at
x
=
0
,
for
example.
Example
1.2.3
Computing
a
Limit
Ho
w
do
w
e
compute
lim
x
→
3
x
2
−
7
x
+
12
x
−
3
?
Solution
f
(
x
)
=
x
2
−
7
x
+12
x
−
3
is
continuous
on
its
domain,
but
x
=
3
is
not
in
the
domain.
Ho
wever,
let
g
(
x
)
=
x
−
4
.
W
e
kno
w
x
2
−
7
x
+12
x
−
3
=
x
−
4
for
every
x
except
x
=
3
.
Sp
ecifically
,
in
any
neighb
orhoo
d
around
x
=
3
,
f
(
x
)
=
g
(
x
)
so
they
have
the
same
limit.
lim
x
→
3
x
2
−
7
x
+
12
x
−
3
=
lim
x
→
3
x
−
4
b
ecause
they
agree
a
round
x
=
3
=
3
−
4
b
ecause
g
(
x
)
=
x
−
4
is
continuous
at
x
=
3
=
−
1
20
Question
1.2.4
What
Is
the
Intermediate
V
alue
Theo
rem?
One
ea
rly
intuition
for
continuity
is
that
the
graph
of
the
function
can
b
e
dra
wn
without
any
breaks.
There
a
re
many
w
a
ys
to
fo
rm
alize
this
idea.
One
of
the
most
imp
ortant
is
the
following
theorem.
Theo
rem
[The
Intermediate
V
alue
Theo
rem]
If
f
is
a
continuous
function
on
[
a,
b
]
and
K
is
a
numb
er
b
etw
een
f
(
a
)
and
f
(
b
)
,
then
there
is
some
numb
er
c
b
et
w
een
a
and
b
such
that
f
(
c
)
=
K
.
This
theorem
essentially
states
that
a
continuous
graph
cannot
get
from
one
side
of
the
line
y
=
K
to
the
other
without
intersecting
y
=
K
.
Notice
that
this
theorem
does
not
say
exactly
where
this
intersection
must
occur,
only
that
it
must
o
ccur
somewhere
in
the
interval
(
a,
b
)
.
It
also
does
not
rule
out
the
p
ossibilit
y
of
mo
re
than
one
such
c
existing.
Example
Sho
w
that
f
(
x
)
=
e
x
−
3
x
has
a
ro
ot
b
et
w
een
0
and
1
.
Solution
A
ro
ot
is
a
numb
er
c
such
that
f
(
c
)
=
0
.
T
o
prove
such
a
ro
ot
exists,
w
e
check
the
conditions
of
the
IVT.
f
(
x
)
is
a
sum
of
continuous
functions,
so
it
is
continuous
on
its
domain.
f
(0)
=
1
f
(1)
=
e
−
3
<
0
0
is
b
et
w
een
f
(0)
and
f
(1)
W
e
conclude
there
is
some
c
b
et
w
een
0
and
1
such
that
f
(
c
)
=
0
.
21
Question
1.2.5
What
Is
a
Limit
at
Infinit
y?
Definition
If
we
can
mak
e
f
(
x
)
arbitra
rily
close
to
some
numb
er
L
b
y
considering
only
x
in
some
interval
(
n,
∞
)
then
we
say
the
limit
of
f
as
x
approaches
∞
is
L
.
W
e
write:
lim
x
→∞
f
(
x
)
=
L
If
f
(
x
)
cannot
b
e
made
a
rbitra
rily
close
to
any
numb
er,
then
this
limit
do
es
not
exist
.
Simila
rly
if
we
can
f
(
x
)
arbitra
rily
close
to
L
by
considering
only
x
in
some
interval
(
−∞
,
n
)
then
w
e
sa
y
the
limit
of
f
as
x
approaches
−∞
is
L
.
We
write:
lim
x
→−∞
f
(
x
)
=
L
If
either
lim
x
→∞
f
(
x
)
=
L
o
r
lim
x
→−∞
f
(
x
)
=
L
,
then
y
=
L
is
a
ho
rizontal
aysmptote
of
the
graph
y
=
f
(
x
)
.
By
observing
graphs
o
r
using
a
rithmetic
intuition,
w
e
a
rrive
at
the
follo
wing
limits
at
infinit
y
.
f
(
x
)
lim
x
→∞
f
(
x
)
lim
x
→−∞
f
(
x
)
Comments
x
n
(
n
o
dd)
∞
−∞
n
>
0
x
n
(
n
even)
∞
∞
n
>
0
n
√
x
(
n
o
dd)
∞
DNE
domain
is
x
≥
0
n
√
x
(
n
even)
∞
−∞
1
x
n
0
0
n
>
0
a
x
(
a
>
1
)
∞
0
a
x
(
0
<
a
<
1
)
0
∞
log
a
x
∞
DNE
a
>
1
,
domain
is
x
>
0
sin
x
DNE
DNE
oscillates
tan
−
1
x
π
2
−
π
2
22
Question
1.2.6
Ho
w
Do
W
e
Measure
the
Change
in
a
F
unction?
Definition
The
average
rate
of
change
of
a
function
f
(
x
)
b
et
w
een
x
=
a
and
x
=
b
is
f
(
b
)
−
f
(
a
)
b
−
a
This
is
also
the
slop
e
of
the
secant
line
from
(
a,
f
(
a
))
to
(
b,
f
(
b
))
on
the
graph
y
=
f
(
x
)
.
Kno
wing
the
average
rate
of
change
over
a
range
of
inputs
(or
times)
doesn’t
tell
us
the
rate
of
change
at
a
sp
ecific
p
oint
(or
moment).
Geometrically
the
is
the
slop
e
of
the
tangent
line
to
y
=
f
(
x
)
at
a
pa
rticula
r
p
oint
(
a,
f
(
a
))
Click to Load Applet
Figure:
A
secant
line
and
a
tangent
line
The
secant
lines
get
closer
and
closer
to
the
tangent
line
(in
slop
e)
as
b
gets
closer
to
a
.
This
suggests
that
w
e
could
tak
e
the
limit
of
these
app
roaching
values
to
get
the
actual
slop
e.
Definition
The
instantaneous
rate
of
change
o
r
derivative
of
a
function
f
(
x
)
at
x
=
a
is
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
p
rovided
that
this
limit
exists.
This
is
also
the
slop
e
of
the
tangent
line
to
y
=
f
(
x
)
at
(
a,
f
(
a
))
.
Two
common
notations
fo
r
the
derivative
a
re
Prime
notation:
f
′
(
a
)
Leibniz
notation:
d
f
dx
x
=
a
23
Question
1.2.6
Ho
w
Do
W
e
Measure
the
Change
in
a
F
unction?
Click to Load Applet
Figure:
A
limit
of
the
slop
es
of
secant
lines
W
e
can
attempt
to
compute
the
derivative
at
any
point
a
.
W
e
can
put
these
values
together
to
create
a
function
f
′
(
x
)
.
Definition
The
derivative
function
of
f
(
x
)
is
the
function
that
tak
es
the
value
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
at
each
x
.
W
e
can
denote
the
derivative
function
as
f
′
(
x
)
o
r
d
f
dx
.
The
second
can
b
e
rewritten
d
dx
f
to
emphasize
that
w
e
a
re
applying
the
differentiation
op
eration
to
the
function
f
.
Example
If
f
(
x
)
=
x
2
+
2
x
,
compute
f
′
(
x
)
.
