MATH 1100 – INTEGRATION / AREA UNDER A CURVE
This is an exposition of section 13.1 that may help you to better understand the material. Hopefully, it
will emphasize what I see as the important ideas and techniques too. First, let us work in a general setting.
Suppose that for a nonnegative function y = f (x) we want to calculate the area under the curve using the
right endpoint method. In other words, we want to calculate
Z b
f (x)dx.
a
First, we divide the interval [a, b] into n-pieces. That gives us a base of ∆ =
b−a
n .
This is the width of
each rectangle, and the base points will be xi = a + ∆ · i. That is the right endpoint of the first rectangle is
a + ∆, the second is a + ∆2, the ith is a + ∆ · i, and the last (nth) is b. Then the height of the ith rectangle
is f (xi ). The total area of all of the rectangles will then be given by the equation
SR =
n
X
f (xi )∆.
i=1
In the case of the left endpoint, everything is the same except that the first endpoint is a, the second is
a+∆, the ith endpoint will be a+∆(i−1) and the last one is b−∆ = a+∆(n−1). So letting yi = a+∆(i−1)
denote the ith endpoint we get
SL =
n
X
f (yi )∆.
i=1
The first two figures illustrate this general case. To calculate the actual area under the given curve, we would
need to simplify the equation for SR or SL to get rid of the summation symbol, then we would take the limit
as n approaches infinity.
Now, let’s work out an example. Suppose that f (x) = x + 2x2 and the interval is [0, 2]. Dividing the
interval into n equal-sized pieces each one will have width n2 . The right endpoints are xi = n2 i, and the left
endpoints are yi =
2
n (i
− 1). Thus, the sum of the rectangles obtained using the right endpoints is
2 !
n
X
2
2
2
SR =
i+2
i
n
n
n
i=1 n
X
4
16 2
i + 3i
=
n2
n
i=1
n
n
4 X
16 X 2
=
i
+
i
n2 i=1
n2 i=1
2n(n + 1) 8n(2n + 1)(n + 1)
=
+
n2
3n3
2
16
= 2 + n + 3 + n8 + 3n8 2 .
Finally, the area is given by the limit
16
2
22
16
8
8
=2+
+ lim
+ lim
+ lim
=
.
2
n→∞ n
n→∞ 3
n→∞ n
n→∞ 3n
3
3
lim SR = lim 2 + lim
n→∞
n→∞
Exercise 1. Calculate the area using the left endpoint method.
1
2
MATH 1100 – INTEGRATION / AREA UNDER A CURVE
Let’s check that our answer is correct by using the fundamental theorem of Calculus. The fundamental
theorem tells us that
2 Z 2
22
16
2x3 x2
− (0 + 0) =
=
2
+
+
.
(x + 2x2 )dx =
2
3
3
3
0
0
So our work checks out to be correct!
Exercise 2. Use rectangles to calculate the area between y = 4x − x2 and the x-axis on the interval [0, 3].
Then check that your answer is correct using the fundamental theorem of Calculus.