SME
Calculus II · Problem type
An integral that blows up at 0
Use this for “Does the integral of 1 ÷ ∜(x³) from 0 to 1 converge? If so, to what?” (p = 3/4).
Chart
Shaded: ∫ from 0.01 to 1 of f(x) dx = 1.8
This is an example. Type your own values to start a new problem.
∫
1
1
x
dx
=
Iε
∫
1
0
1
x
dx
=
I
C
Converges or diverges
worked out (not typed)
Formulas
Iε
=
1 −
ε
1 −
p
1 −
p
1.8
=
1 −
0.01
1 − 0.5
1 − 0.5
C
=
converges
if
p
<
1,
else
diverges
0.5 < 1, so converges
I
=
1
1 −
p
2
=
1
1 − 0.5
Assumptions
•
The integrand 1 ÷ xᵖ grows without end as x → 0⁺, so ∫ from 0 to 1 is the limit of ∫ from ε to 1 as ε → 0⁺.
•
It converges when that limit is a finite number and diverges when the integrals from ε grow without end.
•
An antiderivative of x⁻ᵖ is x¹⁻ᵖ ÷ (1 − p), or ln x when p = 1. The power p is more than 0.
Step-by-step
WE KNOW
p = 0.5, ε = 0.01
FIND
integral from ε to 1 (Iε), converges or diverges (C), limit as ε → 0⁺ (I)
Step 1 · Find integral from ε to 1 (Iε)
Use Iε = (1 − ε¹⁻ᵖ) ÷ (1 − p)
Raise
the
power
−
p
by
one
to
1
−
p
and
divide
by
the
new
power.
Take
the
antiderivative’s
value
at
ε
from
its
value
at
1.
0.5 ≠ 1: the power rule
Iε
=
1 −
0.01
1 − 0.5
1 − 0.5
∫
x
−0.5
dx
=
x
0.5
0.5
+
C
1
∫
0.01
x
−0.5
dx
=
[
x
0.5
0.5
]
from
0.01
to
1
=
1.8
Iε = 1.8
Step 2 · Find converges or diverges (C)
Use C = converges if p < 1, else diverges
With
p
<
1
the
power
1
−
p
is
positive,
so
ε
1 −
p
shrinks
to
0
as
ε
→
0⁺:
the
integrals
from
ε
level
off.
0.5 < 1: converges
C
=
converges
if
0.5
<
1,
else
diverges
C
=
converges
Step 3 · Find limit as ε → 0⁺ (I)
Use I = 1 ÷ (1 − p)
As
ε
→
0⁺,
ε
1 −
p
→
0,
so
the
integral
from
ε
approaches
1 − 0
1 −
p
=
1
1 −
p
.
I
=
1
1 − 0.5
I
=
1
0.5
I
=
2
Check
Put the numbers back in. Both sides should match.
1.8
=
1 −
0.01
1 − 0.5
1 − 0.5
✓
0.5 < 1, so converges ✓
2
=
1
1 − 0.5
✓
Topic check
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