24
Solution
f
′
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
definition
of
derivative
=
lim
h
→
0
(
x
+
h
)
2
+
2(
x
+
h
)
−
x
2
−
2
x
h
plug
in
x
and
x
+
h
=
lim
h
→
0
x
2
+
2
xh
+
h
2
+
2
x
+
2
h
−
x
2
−
2
x
h
distribute
=
lim
h
→
0
2
xh
+
h
2
+
2
h
h
cancel
=
lim
h
→
0
2
x
+
h
+
2
functions
agree
except
at
h
=
0
so
limits
are
equal
=
2
x
+
0
+
2
limit
=
value
on
a
continuous
function
=
2
x
+
2
Theo
rem
If
f
′
(
x
)
>
0
for
all
x
in
some
interval
[
a,
b
]
then
f
(
x
)
is
increasing
on
[
a,
b
]
.
If
f
′
(
x
)
<
0
for
all
x
on
[
a,
b
]
then
f
(
x
)
is
decreasing
on
[
a,
b
]
.
W
e
can
take
higher
order
derivatives
b
y
taking
derivatives
of
derivatives.
The
derivative
function
of
f
in
this
context
is
called
the
first
derivative
.
Its
derivative
function
is
the
second
derivative
.
The
second
derivative’s
derivative
function
is
the
third
derivative
and
so
on.
Notation
The
follo
wing
notations
a
re
used
fo
r
higher
o
rder
derivatives
name
p
rime
notation
Leibniz
notation
first
derivative
f
′
(
x
)
d
f
dx
second
derivative
f
′′
(
x
)
d
2
f
dx
2
third
derivative
f
′′′
(
x
)
d
3
f
dx
3
fourth
derivative
f
(4)
(
x
)
d
4
f
dx
4
fifth
derivative
f
(5)
(
x
)
d
5
f
dx
5
25
Question
1.2.6
Ho
w
Do
W
e
Measure
the
Change
in
a
F
unction?
The
sign
of
a
higher
o
rder
derivative
tells
us
how
the
derivative
of
one
o
rder
low
er
is
changing.
Fo
r
example
if
d
5
f
dx
5
<
0
,
then
d
4
f
dx
4
is
decreasing.
The
sign
of
higher
o
rder
derivatives
is
difficult
to
discern
from
the
shap
e
of
y
=
f
(
x
)
,
with
the
exeption
of
the
second
derivative.
Theo
rem
If
f
′′
(
x
)
>
0
on
some
interval,
then
y
=
f
(
x
)
is
concave
up
on
that
interval.
If
f
′′
(
x
)
<
0
,
then
y
=
f
(
x
)
is
concave
down
.
Definition
A
point
a
such
that
f
(
x
)
is
concave
up
to
one
side
of
a
and
concave
down
to
the
other
side
is
called
an
inflection
p
oint
.
Question
1.2.7
Ho
w
Do
W
e
Compute
Derivatives
The
limit
definition
of
a
derivative
is
to
o
unwieldy
to
use
every
time.
A
b
etter
app
roach
is
to
learn
the
derivatives
of
some
simple
functions,
and
then
use
theorems
to
compute
derivatives
when
those
functions
a
re
combined.
Derivatives
of
Simple
F
unctions
d
dx
c
=
0
(derivative
of
a
constant
is
0
)
d
dx
x
n
=
nx
n
−
1
fo
r
any
n
=
0
(The
Po
wer
Rule)
d
dx
sin
x
=
cos
x
d
dx
cos
x
=
−
sin
x
d
dx
e
x
=
e
x
d
dx
a
x
=
a
x
ln
a
for
a
>
0
d
dx
ln
x
=
1
x
26
Theo
rem
The
follo
wing
rules
allow
us
to
differentiate
functions
made
of
simpler
functions
whose
derivative
we
kno
w.
Sum
Rule
(
f
(
x
)
+
g
(
x
))
′
=
f
′
(
x
)
+
g
′
(
x
)
Constant
Multiple
Rule
(
cf
(
x
))
′
=
cf
′
(
x
)
Pro
duct
Rule
(
f
(
x
)
g
(
x
))
′
=
f
′
(
x
)
g
(
x
)
+
g
′
(
x
)
f
(
x
)
Quotient
Rule
f
(
x
)
g
(
x
)
′
=
f
′
(
x
)
g
(
x
)
−
g
′
(
x
)
f
(
x
)
(
g
(
x
))
2
unless
g
(
x
)
=
0
Chain
Rule
(
f
(
g
(
x
))
′
=
f
′
(
g
(
x
))
g
′
(
x
)
Example
Compute
d
dx
tan(
x
)
Solution
tan
x
=
sin
x
cos
x
.
We
apply
the
quotient
rule
(tan
x
)
′
=
(sin
x
)
′
cos
x
−
(cos
x
)
′
sin
x
cos
2
x
quotient
rule
=
cos
2
x
+
sin
2
x
cos
2
x
=
1
cos
2
x
Pythago
rean
identit
y
=
sec
2
x
27
Application
1.2.8
The
Shap
e
of
a
Graph
What
can
the
first
and
second
derivative
of
f
(
x
)
=
8
x
3
−
x
4
tell
us
ab
out
the
shap
e
of
its
graph?
Solution
W
e
will
compute
the
first
and
second
derivative
using
the
p
ow
er
rule.
Facto
ring
them
will
allo
w
us
to
p
erfo
rm
a
sign
analysis.
f
′
(
x
)
=
24
x
2
−
4
x
3
f
′′
(
x
)
=
48
x
−
12
x
2
=
4
x
2
(6
−
x
)
=
12
x
(4
−
x
)
4
x
2
+
+
+
(6
−
x
)
+
+
−
f
′
(
x
)
+
+
−
0
6
12
x
−
+
+
(4
−
x
)
+
+
−
f
′′
(
x
)
−
+
−
0
4
F
rom
the
sign
of
f
′
(
x
)
w
e
conclude
f
is
increasing
on
(
−∞
,
0)
and
(0
,
6)
but
decreasing
on
(6
,
∞
)
.
F
rom
the
sign
of
f
′′
(
x
)
we
conclude
that
f
is
concave
down
on
(
−∞
,
0)
and
(4
,
∞
)
,
but
concave
up
on
(0
,
4)
.
Figure:
The
graph
of
y
=
8
x
3
−
x
4
28
Section
1.2
Exercises
1.2.1
Q1
Given
the
graph
of
y
=
f
(
x
)
here,
give
the
value
of
each
of
the
following
limits
(if
they
exist).
a
lim
x
→−
3
−
f
(
x
)
b
lim
x
→−
3
+
f
(
x
)
c
lim
x
→−
2
f
(
x
)
d
lim
x
→
0
f
(
x
)
e
lim
x
→
4
−
f
(
x
)
f
lim
x
→
4
+
f
(
x
)
Q2
Given
the
graph
of
y
=
g
(
x
)
here,
give
the
value
of
each
of
the
following
limits
(if
they
exist).
a
lim
x
→
0
−
g
(
x
)
b
lim
x
→
0
+
g
(
x
)
c
lim
x
→
0
g
(
x
)
d
lim
x
→
3
−
g
(
x
)
e
lim
x
→
3
+
g
(
x
)
f
lim
x
→
3
g
(
x
)
g
lim
x
→−
4
−
g
(
x
)
h
lim
x
→−
4
+
g
(
x
)
i
lim
x
→−
1
g
(
x
)
29
Section
1.2
Exercises
1.2.2
Q3
Explain
why
f
(
x
)
=
e
x
x
2
+3
is
continuous
on
R
.
Q4
Explain
why
f
(
x
)
=
p
sin(3
x
2
)
is
continuous
on
its
domain.
Q5
Is
f
(
x
)
=
sin(2
x
)
if
x
<
0
4
if
x
=
0
−
x
2
if
x
>
0
continuous
at
x
=
0
?
