LECTURE 29
Areas under Curves
1. Some Sum Formulas
Before we get into the material in Chapter 5, I need to make introduce some simple summation formulas
that we’ll be using later on.
First, consider the sum
S (n) = 1 + 2 + 3 + 4 + · · · + (n − 1) + n
The result of this sum obviously depends on the number n of terms we use. In fact, is a simple polynomial
function of n.
(1)
1 + 2 + 3 + ··· + n =
n (n + 1)
2
To see this, let’s first look at a few examples
n
1
2
3
4
5
S (n)
1
1+2=3
1+2+3=6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
n(n+1)
2
1(2)
2 =
2(3)
2 =
3(4)
2 =
4(5)
2 =
5(6)
2 =
1
3
6
10
15
However, looking at a few examples does not constitute a proof of the formula (1). To prove it, one needs
to use Mathematical Induction, which means that we check the validity of the formula for the when case
n = 1, and then show that if it’s true for n = N , then it’s also true for n = N + 1. We already confirmed the
validity of formula (1) for n = 1, 2, 3, 4, 5, let’s now carry out the second part of a proof by mathematical
induction.
So we suppose
(2)
1 +2 +···+ N =
N (N + 1)
2
is true. We want to conclude that this implies
(3)
1 + 2 + · · · + N + (N + 1) =
122
(N + 1) (N + 2)
2
2. COMPUTING AREAS UNDER GRAPHS
123
is true. Using (2) we have
1 + 2 + · · · + N + (N + 1) = (1 + 2 + · · · + N ) + (N + 1)
N (N + 1)
=
+ (N + 1)
2
N (N + 1) + 2 (N + 1)
(putting everything over a common denominator)
=
2
2
N + 3N + 2
(expanding the numerator)
=
2
(N + 1) (N + 2)
=
(after factoring the numerator)
2
Comparing the extreme sides of this last equation we see that we have the desired relation (3).
Later in this lecture we’ll also need a formula for
(1)2 + (2)2 + (3)2 + · · · + (n)2
This turns out to be
(1)2 + (2)2 + (3)2 + · · · + (n)2 =
(4)
n (n + 1) (2n + 1)
6
However, rather than prove it (which can be in essentially the same we handled 1 + 2 + · · · + n, except
with more algebraic tedium), let me instead introduce the idea of sigma notation. This idea is very simple.
When we have a bunch of numbers n1 , n2 , . . . , nN that we want to sum up each of which depends on some
counting index i, we write
N
ni
=
N
1 = 1 +1 +···+1
N
i ≡ 1 +2 +3 +··· +N
N
i2 = (1)2 + (2)2 + (3)2 · · · + (N )2
for
n1 + n2 + n3 + · · · + nN
i=1
Thus, we have
N
(as there are N terms in the sum)
i=1
N (N + 1)
2
=
N (N + 1) (2N + 1)
6
=
i=1
i=1
2. Computing Areas Under Graphs
Recall that the notion of derivative was introduced as a means to compute the slopes of tangent lines to
the graph of a function. There is also a prototypical geometric problem underlying the notion of an antiderivative; it is the problem of figuring out the area that lies between the graph of a function and the x-axis.
This we shall now develop.
Let’s begin with the graph of a simple parabola.
f (x) = x2
between x = 0 and x = 1.
2. COMPUTING AREAS UNDER GRAPHS
124
Because of its curvy shape, it is not possible to use any of the formulas of the areas of rectangles or triangles
to figure out the area under this curve. Rather what we shall do is figure out the area under this curve via
a limiting process.
