Math 131.
Area
Larson Section 4.2
Much of the rest of Chapter 4 will be devoted to areas of regions. The connection back to
antiderivatives will be made in Section 4.4 with the Fundamental Theorem of Calculus.
Example 1. Suppose we would like to find the area under the curve y = 0.4x2 + 1.4 for
2 ≤ x ≤ 4 as shaded below.
y
10
8
6
4
2
x
−1
−2
1
2
3
4
5
We could estimate the area of this shaded region with the rectangles as below. The rectangles
from the graph to the left are all below the curve, so this will underestimate the area (these
are the inscribed rectangles), and the sum of the rectangle areas would be called a lower sum.
The rectangles from the graph to the left are all above the curve (these are the circumscribed
rectangles, so this will overestimate the area, and the sum of the rectangle areas would be
called an upper sum.
10
y
10
8
8
6
6
4
4
2
2
y
x
−1
−2
1
2
3
4
5
x
−1
−2
1
2
3
4
5
Estimate (a) the lower sum; (b) the upper sum; and then (c) give a range for the actual area.
Solution: (a) In the lower sum, each of the rectangles has base 0.5, and the heights of
the rectangle are determined by the value of the function f (x) = 0.4x2 + 1.4 where each
rectangle touches the curve. Thus the heights are
(0.4)(22 ) + 1.4 = 3.000,
(0.4)(32 ) + 1.4 = 5.000,
(0.4)(2.52 ) + 1.4 = 3.900
and
(0.4)(3.52 ) + 1.4 = 6.300 respectively.
Therefore, the sum of the area of the rectangles is
A = (0.5)(3.000) + (0.5)(3.900) + (0.5)(5.000) + (0.5)(6.300)
= (0.5)(3.000 + 3.900 + 5.000 + 6.300) = 9.100
Thus the lower sum is 9.100
(b) For the upper sum, each of the rectangles has base 0.5, and the heights of the rectangle
are determined by the value of the function f (x) = 0.4x2 +1.4 where each rectangle touches
the curve. Thus the heights are
(0.4)(2.52 ) + 1.4 = 3.900
(0.4)(3.52 ) + 1.4 = 6.300
(0.4)(32 ) + 1.4 = 5.000,
(0.4)(42 ) + 1.4 = 7.800 respectively.
and
and
Therefore, the sum of the area of the rectangles is
A = (0.5)(3.900) + (0.5)(5.000) + (0.5)(6.300) + (0.5)(7.800)
= (0.5)(3.900 + 5.000 + 6.300 + 7.800) = 11.500
Thus the upper sum is 11.500
(c) The actual area is always between any lower sum and any upper sum, so
9.100 ≤ Actual Area ≤ 11.500
The natural way to improve our estimate (make the above range for the actual area smaller)
would be to use rectangles with smaller bases, and ideally take a limit. The goal of this section
is to do precisely that. First, some tools and notation are needed.
The first thing we introduce is sigma notation, it allows us to write many sums is a more
compact form.
Sigma Notation. The sum of n terms a1 , a2 , a3 , . . . , an is written as
n
X
ai = a1 + a2 + a3 + . . . + an
i=1
where i is the index of summation, ai is the ith term of the sum, and the upper and lower
bounds of the summation are n and 1.
Example 2. (Examples of Sigma Notation)
(a)
4
X
i2 = 12 + 22 + 32 + 42 = 30
i=1
5
X
(b)
(2k + 1) = 5 + 7 + 9 + 11 = 32
k=2
(c)
n
X
f (xi )∆x = f (x1 )∆x + f (x2 )∆x + . . . + f (xn )∆x
i=1
Sometimes you will work from a sum written long-hand to summation notation. For example,
write 3 + 4 + 5 + 6 + 7 + 8 + 9 in summation notation. There are several ways to do this, for
example
(a)
9
X
i
(b)
i=3
6
X
(i + 2)
i=1
(c)
5
X
(i + 3)
i=0
and, in fact, there are infinitely many ways to write the sum!
