SME
Calculus I · Problem type
Riemann sums
Use this for “Estimate the integral of x² from 1 to 4 with 6 right-endpoint rectangles, then find the error.”
Chart
Shaded: 0 ≤ x ≤ 2
4 right-edge rectangles of width 0.5 from x = 0 to 2: their areas add to S = 3.75
This is an example. Type your own values to start a new problem.
S
=
Sₙ
≈
∫
x²
dx
=
8
3
I
Δx
Width of each rectangle
worked out (not typed)
E
Error of the sum, Sₙ − I
worked out (not typed)
Formulas
Δx
=
b
−
a
n
0.5
=
2 − 0
4
Sₙ
=
Δx
×
(
n
×
a
2
+
2
×
a
×
Δx
×
n
×
n
+ 1
2
+
Δx²
×
n
×
(
n
+
1)
×
2 ×
n
+ 1
6
)
3.75
=
0.5
×
(4
×
0
2
+
2
×
0
×
0.5
×
4
×
4 + 1
2
+
0.5
2
×
4
×
(4
+
1)
×
2 × 4 + 1
6
)
I
=
b
3
−
a
3
3
8
3
=
2
3
−
0
3
3
E
=
Sₙ
−
I
1.0833
=
3.75
−
8
3
Assumptions
•
Cut a to b into n strips of width Δx = (b − a) ÷ n. Each rectangle is as tall as f(x) = x² at its right end, x = a + iΔx.
•
The rectangles’ total area Sₙ estimates the integral; the exact value is I = (b³ − a³) ÷ 3, by the Fundamental Theorem.
•
Where x² rises across a strip (x ≥ 0), the right end is the tallest point, so Sₙ is too big; where it falls, too small.
•
The more strips, the closer Sₙ comes to I: the integral is the limit of Sₙ as n → ∞.
Step-by-step
WE KNOW
a = 0, b = 2, n = 4
FIND
width of each rectangle (Δx), right Riemann sum (Sₙ), exact value of the integral (I), error of the sum, Sₙ − I (E)
Step 1 · Find width of each rectangle (Δx)
Use Δx = (b − a) ÷ n
Cut
the
interval
from
a
to
b
into
n
strips
of
equal
width.
Δx
=
2 − 0
4
Δx = 0.5
Step 2 · Find right Riemann sum (Sₙ)
Use Sₙ = Δx × (n × a² + 2 × a × Δx × n × (n + 1) ÷ 2 + Δx² × n × (n + 1) × (2 × n + 1) ÷ 6)
Square each right end to get the rectangle’s height, add the heights, then multiply by the width Δx.
Sₙ
=
(
0.5
2
+
1
2
+
1.5
2
+
2
2
)
×
0.5
Sₙ = (0.25 + 1 + 2.25 + 4) × 0.5
Sₙ = 3.75
Step 3 · Find exact value of the integral (I)
Use I = (b³ − a³) ÷ 3
F(x)
=
x
3
3
is
an
antiderivative
of
x
2
,
so
the
integral
is
F(
b
)
−
F(
a
).
I
=
2
3
−
0
3
3
I
=
8
3
I
=
8
3
Step 4 · Find error of the sum, Sₙ − I (E)
Use E = Sₙ − I
Subtract the exact value from the sum: how far the rectangles miss.
E
=
3.75
−
8
3
E
=
1.0833
→
The
sum
is
too
big;
more
rectangles
bring
it
closer
to
I
Check
Put the numbers back in. Both sides should match.
0.5
=
2 − 0
4
✓
3.75
=
0.5
×
(4
×
0
2
+
2
×
0
×
0.5
×
4
×
4 + 1
2
+
0.5
2
×
4
×
(4
+
1)
×
2 × 4 + 1
6
)
✓
8
3
=
2
3
−
0
3
3
✓
1.0833
=
3.75
−
8
3
✓
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Definite integrals and the Fundamental Theorem
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The area function and the Fundamental Theorem, part 1
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