We begin by dividing the interval between x = 0 and x = 1 into n equal subintervals
1
1 2
2 3
n−1 n
[0, 1] = 0,
∪
,
∪
,
∪ ··· ∪
,
n
n n
n n
n
n
Let’s set
xi =
i
n
i
so that xi is the right-most boundary of the ith subinterval i−1
n , n . For each of these subintervals we then
form a small rectangle of width n1 and height f (xi ). This will look something like
2. COMPUTING AREAS UNDER GRAPHS
125
Now while the area covered by these rectangles does not equal the area under the curve, it is approximately
equal to the area under the curve: each of the rectangles overshoots the curve by a just little bit, and
these overshoots we could try to minimize by using more and more rectangles. Morever, we can explicitly
compute the area covered by these rectangles. We have
area covered by rectangles =
n
width of ith rectangle height of ith rectangle
i=1
where the symbol
n
i=1
means we’re to sum over all the rectangles (as labeled by the index i running from 1 to n). Now
1
n
width of ith rectangle
=
height of ith rectangle
= f
2
i
i
=
n
n
2. COMPUTING AREAS UNDER GRAPHS
126
and so we have
area covered by rectangles =
n 2
1
i
n
n
i=1
2 2 2
1
1
1
1
1
2
3
n 2
=
+
+
+···+
n
n
n
n
n
n
n
n
1 =
1 + 22 + 33 + · · · + n2
n3
Now from formula (4) we have
12 + 22 + · · · + n2 =
n
i2 =
i=1
n (n + 1) (2n + 1)
6
and
n (n + 1) (2n + 1)
1 1 + 22 + 33 + · · · + n2 =
3
n
6n3
This gives a precise formula for the area covered by the rectangles, which should be approximately equal to
the area under the original curve, especially if we take n to be very large. Suppose in fact we consider the
limit n → ∞,
area covered by rectangles=
n (n + 1) (2n + 1)
6n3
3
2n + 3n2 + n
= lim
n→∞
6n3
2
2 + 3 n1 + n1
= lim
n→∞
6
1
=
3
lim (area covered by rectangles) =
n→a
lim
n→∞
However, how can I be so sure that the area that we’re neglecting (the overshoots) is really going to zero
as n → ∞.
Well, let’s consider a different set of rectangles. This time we’ll use the value of f (x) at the beginning of
the subintervals to get the height of the rectangles:
Now we’ll have
area under rectangles
=
2
n 1
i−1
i=1
n
n
n
=
1
n3
=
n
1 2
i − 2i + 1
3
n i=1
=
n
n
1 2
2 1 i
−
i
+
1
n3 i=1
n3
n3 i=1
=
=
(i − 1)2
i=1
1 n (n + 1) (2n + 1)
2 n (n + 1)
1
− 3
+ 3 (n)
n3
6
n
2
n
1 2n2 − 3n + 1
6
n2
2. COMPUTING AREAS UNDER GRAPHS
127
Figure 1
Now since these rectangles all lie strictly below the graph of f (x), their total area has to be less than the
area under the graph. But again
1 2n2 − 3n + 1
=
2
n→∞ 6
n
1
1
1
= lim
2−3 + 2
n→∞ 6
n n
1
=
3
lim (area under the lower rectangles) =
n→∞
lim
But now since
area under the lower rectangles < area under the curve < area under the upper rectangles
and
lim (area under the lower rectangles) =
n→∞
1
= lim (area under the upper rectangles)
3 n→∞
2. COMPUTING AREAS UNDER GRAPHS
128
The Squeeze Theorem tells us that
area under the curve =
1
3
This example generalizes as follows.
Suppose
f (x) is a function between x = a and x = b. Divide the interval [a, b] into n equal subintervals
xi−1 , xi by setting
b−a
i
n
(Note x0 = a, and xn = b). Choose a point x∗i in each subinterval xi−1 , xii motivates the following
definition. Then the width of each subinterval will be
b−a
∆x =
n
and the rectangle of width ∆x and height f (x∗i ) will have approximately the same area as the area under
the curve between xi−1 and xi . Summing over all these thin rectangular areas we get
n
(Area under the graph of f (x) between x = a and x = b) ≈
f (x∗i ) ∆x
xi = a +
i=1
Definition 29.1. The area under the graph of a function f between x = a and x = b, is the limit of the
sum of areas of approximating rectangles as set up above
n
A = lim
f (x∗i ) ∆x
n→∞
i=1
Definition 29.2. :