Justify
your
answer.
Q6
Is
f
(
x
)
=
(
x
3
−
2
x
+
1
if
x
<
0
e
x
if
x
≥
0
continuous
at
x
=
0
?
Justify
your
answer.
Q7
Is
f
(
x
)
=
x
+
5
if
x
<
1
6
if
x
=
1
x
2
+
4
x
+
1
if
x
>
1
continuous
at
x
=
1
?
Justify
your
answer.
Q8
Where
is
f
(
x
)
=
cos(
π
x
)
if
x
<
4
1
if
x
=
4
√
x
−
3
if
x
>
4
continuous?
30
1.2.3
Q9
Compute
lim
x
→
3
x
−
3
x
2
−
9
Q10
Compute
lim
x
→
1
x
2
−
4
x
+
3
x
−
1
Q11
Compute
lim
x
→
9
2
x
−
18
√
x
−
3
Q12
Compute
lim
x
→
4
1
x
2
−
1
16
x
−
4
1.2.4
Q13
Explain
why
sin
x
=
2
x
−
1
has
a
solution
in
[0
,
1]
.
Q14
Explain
why
3
√
x
=
log
2
x
has
a
solution
in
[0
,
8]
.
Q15
What
do
es
the
Intermediate
Value
Theorem
say
ab
out
whether
f
(
x
)
=
1
x
−
1
2
has
a
ro
ot
in
[
−
1
,
1]
?
Q16
Consider
the
equation
sin
x
=
3
4
.
Gloria
computes
sin
π
3
=
√
3
2
and
sin
5
π
6
=
1
2
.
Since
3
4
is
not
b
et
w
een
1
2
and
√
3
2
,
she
concludes
that
sin
x
=
3
4
has
no
ro
ots
in
π
3
,
5
π
6
.
What
do
you
think
of
Glo
ria’s
reasoning?
1.2.5
Q17
Compute
lim
x
→∞
x
2
+
2
x
−
9
3
x
−
6
.
Q18
Compute
lim
x
→∞
4
x
2
−
7
x
+
9
2
x
2
+
11
.
Q19
Compute
lim
x
→∞
p
e
1
/x
.
Q20
Compute
lim
x
→∞
1
ln
x
.
31
Section
1.2
Exercises
Q21
Compute
lim
x
→−∞
e
e
x
.
Q22
Compute
lim
x
→∞
sin(ln
x
)
.
1.2.6
Q23
Let
f
(
x
)
=
x
3
.
a
Compute
the
average
rate
of
change
of
f
from
x
=
2
to
x
=
5
.
b
Give
the
equation
of
the
secant
line
that
meets
y
=
f
(
x
)
at
x
=
2
and
x
=
5
.
c
Use
the
limit
definition
of
the
derivative
to
compute
f
′
(2)
.
Q24
Let
f
(
x
)
=
√
x
Compute
the
average
rate
of
change
of
f
b
etw
een
x
=
4
and
x
=
9
.
Based
on
the
graph
of
y
=
f
(
x
)
,
is
the
instantaneous
rate
of
change
at
x
=
4
greater
or
less
than
this
average?
Q25
Let
f
(
x
)
=
3
x
2
−
7
.
Compute
f
′
(6)
using
the
limit
definition
of
the
derivative.
Q26
Let
f
(
x
)
=
1
x
+2
.
Compute
f
′
(1)
using
the
limit
definition
of
the
derivative.
Q27
Let
f
(
x
)
=
1
x
2
.
Compute
f
′
(
x
)
using
the
limit
definition
of
the
derivative.
Q28
Let
f
(
x
)
=
√
x
.
Compute
f
′
(
x
)
using
the
limit
definition
of
the
derivative.
32
1.2.7
Q29
Use
derivative
rules
to
differentiate
each
of
the
following
functions.
a
5
x
7
−
3
x
2
+
5
x
2
b
4
x
5
−
2
x
2
+
3
x
+
4
x
c
(
x
2
+
2
x
)
sin
x
d
e
x
x
2
e
√
x
−
5
f
cos(4
x
)
g
sin(
e
x
)
h
(
x
2
+
5
x
+
4)
60
i
e
x
2
sin
x
j
ln(
x
2
+
2)
x
2
+
3
x
Q30
Use
derivative
rules
to
differentiate
each
of
the
following
functions.
a
3
x
+
7
x
3
b
5
x
4
+
3
x
3
−
8
x
2
x
2
c
ln
x
x
d
4
x
sin(
x
)
e
tan(2
x
+
7)
f
e
3
x
+2
g
cos(
x
3
+
2
x
)
h
5
(cos
x
)
3
i
e
x
2
sin
3
x
j
ln(
√
x
sin
x
)
Q31
Let
f
(
x
)
=
sin(3
x
)
.
Compute
f
′′′
(
x
)
.
Q32
Let
f
(
x
)
=
e
x
3
.
Compute
f
′′
(
x
)
.
1.2.8
Q33
Where
in
its
domain
is
the
function
f
(
x
)
=
x
3
−
x
2
increasing?
33
Section
1.2
Exercises
Q34
Where
in
its
domain
is
f
(
x
)
=
e
x
−
x
2
concave
up?
Q35
Where
in
its
domain
is
f
(
x
)
=
1024
√
x
−
x
4
increasing?
Q36
Find
the
inflection
p
oint(s)
of
x
4
−
8
x
3
.
34
Section
1.3
Applications
of
Derivatives
Goals:
1
Write
the
equation
of
a
tangent
line.
2
Identify
lo
cal
maximums
and
minimums.
3
Use
the
Extreme
Value
Theorem
to
find
minimums
and
maximums.
4
Use
l’Hˆ
opital’s
rule
to
compute
limits.
This
section
reviews
the
most
imp
o
rtant
applications
of
the
derivative.
Application
1.3.1
The
T
angent
Line
to
a
Graph
Given
a
function
f
(
x
)
,
the
derivative
f
′
(
a
)
is
the
slope
of
the
line
tangent
to
y
=
f
(
x
)
at
(
a,
f
(
a
))
.
F
o
rmula
The
equation
of
the
tangent
line
to
y
=
f
(
x
)
at
x
=
a
in
p
oint-slop
e
form
is:
y
−
f
(
a
)
=
f
′
(
a
)(
x
−
a
)
W
e
can
rewrite
the
tangent
line
as
a
function
of
x
.
We
call
this
a
linearization,
b
ecause
this
function
is
linea
r,
but
it
app
ro
ximates
the
value
of
f
(
x
)
fo
r
x
near
a
.
F
o
rmula
The
linea
rization
of
y
=
f
(
x
)
at
x
=
a
is
the
function:
L
(
x
)
=
f
(
a
)
+
f
′
(
a
)(
x
−
a
)
If
we
want
to
emphasize
the
change
in
x
and
y
instead
of
their
actual
values
we
can
use
differential
notation:
35
Application
1.3.1
The
T
angent
Line
to
a
Graph
Notation
If
y
=
f
(
x
)
is
appro
ximated
by
a
tangent
line
at
x
=
a
then
we
let
dx
=
x
−
a
represents
a
change
in
x
from
a
.
Since
x
is
an
indep
endent
variable,
so
is
dx
.
dy
=
f
′
(
a
)
dx
is
equal
to
the
change
in
y
corresponding
to
dx
,
if
we
travel
along
the
tangent
line.
This
appro
ximates
the
actual
change
in
f
(
x
)
if
x
increases
by
dx
.
