SME
Grade 12 Math · Precalculus
Limits of sequences and the area under a curve
Use this for “Estimate the area under y = x² from 0 to 3 with 6 rectangles, then find the exact area as a limit.”
Chart
Shaded: 0 ≤ x ≤ 3
6 right-edge rectangles of width 0.5 from x = 0 to 3: their areas add to S = 11.375
This is an example. Type your own values to start a new problem.
c
Number before x²
example
b
Right end
example
n
Number of rectangles
example
w
Width of each rectangle
worked out (not typed)
S
Sum of the rectangles
worked out (not typed)
A
Exact area
calculated
Formulas
w
=
b
n
0.5
=
3
6
S
=
c
×
w
3
×
n
×
(
n
+
1)
×
2 ×
n
+ 1
6
11.375
=
1
×
0.5
3
×
6
×
(6
+
1)
×
2 × 6 + 1
6
A
=
c
×
b
3
3
9
=
1
×
3
3
3
Assumptions
•
Cut 0 to b into n strips of width w = b ÷ n; each rectangle is as tall as the curve at its right edge, x = iw.
•
The heights add with Σi² = n(n + 1)(2n + 1)/6, so S = cw³ × n(n + 1)(2n + 1)/6.
•
As n → ∞, S → cb³/3: the exact area under y = cx² from 0 to b.
Step-by-step
WE KNOW
c = 1, b = 3, n = 6
FIND
width of each rectangle (w), sum of the rectangles (S), exact area (A)
Step 1 · Find width of each rectangle (w)
Use w = b ÷ n
Cut
0
to
b
into
n
strips
of
equal
width.
w
=
3
6
w
=
0.5
Step 2 · Find sum of the rectangles (S)
Use S = c × w³ × n × (n + 1) × (2 × n + 1) ÷ 6
Each
rectangle
is
w
wide
and
c
(iw)
2
tall;
adding
them
gives
cw³(
1
2
+
2
2
+
…
+
n
2
).
S
=
1
×
0.5
3
×
6
×
(6
+
1)
×
2 × 6 + 1
6
1
2
+
2
2
+
…
+
6
2
=
6
×
7
×
13
6
=
91
S
=
1
×
0.5
3
×
91
S
=
0.125
×
91
S
=
11.375
Step 3 · Find exact area (A)
Use A = c × b³ ÷ 3
The
limit
as
n
→
∞:
n
(
n
+
1)
2
n
+ 1
n
3
goes
to
2,
so
S
goes
to
cb³
×
2
6
.
A
=
1
×
3
3
3
A
=
27
3
A
=
9
→
S
=
11.375
is
above
the
exact
area;
more
rectangles
bring
S
closer
to
it
Check
Put the numbers back in. Both sides should match.
0.5
=
3
6
✓
11.375
=
1
×
0.5
3
×
6
×
(6
+
1)
×
2 × 6 + 1
6
✓
9
=
1
×
3
3
3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.