SME
Calculus II · Problem type
Integrating a series term by term
Use this for “Estimate ∫ from 0 to 0.6 of e−x² dx with three terms of a series, to within how much?”
Chart
Dashed: the degree-8 Taylor polynomial of e−x² about x = 0
∫ from 0 to 1 of P dx = 0.7475, term by term
This is an example. Type your own values to start a new problem.
∫
0
e
−x²
dx
≈
I
=
Iₘ
B
Error bound (next term)
worked out (not typed)
I
True integral
worked out (not typed)
E
Actual error |I − Iₘ|
worked out (not typed)
Formulas
Iₘ
=
Σ
from
k
=
0
to
m
−
1
of
(−1)
k
×
b
2k + 1
(2k + 1) × k!
0.7475
=
Σ
from
k
=
0
to
5
−
1
of
(−1)
k
×
1
2k + 1
(2k + 1) × k!
B
=
b
2 ×
m
+ 1
(2 ×
m
+ 1) ×
m
!
0.0007576
=
1
2 × 5 + 1
(2 × 5 + 1) × 5!
I
=
b
∫
0
(
e
−
x
2
) dx
0.7468
=
1
∫
0
(
e
−
x
2
) dx
E
=
|
I
−
Iₘ|
0.0006626 = |0.7468 − 0.7475|
Assumptions
•
Put −x² into eᵘ = 1 + u + u²/2! + …: e−x² = 1 − x² + x⁴/2! − x⁶/3! + …, true for every x.
•
A power series can be integrated term by term inside its interval: xᵏ integrates from 0 to b to bᵏ⁺¹ ÷ (k + 1).
•
For b up to 1.7 the terms shrink in size and alternate, so the error is at most the first term left out.
Step-by-step
WE KNOW
b = 1, m = 5
FIND
estimate from m terms (Iₘ), error bound (next term) (B), true integral (I), actual error |I − Iₘ| (E)
Step 1 · Find estimate from m terms (Iₘ)
Use Iₘ = Σ from k = 0 to m − 1 of (−1)ᵏ × b²ᵏ⁺¹ ÷ ((2k + 1) × k!)
The
k-th
term,
(−1)ᵏx²ᵏ
divided
by
k!,
integrates
from
0
to
b
to
(−1)ᵏb²ᵏ⁺¹
divided
by
(2k
+
1)k!.
Add
the
first
m
of
them.
Iₘ
=
b
−
b
3
3
+
b
5
10
−
b
7
42
+
b
9
216
Iₘ
=
1
−
1
3
3
+
1
5
10
−
1
7
42
+
1
9
216
Iₘ
=
1
−
1
3
+
1
10
−
1
42
+
1
216
Iₘ = 1 − 0.3333 + 0.1 − 0.02381 + 0.00463
Iₘ = 0.747487
Step 2 · Find error bound (next term): B
Use B = b2 × m + 1 ÷ ((2 × m + 1) × m!)
The
first
term
left
out
is
number
m
(counting
from
k
=
0).
Its
size
bounds
the
error
of
an
alternating
series
whose
terms
shrink.
B
=
b
11
1,320
B
=
1
11
1,320
B
=
1
1,320
B
=
0.0007576
Step 3 · Find true integral (I)
Use I = ∫ from 0 to b of (e−x²) dx
The
true
value,
to
compare
with
the
estimate.
e
−
x
2
has
no
antiderivative
in
elementary
terms,
which
is
why
the
series
is
used.
I
=
1
∫
0
(
e
−
x
2
) dx
I
=
0.746824
Step 4 · Find actual error |I − Iₘ| (E)
Use E = |I − Iₘ|
How
far
the
estimate
really
is
from
the
integral.
It
is
under
the
bound
B
.
E
=
|0.746824
−
0.747487|
E
=
|−0.000663|
E
=
0.0006626
Check
Put the numbers back in. Both sides should match.
0.7475
=
1
−
1
3
3
+
1
5
10
−
1
7
42
+
1
9
216
✓
0.0007576
=
1
11
1,320
✓
0.746824
=
1
∫
0
(
e
−
x
2
) dx
✓
0.0006626 = |0.746824 − 0.747487| ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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