Click to Load Applet
Figure:
The
differentials
dx
and
dy
on
the
tangent
line
to
y
=
f
(
x
)
Application
1.3.2
Maximum
and
Minimum
V
alues
of
a
F
unction
Definition
A
numb
er
a
is
a
maximum
of
a
function
f
(
x
)
if
f
(
a
)
≥
f
(
x
)
for
all
x
in
the
domain
of
f
.
a
is
a
minimum
if
f
(
a
)
≤
f
(
x
)
for
all
x
in
the
domain
of
f
.
36
Definition
A
numb
er
a
is
a
lo
cal
maximum
of
a
function
f
(
x
)
if
f
(
a
)
≥
f
(
b
)
for
all
b
in
some
neighb
orhoo
d
of
a
.
a
is
a
lo
cal
minimum
if
f
(
a
)
≤
f
(
b
)
for
all
b
in
some
neighb
orhoo
d
of
a
.
T
o
distinguish
o
rdinary
maximums
from
the
lo
cal
variet
y
,
we
sometimes
call
them
global
maximums
o
r
absolute
maximums
.
Every
global
maximum
is
a
lo
cal
maximum,
but
lo
cal
maximums
need
not
b
e
global
maximums.
If
f
′
(
a
)
>
0
then
there
a
re
larger
values
of
f
(
a
)
to
the
right
of
a
and
low
er
values
to
the
left.
Thus
a
cannot
b
e
a
lo
cal
maximum
or
minimum.
The
same
argument
applies
if
f
′
(
a
)
<
0
.
Figure:
Maximum
and
minimum
values
of
f
(
x
)
Definition
A
critical
p
oint
of
f
(
x
)
is
a
value
a
in
the
domain
of
f
such
that
either
f
′
(
a
)
=
0
o
r
f
′
(
a
)
does
not
exist.
Theo
rem
[The
First
Derivative
T
est]
Lo
cal
maximums
and
minimums
of
f
(
x
)
can
only
o
ccur
at
critical
p
oints.
W
e
can
use
concavit
y
as
a
wa
y
to
classify
critical
p
oints.
Kno
wing
whether
a
graph
is
concave
up
o
r
concave
do
wn
at
a
p
oint
where
f
′
(
x
)
=
0
allo
ws
us
to
visualize
a
small
neighb
o
rho
o
d
of
that
p
oint.
37
Application
1.3.2
Maximum
and
Minimum
V
alues
of
a
F
unction
Theo
rem
[The
Second
Derivative
T
est]
Let
a
b
e
a
critical
p
oint
of
f
.
If
f
′′
(
a
)
<
0
then
a
is
a
lo
cal
maximum.
If
f
′′
(
a
)
>
0
then
a
is
a
lo
cal
minimum.
If
f
′′
(
a
)
=
0
or
do
es
not
exist,
then
the
test
is
inconclusive.
a
could
b
e
a
lo
cal
maximum,
a
lo
cal
minimum,
o
r
neither.
Example
What
do
es
the
second
derivative
test
tell
you
ab
out
the
critical
p
oints
of
f
(
x
)
=
8
x
3
−
x
4
?
Solution
First
w
e
compute
the
critical
p
oints.
f
′
(
x
)
=
24
x
2
−
4
x
3
compute
first
derivative
0
=
24
x
2
−
4
x
3
set
equal
to
0
0
=
4
x
2
(6
−
x
)
facto
r
x
=
0
or
x
=
6
No
w
w
e
compute
the
second
derivative
and
evaluate
it
at
each
critical
p
oint.
f
′′
(
x
)
=
48
x
−
12
x
2
f
′′
(0)
=
0
f
′′
(6)
=
(48)(6)
−
(12)(36)
=
−
144
f
′′
(6)
<
0
so
x
=
6
is
a
lo
cal
maximum.
f
′′
(0)
=
0
so
the
second
derivative
test
cannot
tell
whether
x
=
0
is
a
lo
cal
maximum
or
lo
cal
minimum
(in
fact
it
is
neither).
38
Question
1.3.3
Do
es
a
F
unction
Alw
a
ys
Have
a
Maximum?
No.
Many
functions
don’t
have
maximums,
b
ecause
as
x
gets
larger
and
larger
the
values
of
f
(
x
)
increase
or
decrease
without
b
ound.
How
ever,
if
we
restrict
the
domain,
w
e
can
sometimes
guarantee
a
maximum
Theo
rem
[The
Extreme
V
alue
Theo
rem]
If
f
(
x
)
is
a
continuous
function
on
a
closed
domain
[
a,
b
]
then
f
has
an
absolute
maximum
and
an
absolute
minimum
on
[
a,
b
]
.
Rema
rk
When
the
EVT
applies,
w
e
can
find
the
absolute
maximum
and
minimum
by
process
of
elimination.
A
maximum
exists,
so
it
must
o
ccur
at
a
critical
p
oint.
W
e
can
find
the
critical
points
and
evaluate
f
at
each
of
them.
Whichever
has
the
greatest
value
is
the
maximum.
Note
that
a
and
b
are
alwa
ys
critical
p
oints
b
ecause
the
derivative
do
es
not
exist
there.
There
is
no
limit
from
the
left
at
a
b
ecause
those
points
a
re
outside
the
domain
of
f
.
Similarly
,
there
is
no
limit
from
the
right
at
b
.
Example
Compute
the
maximum
and
minimum
value
of
f
(
x
)
=
8
x
3
−
x
4
on
the
domain
[2
,
8]
,
if
they
exist.
Solution
f
(
x
)
is
continuous
and
[2
,
8]
is
closed,
so
the
EVT
guarantees
that
a
maximum
and
minimum
exist.
The
first
derivative
test
sa
ys
that
they
can
only
o
ccur
at
critical
p
oints.
f
′
(
x
)
=
24
x
2
−
4
x
3
compute
first
derivative
0
=
24
x
2
−
4
x
3
set
equal
to
0
0
=
4
x
2
(6
−
x
)
facto
r
x
=
0
or
x
=
6
x
=
0
is
not
in
the
domain,
so
we
discard
it.
On
the
other
hand
x
=
2
and
x
=
8
are
also
critical
p
oints
b
ecause
the
derivative
do
es
not
exist
there.
T
o
find
which
critical
point
is
the
maximum
and
which
is
the
minimum,
w
e
plug
each
into
f
and
compare.
f
(2)
=
(8)(8)
−
16
=
48
f
(6)
=
(8)(216)
−
1296
=
436
(maximum)
f
(8)
=
(8)(512)
−
4096
=
0
(minimum)
39
Application
1.3.4
L’Hˆ
opital’s
Rule
The
limit
rules
tell
us
ho
w
to
take
limits
of
quotients,
products,
sums
and
differences.
What
happ
ens
if
one
of
the
functions
b
eing
divided
go
es
to
∞
,
o
r
if
the
denominator
of
a
quotient
go
es
to
0
?
In
some
cases
w
e
can
reason
this
out
using
our
intuition
of
arithmetic.
Example
Consider
lim
x
→∞
tan
−
1
(
x
)
ln
x
.
lim
x
→∞
tan
−
1
x
=
π
2
lim
x
→∞
ln
x
=
∞
Since
the
numerators
are
approaching
π
/
2
and
the
denominators
a
re
increasing
without
b
ound,
we
conclude
that
this
ratio
get
smaller
and
smaller
and
will
limit
to
0
.
In
other
cases,
intuition
cannot
help
us.