Some basic properties of sums are
1.
n
X
kai = k
i=1
n
X
ai
i=1
n
n
n
X
X
X
2.
(ai ± bi ) =
ai ±
bi
i=1
i=1
i=1
We will use some specific formulas when we compute areas. These are given in the following
theorem.
Theorem. (Summation Formulas)
(a)
n
X
c = cn
i=1
(b)
n
X
i=1
(c)
n
X
i=
n(n + 1)
2
i2 =
n(n + 1)(2n + 1)
6
i3 =
n2 (n + 1)2
4
i=1
(d)
n
X
i=1
42
X
Example 3. Evaluate the sum
(10 − 2n).
n=0
Solution: We will use the formula
n
X
k=
k=1
42
X
n(n + 1)
with n = 42 to get
2
42
X
(10 − 2n) = 10 +
(10 − 2n)
n=0
n=1
= 10 +
42
X
10 − 2
n=1
42
X
= (42 + 1)(10) − 2
Example 4.
Evaluate the sum
36
X
n
n=1
(42)(42 + 1)
2
= −1376
(5n − 5)(2n − 3).
n=1
Solution: First, use FOIL on the summand, so
36
X
36
X
(5n − 5)(2n − 3) =
(10n2 − 25n + 15)
n=1
Then using the formulas
n=1
N
X
n=
n=1
N
X
N (N + 1)
N (N + 1)(2N + 1)
n2 =
and
where N =
2
6
n=1
36 one has
36
X
2
10n − 25n + 15 = 10
n=1
36
X
n=1
=
2
n − 25
36
X
n=1
n + 15
36
X
1
n=1
10(36)(36 + 1)(2(36) + 1) 25(36)(36 + 1)
−
+ (15)(36)
6
2
= (10)(16206) − (25)(666) + 540 = 145950
Finding Areas of a Region in the Plane. Let f be a nonnegative continuous function
on an interval [a, b]. Define the region R as the region bounded above by the graph of f (x),
bounded below by the x-axis and to the left and right by x = a and x = b respectively.
For example, the shaded region below (in the left graph) is the region bounded above by the
18
curve y =
, and below by the x-axis and to the left and right by x = 1 and x = 3
x+4
respectively, the graph to the right shows a generic region
y
4
3
2
1
x
1
2
3
In general, we will divide the interval [a, b] into n equal subintervals. Thus each subinterval has
width ∆x = (b − a)/n the endpoints of the intervals will be a = x0 < x1 < x2 < . . . < xn = b
where xi = a + i(∆x), that is x0 = a + 0(∆x) = a, x1 = a + 1(∆x), x2 = a + 2(∆x), . . . , and
xn = a + n(∆x) = b (you should check this really is b). The ith-subinterval is [xi−1 , xi ] for
i = 1, 2, . . . , n. Because f is continuous, it has a maximum f (Mi ) and a minimum f (mi ) in
the ith-subinterval (see the diagram below).
The upper and lower sums are as follows:
Lower sum = s(n) =
n
X
f (mi )∆x (Area of inscribed rectangles)
i=1
Upper sum = S(n) =
n
X
f (mi )∆x (Area of circumscribed rectangles)
i=1
It will always be true that s(n) ≤ (Area of region) ≤ S(n). A diagram of this is given below.
Example 5. Use upper and lower sums to approximate the area of the region under the curve
28
y =
for 1 ≤ x ≤ 2 with five rectangles of equal width. The lower sum rectangles are
x+6
shown below to the left, and the upper sum rectangles are below to the right.
