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18.01 Single Variable Calculus
Fall 2006
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IMPROPER
INTEGRALS
In deciding whether
an
improper integral converges or diverges,
it is often awkward or
impossible to try to decide this
by
actually carrying out the integration, i.e., finding an
antiderivative explicitly. For example both of these two improper integrals converge:
0"
daz
z+3x
s
+22
+
1'
and
1
z
but there is no explicit antiderivative for the second integral, and finding one for the first
would be
a hairy exercise in partial fractions, even ifone were able to factor the denominator.
Instead of explicit integration, therefore, we show they converge
by
using estimation
instead, comparing them with simpler integrals which are known to converge. Thus, for
the
first,
1
1
s6 +
3Zs
+ 2
+1

~'
X>
0,
so
that
J
d0
1
"
dsz
+
<
z*=E.
x

z+3X3+2X2+1
Since the right hand integral converges, so does the left, which is smaller (but still positive).
In a
similar
way, for the second integral, we estimate
1
1
~P,'FF
5
X>
0,
so that
I
F
v&
d
<
<i
F•/
=
2.
In the same way we can show the divergence say of
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 Spring '11
 Blake
 Math, Improper Integrals, Integrals

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