Definition
A
limit
of
the
fo
rm
lim
x
→
a
f
(
x
)
g
(
x
)
is
of
indeterminate
fo
rm
if
either
lim
x
→
a
f
(
x
)
=
±∞
and
lim
x
→
a
g
(
x
)
=
±∞
or
lim
x
→
a
f
(
x
)
=
0
and
lim
x
→
a
g
(
x
)
=
0
This
definition
also
applies
to
one-sided
limits
or
to
limits
at
±∞
.
Limits
of
products
and
sums
can
sometimes
b
e
rewritten
as
quotients
of
indeterminate
form
as
well.
Theo
rem
[L’Hˆ
opital’s
Rule]
If
lim
x
→
a
f
(
x
)
g
(
x
)
is
of
indeterminate
fo
rm,
then
it
is
equal
to
lim
x
→
a
f
′
(
x
)
g
′
(
x
)
assuming
this
limit
exists.
Often
L’Hˆ
opital’s
Rule
converts
a
limit
of
indeterminate
fo
rm
to
one
w
e
can
evaluate
through
intuition
o
r
direct
computation.
Sometimes,
we
need
to
apply
L’Hˆ
opital’s
Rule
more
than
once.
W
a
rning
If
a
limit
is
not
of
indeterminate
fo
rm,
then
L’Hˆ
opital’s
Rule
do
es
not
apply
.
Attempting
to
apply
it
will
usually
give
an
inco
rrect
value
fo
r
the
limit.
40
Example
1.3.5
A
Limit
of
Indeterminate
F
o
rm
Evaluate
lim
x
→
0
e
x
−
x
−
1
x
2
Solution
lim
x
→
0
e
x
−
x
−
1
x
2
0
0
fo
rm
=
lim
x
→
0
e
x
−
1
2
x
L’Hˆ
opital’s
Rule,
still
0
0
fo
rm
=
lim
x
→
0
e
x
2
L’Hˆ
opital’s
Rule
again
=
1
2
Section
1.3
Exercises
1.3.1
Q1
W
rite
the
equation
of
the
tangent
line
to
y
=
√
x
at
(4
,
2)
.
Q2
W
rite
the
equation
of
the
tangent
line
to
y
=
1
x
2
at
5
,
1
25
.
Q3
Let
f
(
x
)
=
sin(
x
)
a
W
rite
the
equation
of
the
linea
rization
y
=
f
(
x
)
at
x
=
π
3
.
b
If
we
wanted
to
use
a
to
appro
ximate
sin(1)
by
hand,
what
numb
er(s)
would
w
e
need
decimal
app
ro
ximations
of
?
c
Use
a
calculator
to
get
decimal
appro
ximations
of
those
numb
ers,
then
show
how
to
app
rox-
imate
sin(1)
.
Q4
W
rite
a
linea
rization
of
f
(
x
)
=
1
x
at
x
=
3
and
use
it
to
appro
ximate
1
2
.
93
.
41
Section
1.3
Exercises
Q5
A
baterical
culture
has
mass
3
g
after
t
=
5
hours
of
growth.
At
that
time,
its
instantaneous
rate
of
gro
wth
is
0
.
2
g
/hr
.
a
W
rite
a
linea
r
function
to
app
ro
ximate
m
(
t
)
the
mass
of
the
culture
at
hour
t
.
b
App
ro
ximate
the
mass
at
time
8
hours.
c
Given
that
m
′′
(
t
)
>
0
,
is
your
answer
to
b
is
overestiamte
o
r
an
underestimate?
Q6
A
space
capsule
is
descending
from
orbit.
After
90
seconds,
it
is
10
,
000
m
ab
ove
sea
level
and
falling
at
400
m
p
er
second.
a
W
rite
a
linea
rization
fo
r
h
(
t
)
,
the
height
of
the
capsule
at
time
t
.
b
Use
a
to
p
redict
when
the
capsule
will
splash
do
wn
into
the
o
cean.
c
Do
y
ou
exp
ect
that
y
our
answ
er
to
b
is
an
overestimate
o
r
underestimate?
Explain.
1.3.2
Q7
Find
the
critical
p
oints
of
f
(
x
)
=
12
x
2
/
3
−
x
.
Q8
Find
the
critical
p
oints
of
g
(
x
)
=
x
4
−
18
x
2
+
5
.
Apply
the
second
derivative
test
to
each.
Q9
Find
the
critical
p
oints
of
f
(
x
)
=
x
3
−
75
x
.
Apply
the
second
derivative
test
to
each.
Q10
Find
the
critical
p
oints
of
g
(
x
)
=
e
x
−
2
x
.
Apply
the
second
derivative
test
to
each.
1.3.3
Q11
Find
the
maximum
and
minimum
values
of
f
(
x
)
=
x
2
/
3
on
[
−
8
,
1]
.
Q12
Find
the
maximum
and
minimum
values
of
f
(
x
)
=
x
3
−
75
x
on
[
−
10
,
10]
.
42
1.3.4
Q13
Evaluate
lim
x
→
0
+
x
cos(
x
−
π
)
e
x
−
1
.
Q14
Evaluate
lim
x
→
0
+
e
−
3
x
+
3
x
−
1
sin(
x
2
)
.
Q15
Evaluate
lim
x
→∞
x
ln
x
x
5
/
2
+
3
.
Q16
Evaluate
lim
x
→−∞
e
x
x
2
.
43
Section
1.4
Definite
Integrals
Goals:
1
Express
areas
under
a
graph
and
antiderivatives
using
integral
notation.
2
Derive
antiderivatives
from
known
derivatives.
3
Compute
general
antiderivatives.
4
Compute
definite
integrals
using
the
Fundamental
Theorem
of
Calculus.
5
Use
u
-substitution
to
compute
integrals
where
necessary
.
By
definition,
integrals
compute
area
under
a
graph.
The
F
undamental
Theorem
of
Calculus
connects
integrals
to
antiderivatives,
meaning
that
integrals
can
also
b
e
used
to
compute
total
change,
given
a
rate
of
change
function.
Question
1.4.1
What
Is
an
Antiderivative?
Definition
F
(
x
)
is
antiderivative
of
a
function
f
(
x
)
,
if
F
′
(
x
)
=
f
(
x
)
.
Every
derivative
w
e
kno
w
also
tells
us
an
antiderivative.
Example
d
dx
x
2
2
+
5
=
x
so
F
(
x
)
=
x
2
2
+
5
is
an
antiderivative
of
f
(
x
)
=
x
.
Notice
that
x
2
2
+
2
,
x
2
2
−
6
,
and
x
2
2
a
re
also
antiderivatives
of
f
(
x
)
=
x
.
F
unctions
have
infinitely
many
antiderivatives.
Adding
a
constant
to
one
antiderivative
produces
another,
since
the
derivative
of
a
constant
is
0
.
In
fact,
this
is
the
only
relationship
b
etw
een
antideriva-
tives.
Theo
rem
If
F
(
x
)
and
G
(
x
)
are
antideriavatives
of
f
(
x
)
,
then
there
is
a
constant
c
such
that
F
(
x
)
=
G
(
x
)
+
c.
Since
the
antiderivatives
are
related
this
wa
y
,
it
is
easy
to
exp
ress
all
of
the
antiderivatives
of
a
function
at
once.
44
Definition
If
F
(
x
)
is
an
antiderivative
of
f
(
x
)
,
then
the
general
antiderivative
of
f
(
x
)
is
the
family
of
functions:
F
(
x
)
+
c
where
c
can
b
e
any
constant.
Here
is
a
table
of
antiderivatives
that
we
can
compute
just
by
reverse
engineering
the
derivatives
we
already
kno
w.
f
(
x
)
general
antiderivative
of
f
(
x
)
x
n
x
n
+1
n
+1
+
c
e
x
e
x
+
c
a
x
a
x
ln
a
+
c
1
x
ln
x
+
c
sin
x
−
cos
x
cos
x
sin
x
Rema
rk
Many
familiar
functions
are
missing
from
this
list.
