7.4 - MATH
1081
 Wednesday,
April
13
 


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Unformatted text preview: MATH
1081
 Wednesday,
April
13
 
 Chapter
7
–
Section
4
 
 THE
FUNDAMENTAL
THEOREM
OF
 CALCULUS
 Homework
#12
(due
4/18):
 








 
 
 Section
7.3
#8,
28,
36
 Section
7.4
#4,
12,
26,
34,
56
 Clicker
Check­in:

Choose
any
letter
to
check
in
now.
 Last
time,
we
defined
 b a n→ ∞ 
 Geometrically,
this
represented
the
area
of
the
region
between
 the
function
 f ( x ) 
and
the
 x ‐axis
from
 x = a 
to
 x = b .
 
 We
can
approximate
the
value
of
this
definite
integral
by
using
 some
finite
sum
of
areas
of
rectangles.

However,
to
find
the
 exact
value
we
must
evaluate
the
limit.
 
 The
Fundamental
Theorem
of
Calculus
tells
us
that
actually
this
 limit
is
equal
to
the
difference
in
the
value
of
the
 antiderivative
of
 f ( x ) 
at
 x = a 
and
 x = b .
 ⎣ ⎦ ∫ f ( x ) dx = lim ⎡ f ( x ) + f ( x ) + f ( x ) + … + f ( x )⎤ Δx .
 1 2 3 n 
 So,
we
only
need
to
find
an
antiderivative
of
 f ( x ) ,
evaluate
it
at
 x = a 
and
 x = b ,
and
then
subtract
(always
the
value
at
the
 “top”
bound
minus
the
value
at
the
“bottom”
bound).
 
 Note
that
we
can
use
any
antiderivative.

So,
in
particular,
we
 can
chose
the
one
where
 C = 0 
or
whatever
value
we
like.


 Why
is
the
 C 
unimportant
here? 
 
Example:

Evaluate
the
definite
integral.
 
 4⎛ 12 ⎞ 
 
 
 1.

 ∫ ⎜ − x + 4 ⎟ dx 
 0⎝ ⎠ 4 
 
 
 31 
 
 
 2.

 ∫ dx 
 1x 
 
 
 1 e2 x 
 
 
 3.

 ∫ dx 
 2x 0 1+ e 
 In
the
last
clicker
question,
we
found
the
value
of
the
definite
 integral
to
be
negative.

If
this
is
meant
to
represent
the
“area
 under
the
curve”,
how
can
it
be
negative?
 
 In
this
case,
we
were
considering
an
interval
where
the
 function
was
actually
below
the
 x ‐axis.

So,
in
the
sum
 ⎡ f ( x1 ) + f ( x2 ) + f ( x3 ) + … + f ( xn ) ⎤ Δx ,
every
 f ( x* )
is
negative,
 ⎣ ⎦ making
the
result
negative.
 
 
 In
fact,
a
better
interpretation
of
the
definite
integral
is
the
 signed
area
or
net
area
of
the
region
between
the
curve
and
the
 x ‐axis.

Regions
above
the
 x ‐axis
contribute
a
positive
amount
 and
regions
below
it
contribute
a
negative
amount.
 In
general,
we
want
the
definite
integral
to
represent
the
 signed
(or
net)
area.

Thinking
in
terms
of
a
velocity
function,
a
 negative
velocity
would
imply
that
we
are
travelling
in
reverse
 somehow.


So,
when
computing
a
total
distance,
we
want
the
 “backward”
moments
to
be
subtracted
from
the
“forward”
 moments.
 
 In
some
cases,
you
will
be
asked
to
find
the
total
area
between
 the
curve
and
the
 x ‐axis.

In
this
case,
we
do
NOT
want
those
 regions
below
the
 x ‐axis
to
deduct
from
the
total
area.

We
 want
everything
to
be
positive.
 
 Example:

Determine
the
total
area

 
 1. of
the
shaded
region.
 
 
 2. between
the
 x ‐axis
and
the
curve
 f ( x ) = e x − 1
on

 the
interval
[ −1, 2 ] 
 
 Note
that
the
Fundamental
Theorem
of
Calculus
gives
us
a
 method
to
evaluate
the
definite
integral.

It
does
not
define
it.

 The
definite
integral
is
defined
to
be
the
limit
of
a
sum.
 
 This
is
similar
to
how
the
rulesof
differentiation
simply
give
us
 a
method
to
find
a
derivative.

However,
a
derivative
is
defined
 as
a
limit
of
a
difference
quotient.
 
 
 Clicker
Check­out:

Choose
any
letter
to
check
out.
 
 Tomorrow
in
recitation:

Quiz
on
Sections
9.2,
9.3,
7.1,
7.2
 
 Next
time
–
Section
7.5
 
 
 ...
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This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

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