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Lecture 35
18.01 Fall 2006
Lecture
35:
Improper
Integrals
Defnition.
An
improper
integral
, defned by
∞
±
M
f
(
x
)
dx
=
lim
f
(
x
)
dx
a
M
→∞
a
is said to converge i± the limit exists (diverges i± the limit does not exist).
∞
e

kx
dx
= 1
/k
(
k >
0)
Example
1.
0
±
M
M
e

kx
dx
= (

1
/k
)
e

kx
=
(1
/k
)(1

e

kM
)
0
0
Taking the limit as
M
→ ∞
, we fnd
e

kM
→
0
and
∞
e

kx
dx
= 1
/k
0
We rewrite this calculation more in±ormally as ±ollows,
0
∞
e

kx
dx
= (

1
/k
)
e

kx
∞
0
=
(1
/k
)(1

e

k
∞
) = 1
/k
(since
k >
0
)
∞
e

kx
dx
= 1
/k
has an easier ±ormula than the
Note that the integral over the infnite interval
±
M
0
corresponding fnite integral
e

kx
dx
=
(1
/k
)(1

e

kM
)
. As a practical matter, ±or large
M
, the
0
term
e

kM
is negligible, so even the simpler ±ormula
1
/k
serves as a good approximation to the fnite
integral. Infnite integrals are o±ten easier than fnite ones, just as infnitesimals and derivatives are
easier than di²erence quotients.
Application:
Replace
x
by
t
= time in seconds in Example 1.
R
=
rate o± decay = number o± atoms that decay per second at time
0
.
At later times
t >
0
the decay rate is
Re

kt
(smaller by an exponential ±actor
e

kt
)
Eventually (over time
0
≤
t <
∞
) every atom decays.
So the total number o± atoms
N
is
calculated using the ±ormula we ±ound in Example 1,
∞
Re

kt
dt
=
R/k
N
=
0
The hal± li±e
H
o± a radioactive element is the time
H
at which the decay rate is hal± what it was at
the start. Thus
e

kH
= 1
/
2
=
⇒

kH
=
ln(1
/
2)
=
⇒
k
=
(ln
2)
/H
1
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Lecture 35
18.01 Fall 2006
Hence
R
=
Nk
=
N
(ln 2)
/H
Let us illustrate with Polonium 210, which has been in the news lately. The half life is 138 days
or
H
=
(138
days
)(24
hr/day
)(60
2
sec/hr
)
=
(138)(24)(60)
2
seconds
Using this value of
H
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 Fall '08
 Staff
 Improper Integrals, Integrals, Radioactive Decay, Limit, dx, 1 m, 1 gram

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