This
is
b
ecause
we
just
haven’t
come
across
them
as
derivatives
of
some
other
function.
F
or
instance,
we
do
not
yet
know
a
function
F
(
x
)
whose
derivative
is
ln
x
or
tan
x
.
Question
1.4.2
Ho
w
Do
W
e
Compactly
Denote
a
Sum
of
Many
T
erms
Defining
the
definite
integral
requires
us
to
add
up
many
numb
ers.
The
problem
is
not
just
that
the
numb
er
of
summands
is
la
rge.
We
need
to
b
e
flexible
ab
out
how
many
terms
a
re
in
the
sum.
The
notation
that
gives
us
this
flexibilit
y
is
Σ
notation.
Notation
Σ
(‘sigma’)
notation
allows
us
to
sum
many
different
values
of
an
expression
using
an
index
va
riable
.
The
index
variable
will
b
e
replaced
by
each
integer
b
etw
een
an
initial
and
final
value,
and
the
resulting
outputs
a
re
added
together.
n
X
k
=1
f
(
k
)
=
f
(1)
+
f
(2)
+
f
(3)
+
·
·
·
+
f
(
n
)
W
e
may
cho
ose
any
variable
as
the
index
variable.
The
index
variable
could
also
have
a
different
initial
value,
if
that
is
mo
re
convenient.
45
Question
1.4.2
Ho
w
Do
W
e
Compactly
Denote
a
Sum
of
Many
T
erms
Example
7
X
j
=3
j
2
j
+
1
=
9
4
+
16
5
+
25
6
+
36
7
+
49
8
P
a
rt
of
the
challenge
of
writing
a
sum
in
P
notation
is
cho
osing
an
f
that
will
produce
all
the
terms
of
y
our
sum.
Example
1.4.3
W
riting
a
Sum
in
Σ
Notation
W
rite
each
of
the
follo
wing
sums
in
Σ
notation.
a
4
+
7
+
10
+
13
+
16
+
19
+
22
b
2
+
6
+
18
+
54
+
162
+
486
c
−
3
+
4
−
5
+
6
−
7
+
8
−
9
+
10
d
1
4
+
√
2
9
+
√
3
16
+
2
25
+
√
5
36
Solution
a
The
terms
increase
by
3
each
time.
Rep
eated
addition
is
multiplication,
in
this
case
3
k
plus
some
sta
rting
value.
Starting
with
index
k
=
0
is
convenient,
b
ecause
3(0)
=
0
at
the
starting
value.
4
+
7
+
10
+
13
+
16
+
19
+
22
=
6
X
k
=0
4
+
3
k
b
The
terms
are
multiplied
b
y
3
each
time.
Rep
eated
multiplication
is
exp
onentiation,
in
this
case
3
k
times
some
starting
value.
Sta
rting
with
index
k
=
0
is
convenient,
b
ecause
3
0
=
1
at
the
sta
rting
value.
2
+
6
+
18
+
54
+
162
+
486
=
5
X
k
=0
(2)(3
k
)
46
c
The
absolute
values
of
this
sum
could
just
b
e
the
values
of
the
index
va
riable.
T
o
create
an
alternating
+
and
−
pattern,
w
e
can
multiply
b
y
(
−
1)
k
.
−
3
+
4
−
5
+
6
−
7
+
8
−
9
+
10
=
10
X
k
=3
(
−
1)
k
k
d
In
a
fraction,
w
e
can
mo
del
the
numerato
r
and
denominato
r
sepa
rately
.
1
4
+
√
2
9
+
√
3
16
+
2
25
+
√
5
36
=
5
X
k
=1
√
k
(
k
+
1)
2
Question
1.4.4
Ho
w
Do
W
e
Compute
the
Area
Under
a
Graph?
Supp
ose
we
would
like
to
kno
w
the
area
b
elow
the
graph
y
=
f
(
x
)
b
et
w
een
x
=
a
and
x
=
b
.
We
app
ro
ximate
this
area
b
y
rectangles.
We
can
imp
rove
these
appro
ximations
and
take
a
limit
of
such
imp
rovements
to
compute
the
actual
a
rea.
Here
is
the
procedure.
1
Divide
[
a,
b
]
into
n
subintervals,
of
lengths
∆
x
i
.
2
Pick
a
p
oint
x
∗
i
in
each
subinterval.
3
Evaluate
f
(
x
∗
i
)
,
which
is
the
height
of
the
graph
ab
ove
x
∗
i
.
4
Pro
duce
a
rectangle
of
height
f
(
x
∗
i
)
and
width
∆
x
i
over
each
subinterval.
5
Sum
the
areas
of
these
rectangles.
This
is
an
appro
ximation
of
the
actual
area.
6
T
ake
a
limit
of
such
appro
ximations
as
|
∆
x
|
,
the
largest
of
the
∆
x
i
go
es
to
0
.
Click to Load Applet
Figure:
The
area
under
y
=
f
(
x
)
appro
ximated
by
rectangles
47
Question
1.4.4
Ho
w
Do
W
e
Compute
the
Area
Under
a
Graph?
Defintion
W
e
define
the
definite
integral
of
f
(
x
)
over
[
a,
b
]
to
b
e
Z
b
a
f
(
x
)
dx
=
lim
|
∆
x
|→
0
X
f
(
x
∗
i
)∆
x
i
where
the
limit
is
taken
over
all
divisions
of
[
a,
b
]
,
∆
x
i
is
the
length
of
the
i
th
subinterval,
x
∗
i
is
a
p
oint
in
the
i
th
subinterval
and
|
∆
x
|
is
the
la
rgest
∆
x
i
.
Notice
there
is
no
requirement
that
the
subintervals
be
the
same
length.
Because
of
this,
we
don’t
tak
e
a
limit
as
n
approaches
∞
.
Fo
r
instance,
using
a
large
numb
er
of
rectangles
from
a,
a
+
b
2
and
only
a
single
rectangle
over
a
+
b
2
,
b
will
not
give
us
a
go
o
d
app
roximation,
no
matter
how
many
rectangles
w
e
use.
Instead
we
take
a
limit
as
the
largest
∆
x
i
app
roaches
0
.
In
practice,
we
get
the
same
limit
whether
the
subintervals
are
equal
length
or
not
not.
It
is
common
to
use
the
same
∆
x
=
b
−
a
n
fo
r
each
subinterval.
The
definite
integral
almost
solves
our
a
rea
problem,
but
wherever
f
(
x
)
<
0
,
the
product
f
(
x
∗
i
)∆
x
i
will
b
e
negative.
Theo
rem
If
f
(
x
)
>
0
on
[
a,
b
]
then
Z
b
a
f
(
x
)
dx
computes
the
a
rea
under
y
=
f
(
x
)
over
[
a,
b
]
.
In
general
Z
b
a
f
(
x
)
dx
computes
the
signed
area
b
etw
een
y
=
f
(
x
)
and
the
x
-axis,
where
a
rea
above
the
axis
counts
as
p
ositive,
and
a
rea
b
elo
w
the
axis
counts
a
negative.
Since
integrals
are
limits,
they
inherit
tw
o
laws
from
limits.
The
third
can
b
e
taken
from
geometry
,
setting
the
a
rea
of
a
region
equal
to
the
sum
of
the
areas
of
tw
o
sub
regions.
