This document provides solutions to a calculus worksheet involving estimating the area under the curve of the function f(x) = ex on the interval [0,1]. [1] The solution estimates the area at increasing levels of subdivision (n = 2, 4, 8) and finds the area is approaching a limit. [2] Closed form expressions are derived for the left and right approximations Ln and Rn to the area, showing Rn = e1/nLn. [3] The common limit of Ln and Rn as n approaches infinity is shown to be 1-e, which is the exact area under the curve on the interval.

1. Solutions to Worksheet for Section 5.1
Areas and Distances
V63.0121, Calculus I
Summer 2010
1. Draw the graph of f (x) = ex on the interval [0, 1]. We are going to find the area below the
curve on this interval.
Solution. Here is the region:
y
3
2
1
x
1
2. Estimate the area by computing with a calculator.
(i) L2 and R2
(ii) L4 and R4
(iii) L8 and R8
Solution. We have
L2 = e0 (0.5) + e0.5 (0.5) = 1.32436
R2 = e0.5 (0.5) + e1.0 (0.5) = 2.18350
And so on.
n Rn Ln
2 2.18350 1.32436
4 1.94201 1.51244
8 1.82791 1.61313
1

2. 3. Show that
n
1 e(i−1)/n
Ln = 1 + e1/n + e2/n + · · · + e(n−1)/n =
n i=1
n
n
1 1/n ei/n
Rn = e + e2/n + · · · + e = .
n i=1
n
4. Find “closed form” expressions (i.e., without ellipses or sigmas) for Ln and Rn .
Hint. Remember that for any number r that
1 − rn
1 + r + r2 + r3 + · · · + rn−1 = .
1−r
Solution. Using the formula from on Ln with r = e1/n , we have
n
1 i−1 1−e
Ln = e1/n =
n i=1
n(1 − e1/n )
If you look carefully, you notice Rn = e1/n Ln , so
e1/n (1 − e)
Rn =
n(1 − e1/n )
5. Find lim Ln and lim Rn . These limits should be the same, and their common value is the
n→∞ n→∞
area of the region.
Solution.
1−e
lim Ln = lim
n→∞ n→∞ n(1 − e1/n )
The numerator tends to 1 − e, but the denominator is of the indeterminate form ∞ · 0. So we
apply L’Hˆpital’s Rule:
o
1 − e1/x
lim x(1 − e1/x ) = lim
x→∞ x→∞ 1/x
H −e1/x (−1/x2 )
= lim
x→∞ −1/x2
= lim −e1/x = −1.
x→∞
2