A Careful Area Computation
We are going to carefully compute the exact area of the region 0 ≤ y ≤ ex ≤ 1,
0 ≤ x ≤ 1. There will be no uncontrolled approximations.
Because derivative ddx ex = ex is always positive, the function ex increases as x
increases. Consequently, the smallest and largest values of ex on the interval a ≤ x ≤ b are
ea and eb , respectively. In particular, for 0 ≤ x ≤ N1 , ex takes values only between e0 and
e1/N . As a result, the set
(x, y) 0 ≤ x ≤ N1 , 0 ≤ y ≤ ex
contains the rectangle of 0 ≤ x ≤ N1 , 0 ≤ y ≤ e0 (the lighter rectangle in the figure on the
left below) and is contained in the rectangle 0 ≤ x ≤ N1 , 0 ≤ y ≤ e1/N (the largest rectangle
in the figure on the left below). Hence
1 0
e ≤ Area (x, y) 0 ≤ x ≤ 1 , 0 ≤ y ≤ ex ≤ 1 e1/N
(1)
N
N
N
y = ex
y
y = ex
y
e2/N
e1/N
e0
e1/N
e0
x
1
N
1
N
Similarly, as in the figure on the right above,
1 1/N
≤ Area (x, y) N1 ≤ x ≤
Ne
1 2/N
e
≤ Area (x, y) 2 ≤ x ≤
N
N
..
.
1 (N−1)/N
e
N
2
N,
3
,
N
2
N
···
0 ≤ y ≤ ex
≤
0 ≤ y ≤ ex
≤
N
N
1 2/N
Ne
1 3/N
e
N
..
.
≤ Area (x, y) N−1
N
..
.
≤x≤
N
,
N
0 ≤ y ≤ ex
=
c Joel Feldman.
2002. All rights reserved.
1
1
N
N e
1
1 N
Ne
December 13, 2002
N
2
+ eN + · · · + eN
1
(2)
1 N/N
e
N
≤
Adding (1) and all of the lines of (2) together gives
N −1 1
1
N + ··· + e N
N 1+e
≤ Area (x, y) 0 ≤ x ≤ 1, 0 ≤ y ≤ ex
≤
x
1 + eN + · · · + e
N −1
N
A Careful Area Computation
1
Using 1 + r + · · · + r m =
1−rm+1
1−r
with r = e1/N and m = N − 1, so that r m+1 = e1/N
N
= e,
1 1−e
0 ≤ x ≤ 1, 0 ≤ y ≤ ex ≤ 1 e1/N 1 − e
≤
Area
(x,
y)
N 1 − e1/N
N
1 − e1/N
Thus the exact area must be at least as large as
the exact area must also be at least as large as
lim 1 1−e
1/N
N→∞ N 1−e
= (1 − e)
lim
1
x= N
→0
1−e
1
N 1−e1/N
x
1−ex
for every single integer N ≥ 1. So
1
x
x→0 −e
= (1 − e) lim
=e−1
1
1−e
by L’Hôpital’s rule. Similarly, the exact area must be smaller than (or equal to) N1 e N 1−e
1/N
for every single natural number N . So the exact area must also be smaller than or equal to
1
1−e
1 N
e 1−e
1/N
N
N→∞
lim
x
x
= (1 − e) lim ex 1−e
lim
x = (1 − e) lim e
x→0
x→0
x
x
x→0 1−e
=e−1
We have now shown that
e − 1 ≤ Area (x, y) 0 ≤ y ≤ ex , 0 ≤ x ≤ 1 ≤ e − 1
so that the area must be exactly e − 1.
c Joel Feldman.
2002. All rights reserved.
December 13, 2002
A Careful Area Computation
2