Integral
La
ws
Z
b
a
f
(
x
)
+
g
(
x
)
dx
=
Z
b
a
f
(
x
)
dx
+
Z
b
a
g
(
x
)
dx
(Sum
Rule)
Z
b
a
cf
(
x
)
dx
=
c
Z
b
a
f
(
x
)
dx
(Constant
Multiple
Rule)
Z
b
a
f
(
x
)
dx
=
Z
c
a
f
(
x
)
dx
+
Z
b
c
f
(
x
)
dx
(Union
Rule)
48
Question
1.4.5
Ho
w
Do
W
e
Evaluate
an
Integral?
The
limit
fo
rm
of
an
integral
is
usually
impossible
to
evaluate
directly
.
Instead
w
e
use
a
p
ow
erful
pair
of
theo
rems.
Theo
rem
[The
First
F
undamental
Theo
rem
of
Calculus]
Given
a
function
f
(
x
)
,
let
g
(
x
)
=
Z
x
a
f
(
t
)
dt
.
At
any
x
where
f
is
continuous,
g
′
(
x
)
=
f
(
x
)
.
T
o
p
rove
this,
w
e
use
the
definition
of
a
derivative.
g
′
(
x
)
=
lim
h
→
0
g
(
x
+
h
)
−
g
(
x
)
h
=
lim
h
→
0
R
x
+
h
a
f
(
t
)
dt
−
R
x
a
f
(
t
)
dt
h
=
lim
h
→
0
R
x
+
h
x
f
(
t
)
dt
h
union
rule
As
the
interval
[
x,
x
+
h
]
shrinks,
the
values
of
f
over
that
interval
can
b
e
made
a
rbitrarily
close
to
f
(
x
)
,
since
f
is
continuous.
Thus
R
x
+
h
x
f
(
t
)
dt
approaches
the
area
of
a
rectangle
of
height
f
(
x
)
and
width
h
.
Thus
lim
h
→
0
R
x
+
h
x
f
(
t
)
dt
h
=
f
(
x
)
Click to Load Applet
Figure:
g
(
x
+
h
)
−
g
(
x
)
represented
as
an
area
The
main
use
of
the
First
Fundamental
Theo
rem
of
Calculus
is
to
prove
the
Second
Fundamental
Theo
rem
of
Calculus.
Theo
rem
[The
Second
F
undamental
Theo
rem
of
Calculus]
Let
f
(
x
)
b
e
a
continuous
function
on
[
a,
b
]
.
If
F
(
x
)
an
antiderivative
of
f
(
x
)
,
then
Z
b
a
f
(
x
)
dx
=
F
(
b
)
−
F
(
a
)
49
Question
1.4.5
Ho
w
Do
W
e
Evaluate
an
Integral?
This
follows
immediately
from
the
First
F
undamental
Theorem.
If
we
continue
to
define
g
(
x
)
=
R
x
a
f
(
t
)
dt
,
then
Z
b
a
f
(
x
)
dx
=
Z
b
a
f
(
x
)
dx
−
Z
a
a
f
(
x
)
dx
=
g
(
b
)
−
g
(
a
)
W
e
know
that
g
(
x
)
is
an
antiderivative
of
f
(
x
)
.
If
we
instead
pick
a
different
antiderivative
F
(
x
)
,
then
F
(
x
)
=
g
(
x
)
+
c
,
and
F
(
b
)
−
F
(
a
)
=
g
(
b
)
+
c
−
(
g
(
a
)
+
c
)
=
g
(
b
)
−
g
(
a
)
=
Z
b
a
f
(
x
)
dx
Because
w
e
will
b
e
computing
F
(
b
)
−
F
(
a
)
frequently
,
we
will
develop
the
following
shorthand.
Notation
The
quantit
y
F
(
b
)
−
F
(
a
)
can
b
e
denoted
F
(
x
)
b
a
This
relationship
b
et
w
een
integrals
and
antiderivatives
motivates
the
follo
wing
vo
cabula
ry
.
Notation
The
general
antiderivative
of
f
(
x
)
is
also
called
an
indefinite
integral
and
is
denoted
Z
f
(
x
)
dx.
50
Example
1.4.6
A
Definite
Integral
Compute
Z
5
2
x
2
dx
Solution
Z
5
2
x
2
dx
=
x
3
3
5
2
=
5
3
3
−
2
3
3
=
125
−
8
3
=
39
Question
1.4.7
Ho
w
Do
W
e
Apply
the
Chain
Rule
in
an
Antiderivative?
The
chain
rule
states
that
(
f
(
g
(
x
)))
′
=
f
′
(
g
(
x
))
g
′
(
x
)
.
The
k
ey
insight
here
is
to
rewrite
think
of
g
as
a
variable,
in
addition
to
b
eing
a
function
of
x
.
T
ypically
w
e
rename
it
with
a
letter
closer
to
the
end
of
the
alphabet,
lik
e
u
.
The
follo
wing
substituion
theorem
uses
the
chain
rule
to
sa
y
that
w
e
can
integrate
with
resp
ect
to
u
instead
of
x
.
Theo
rem
If
u
(
x
)
is
a
function
of
x
,
then
Z
b
a
f
(
u
(
x
))
u
′
(
x
)
dx
=
Z
u
(
b
)
u
(
a
)
f
(
u
)
du
This
allows
us
to
replace
a
complicated
integrand
in
x
with
a
simpler
one
in
u
.
T
o
correctly
rewrite
the
integral,
the
b
ounds
must
b
e
up
dated
to
the
co
rresp
onding
values
of
u
.
W
e
can
also
apply
this
to
indefinite
integrals.
If
F
is
an
antiderivative
of
f
,
then
Z
f
(
u
(
x
))
u
′
(
x
)
dx
=
Z
f
(
u
)
du
=
F
(
u
)
+
c
=
F
(
u
(
x
))
+
c
The
most
common
u
substitutions
a
re
linea
r,
where
u
=
ax
.
51
Question
1.4.7
Ho
w
Do
W
e
Apply
the
Chain
Rule
in
an
Antiderivative?
Example
Compute
Z
sin
3
x
dx
Solution
W
e
will
p
erfo
rm
a
u
substitution,
using
u
=
3
x
.
Z
sin(3
x
)
dx
=
Z
1
3
sin
u
du
=
−
1
3
cos
u
+
c
=
−
1
3
cos(3
x
)
+
c
u
=
3
x
du
=
3
dx
1
3
du
=
dx
u
-substitution
Note
that
we
should
express
our
antiderivatives
in
terms
of
the
o
riginal
variable
(often
x
),
not
in
terms
of
u
.
Figure:
The
graphs
y
=
sin
x
,
y
=
sin
3
x
their
related
tangent
lines
Example
1.4.8
A
u
-substitution
Compute
the
integral:
Z
3
0
xe
x
2
dx
52
Solution
W
e
start
b
y
lo
oking
fo
r
a
candidate
for
u
(
x
)
.
Since
we
w
ant
the
integrand
to
b
e
f
(
u
(
x
))
u
′
(
x
)
,
we
note
u
(
x
)
should
b
e
the
inner
function
in
some
comp
osition.
x
2
is
the
natural
target.
We
attempt
the
substituion,
and
hope
that
the
remaining
factors
in
the
integrand
can
b
e
expressed
in
terms
of
u
′
(
x
)
.
W
e
see
that
our
u
′
(
x
)
dx
is
2
x
dx
.
Since
we
only
have
an
x
dx
in
our
integrand,
we
divide
by
2
.
Z
3
0
xe
x
2
dx
=
Z
9
0
1
2
e
u
du
=
1
2
e
u
9
0
=
1
2
(
e
9
−
1)
u
=
x
2
x
=
0
⇒
u
=
0
du
=
2
x
dx
x
=
3
⇒
u
=
9
1
2
du
=
x
dx
u
-substitution
Section
1.4
Exercises
1.4.1
Q1
W
rite
t
w
o
different
anti-derivatives
of
f
(
x
)
=
x
+
5
.
