SME
Calculus II · Problem type
The comparison test
Use this for “Show that the integral of 1 ÷ (x² + 9) from 1 to ∞ converges, and find its value” (c = 3).
Chart
Shaded: ∫ from 1 to 10 of f(x) dx = 0.6857
This is an example. Type your own values to start a new problem.
∫
1
1
x²
+
c
2
dx
=
Ib
∫
∞
1
1
x²
+
c
2
dx
=
I
θ₁
Angle tan⁻¹(1 ÷ c)
worked out (not typed)
θb
Angle tan⁻¹(b ÷ c)
worked out (not typed)
Jb
Integral of 1 ÷ x² from 1 to b
worked out (not typed)
C
Converges or diverges
worked out (not typed)
Formulas
θ₁
=
tan⁻¹(
1
c
)
0.785398
=
tan⁻¹(
1
1
)
θb
=
tan⁻¹(
b
c
)
1.47113
=
tan⁻¹(
10
1
)
Ib
=
θb −
θ₁
c
0.6857
=
1.47113 − 0.785398
1
Jb
=
1
−
1
b
0.9
=
1
−
1
10
C
=
converges,
as
1
x
2
+
c
2
≤
1
x
2
and
∫
from
1
to
∞
of
1
x
2
dx
=
1
1 > 0, so converges
I
=
π ÷ 2 −
θ₁
c
0.7854
=
π ÷ 2 − 0.785398
1
Assumptions
•
For x ≥ 1, x² + c² is more than x², so 0 < 1 ÷ (x² + c²) < 1 ÷ x²: the curve stays under y = 1 ÷ x².
•
The comparison test: if 0 ≤ f ≤ g and ∫ g converges, then ∫ f converges too, to no more than ∫ g. Here ∫ from 1 to ∞ of 1 ÷ x² dx = 1.
•
An antiderivative of 1 ÷ (x² + c²) is (1 ÷ c) × tan⁻¹(x ÷ c), with angles in radians. The constant c is more than 0.
Step-by-step
WE KNOW
c = 1, b = 10
FIND
angle tan⁻¹(1 ÷ c) (θ₁), angle tan⁻¹(b ÷ c) (θb), integral from 1 to b (Ib), integral of 1 ÷ x² from 1 to b (Jb), converges or diverges (C), limit as b → ∞ (I)
Step 1 · Find angle tan⁻¹(1 ÷ c): θ₁
Use θ₁ = tan⁻¹(1 ÷ c)
The
antiderivative
(
1
c
)
×
tan⁻¹(
x
c
)
at
the
lower
limit
x
=
1
needs
the
angle
whose
tangent
is
1
c
,
in
radians.
θ₁
=
tan⁻¹(
1
1
)
θ₁
=
0.785398
Step 2 · Find angle tan⁻¹(b ÷ c): θb
Use θb = tan⁻¹(b ÷ c)
At
the
cut-off
x
=
b
the
angle
is
the
one
whose
tangent
is
b
c
.
It
approaches
π
2
as
b
grows.
θb
=
tan⁻¹(
10
1
)
θb = 1.47113
Step 3 · Find integral from 1 to b (Ib)
Use Ib = (θb − θ₁) ÷ c
Take
the
antiderivative’s
value
at
1
from
its
value
at
b
:
(
1
c
)
×
tan⁻¹(
b
c
)
−
(
1
c
)
×
tan⁻¹(
1
c
).
Ib
=
1.47113 − 0.785398
1
∫
1
x
2
+ 1
dx
=
(
1
1
)
×
tan⁻¹(
x
1
)
+
C
∫
from
1
to
10
of
1
x
2
+ 1
dx
=
0.6857
Ib = 0.6857
Step 4 · Find integral of 1 ÷ x² from 1 to b (Jb)
Use Jb = 1 − 1 ÷ b
The
curve
to
compare
with
is
y
=
1
x
2
,
whose
antiderivative
is
−
1
x
.
From
1
to
b
that
gives
1
−
1
b
,
which
approaches
1.
Jb
=
1
−
1
10
∫
from
1
to
10
of
1
x
2
dx
=
[−
1
x
]
from
1
to
10
=
0.9
Jb = 0.9
Step 5 · Find converges or diverges (C)
Use C = converges, as 1 ÷ (x² + c²) ≤ 1 ÷ x² and ∫ from 1 to ∞ of 1 ÷ x² dx = 1
The
curve
stays
under
1
x
2
,
whose
integral
from
1
to
∞
is
1,
so
the
integrals
to
b
grow
as
b
grows
but
stay
under
1:
they
level
off.
At
b
=
10:
0.6857
≤
0.9.
1 > 0: converges
C
=
converges,
as
1
x
2
+
1
2
≤
1
x
2
and
∫
from
1
to
∞
of
1
x
2
dx
=
1
C
=
converges
Step 6 · Find limit as b → ∞ (I)
Use I = (π ÷ 2 − θ₁) ÷ c
As
b
→
∞,
tan⁻¹(
b
c
)
→
π
2,
so
the
integrals
to
b
approach
π ÷ 2 −
θ₁
c
.
The
comparison
says
this
is
at
most
1.
I
=
π ÷ 2 − 0.785398
1
I
=
1.5708
−
0.7854
I
=
0.7854
Check
Put the numbers back in. Both sides should match.
0.785398
=
tan⁻¹(
1
1
)
✓
1.47113
=
tan⁻¹(
10
1
)
✓
0.6857
=
1.47113 − 0.785398
1
✓
0.9
=
1
−
1
10
✓
1 > 0, so converges ✓
0.7854
=
π ÷ 2 − 0.785398
1
✓
Topic check
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