y
y
4
4
3
3
2
2
1
1
x
1
2
x
1
2
(a) The lower sum is
(b) The upper sum is
(c) Therefore,
≤ Actual Area ≤
Solution: (a) Each of the rectangles has base 0.2, and so the lower sum is
28
28
28
28
28
L = (0.2)
+
+
+
+
1.2 + 6 1.4 + 6 1.6 + 6 1.8 + 6 2 + 6
≈ 3.689
(b) Each of the rectangles has base 0.2, and so the upper sum is
28
28
28
28
28
+
+
+
+
L = (0.2)
1 + 6 1.2 + 6 1.4 + 6 1.6 + 6 1.8 + 6
≈ 3.789
(c) The actual area is always between any lower sum and any upper sum, so
3.689 ≤ Actual Area ≤ 3.789
Our ultimate goal is to find the exact area under the graph of (many) functions An important
theorem is that the limit of upper sums and lower sums of a continuous nonnegative function
always exist and are equal. This common limit is what we define to be the area of that region:
Definition of the Area of a Region in the Plane. Let f be a continuous and nonnegative
function on the interval [a, b]. The area of the region bounded by the graph of f , the x-axis,
and the vertical lines x = a and x = b is
n
X
Area = lim
f (ci )∆x, xi−1 ≤ ci ≤ xi
n→∞
i=1
where ∆x = (b − a)/n.
In using this definition, we usually choose a convenient ci , such as the right end point of the
ith interval.
The next two examples use the limit process to compute an area under a nonnegative continuous
function.
Example 6. This problem will go through a step-by-step calculation of the area under the
graph of f (x) = x2 + 4x + 7 for 0 ≤ x ≤ 2.
(a) Divide the interval [0, 2] into n subintervals of equal length ∆x. What is ∆x?
(b) Because f (x) is increasing, the maximum f (Mk ) occurs when Mk chosen as the right-hand
endpoint of the k-th subinterval. Express Mk in terms of k and n.
(c) Express f (Mk )∆x in terms of k and n.
(d) With the help of appropriate summation formula(s), express
n
X
f (Mk )∆x in terms of n.
k=1
(e) Because f is continuous and nonnegative on [0, 2], taking the limit of the expression in (d)
will provide the area. That is
" n
#
X
Area = lim
f (Mk )∆x =
n→∞
Solution: (a) ∆x =
k=1
2−0
2
=
n
n
2
2k
=
n
n
" # 2
2k
2k
2
8k 2 16k 14
(c) f (Mk )∆x =
+4
+7
= 3 + 2 +
n
n
n
n
n
n
(b) Mk = 0 + k(∆x) = k ·
(d)
n
X
f (Mk )∆x =
k=1
n X
8k 2
k=1
16k 14
+ 2 +
3
n
n
n
n
n
n
8 X 2 16 X
14 X
k
+
k
+
1
n3 k=1
n2 k=1
n k=1
16 n(n + 1)
14
8 n(n + 1)(2n + 1)
+
+
(n)
=
n3
6
n2
2
n
3
1
1
8
2+ + 2 +8 1+
+ 14
=
6
n n
n
=
(e) Using part (d), the area is given by
" n
#
X
Area = lim
f (x∗k )∆x
n→∞
k=1
3
1
1
8
= lim
2+ + 2 +8 1+
+ 14
n→∞ 6
n n
n
=
74
8
− 8 + 14 =
3
3
Example 7. Use a limit process to find area of the region that lies between the x-axis and
the curve y = 9 − x2 for 0 ≤ x ≤ 3.
Solution: To partition the interval [0, 3] into n-subintervals we find
∆x =
3
3i
3−0
= , and xi =
n
n
n
Using right end points, we have that the area, A, is given by
A =
=
=
=
=
=
=
Thus the area is 18 units2 .
lim
n→∞
n
X
f (xi )∆x
i=1
2 ! 3
3i
lim
9−
n→∞
n
n
i=1
n X
3
9i2
lim
9− 2
n→∞
n
n
i=1
!
n
n
27 X
27 X 2
lim
1− 3
i
n→∞
n i=1
n i=1
27 n(n + 1)(2n + 1)
lim 27 − 3 ·
n→∞
n
6
27
27 27
−
−
lim 27 −
n→∞
3
2n 6n2
27
27 −
= 18
3
n
X