Q2
W
rite
t
w
o
different
anti-derivatives
of
f
(
x
)
=
x
3
−
6
x
2
+
2
x
.
Q3
W
rite
a
general
antiderivative
of
f
(
x
)
=
4
cos
x
+
6
x
2
.
Q4
Supp
ose
x
4
−
sin(
x
3
)
is
an
antiderivative
of
f
(
x
)
.
Write
three
other
antiderivatives
of
f
(
x
)
.
Y
ou
should
do
this
without
computing
what
f
is.
Q5
If
F
(
x
)
and
G
(
x
)
are
b
oth
antiderviatives
of
f
(
x
)
,
find
the
value
b
such
that
3
F
(
x
)
−
bG
(
x
)
is
also
an
antiderivative
of
f
(
x
)
.
Q6
Supp
ose
F
and
G
are
b
oth
antiderivatives
of
f
(
x
)
.
Supp
ose
further
that
F
is
an
antiderivative
of
F
and
G
is
an
antiderivative
of
G
.
Describ
e
the
p
ossible
values
of
F
(
x
)
−
G
(
x
)
.
53
Section
1.4
Exercises
1.4.2
Q7
Evaluate
5
X
k
=2
3
k
−
2
Q8
Evaluate
4
X
j
=
−
1
j
2
−
j
Q9
W
rite
a
fo
rmula
fo
r
the
value
of
b
X
k
=
a
c
.
Q10
W
e
do
not
need
to
write
a
constant
multiple
rule
fo
r
Σ
notation
because
we
already
have
one.
Explain
what
rules
of
mathematics
tell
us
that
b
X
k
=
a
cf
(
k
)
=
c
b
X
k
=
a
f
(
k
)
.
Q11
Explain
what’s
wrong
with
the
follo
wing
notation:
k
X
k
=1
3
k
2
+
1
k
Q12
Consider
the
sum
n
X
k
=1
1
2
k
fo
r
a
few
different
values
of
n
.
Can
you
conjecture
a
formula
for
this
sum
(it
will
dep
end
on
n
).
1.4.3
Q13
W
rite
the
follo
wing
sums
in
Σ
notation.
a
3
+
7
+
11
+
15
+
19
b
6
+
12
+
24
+
48
+
96
+
192
c
3
4
−
4
5
+
5
6
−
6
7
+
7
8
−
8
9
.
Q14
W
rite
the
follo
wing
sums
in
Σ
notation.
a
5
−
15
+
25
−
35
+
45
−
55
+
65
−
75
+
85
−
95
54
b
1
4
+
4
16
+
9
64
+
16
256
+
25
1024
c
√
2
+
√
6
+
√
12
+
√
20
+
√
30
+
√
42
+
√
56
.
1.4.4
Q15
Do
es
R
1
1
/
2
ln
x
dx
compute
the
area
under
y
=
ln
x
over
1
2
,
1
?
Explain.
Q16
Supp
ose
R
b
a
f
(
x
)
dx
<
0
.
What
do
es
this
tell
you
ab
out
the
graph
y
=
f
(
x
)
?
Be
sp
ecific.
Q17
Dra
w
a
ca
reful
graph
of
y
=
√
x
.
Use
5
subintervals
of
[1
,
11]
to
estimate
the
area
b
eneath
the
graph
over
[1
,
11]
.
Use
the
left
endp
oints
of
each
subinterval
as
the
test
p
oints
x
∗
i
.
Q18
Dra
w
a
ca
reful
graph
of
y
=
3
x
.
Use
3
subintervals
of
[2
,
8]
to
estimate
the
area
b
eneath
the
graph,
with
the
test
p
oints
x
∗
i
b
eing
the
left
endp
oints
of
each
subinterval.
Q19
Dra
w
the
graph
of
y
=
7
.
Use
geometry
to
evaluate
R
3
87
dx
.
Q20
Dra
w
the
graph
of
y
=
x
3
+
1
.
Use
geometry
to
evaluate
R
−
3
9
x
3
+
1
dx
.
1.4.5
Q21
Let
g
(
x
)
=
R
x
5
f
(
t
)
dt
.
What
is
g
′
(8)
?
Q22
Let
g
(
x
)
=
R
x
2
cos
t
dt
.
Is
g
(
x
)
increasing
or
decreasing
at
x
=
3
?
Explain.
Q23
Supp
ose
f
(
x
)
is
an
increasing
function.
Is
R
31
22
f
′
(
x
)
dx
p
ositive
o
r
negative?
Q24
Supp
ose
F
(
x
)
and
G
(
x
)
are
b
oth
antiderivatives
of
f
(
x
)
.
Given
the
following
incomplete
table
of
values,
compute
R
4
1
f
(
x
)
dx
.
x
1
2
3
4
5
6
F
(
x
)
−
7
−
13
−
9
G
(
x
)
3
−
9
−
10
5
55
Section
1.4
Exercises
Q25
Explain
the
difference
b
et
w
een
Z
f
(
x
)
dx
and
Z
b
a
f
(
x
)
dx
in
a
few
sentences.
Q26
Compute
Z
π
0
cos(
x
)
dx
.
Explain
the
geometric
meaning
of
your
answer
in
a
sentence
or
tw
o.
1.4.6
Q27
Compute
Z
8
1
x
−
3
x
dx
.
Q28
Compute
Z
4
1
1
t
3
/
2
dt
.
Q29
Compute
Z
e
x
−
6
x
2
dx
.
Q30
Compute
Z
0
t
1
3
e
x
+
5
dx
.
Q31
Compute
Z
√
t
dt
.
Q32
Compute
Z
2
10
x
2
+
2
5
x
dx
.
Q33
Compute
Z
3
5
sin
y
dy
.
Q34
Compute
Z
2
0
x
4
−
3
x
+
2
dx
.
Q35
Compute
Z
3
π
/
4
π
/
6
2
cos
v
dv
.
Q36
Compute
Z
π
0
2
sin
t
+
cos
t
dt
.
56
1.4.7
Q37
W
rite
some
general
rules.
Supp
ose
F
(
x
)
+
c
is
the
antiderivative
of
f
(
x
)
a
What
is
the
antiderivative
of
f
(
x
+
a
)
?
b
What
is
the
antiderivative
of
f
(
ax
)
?
Q38
Assuming
that
Z
b
a
f
(
x
)
dx
exists,
argue
that
it
is
equal
to
Z
2
b
2
a
1
2
f
x
2
dx
,
in
the
following
tw
o
w
a
ys
:
a
By
app
ealing
to
an
integration
rule.
b
By
describing
the
relationship
b
etw
een
the
graphs
of
y
=
f
(
x
)
and
y
=
f
x
2
.
A
picture
might
help.
1.4.8
Q39
Compute
Z
e
7
x
dx
.
Q40
Compute
Z
√
5
x
+
3
dx
.
Q41
Compute
Z
cos
θ
3
dθ
.
Q42
Compute
Z
(
t
−
2)
6
dt
.
Q43
Compute
Z
1
/
4
0
sin(
π
t
)
dt
.
Q44
Compute
Z
3
0
x
2
e
x
3
dx
.
Q45
Compute
Z
(
x
5
−
2
x
)(5
x
4
−
2)
dx
.
Q46
Compute
Z
3
π
/
4
π
/
4
cos(
x
)
1
sin
2
x
dx
.
57
Section
1.4
Exercises
58
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