Chapter 5
1.
Section 5.1
1. The velocity
v
(
t
) in the table below is increasing. Use
n
= 4 subdivisions to approximate the total
distance traveled. Also, ﬁnd
t
3
6
9
12
15
f
(
t
)
20
21
25
30
40
(a) An upper estimate.
(b) A lower estimate.
Solution:
(a) The thing to notice is that
f
(
t
) is increasing and getting progressively steeper. This means that
an upper estimate can be obtained by using right sums. The right point sum in this case is
given by
distance
≈
3(
f
(6) +
f
(9) +
f
(12) +
f
(15)) = 3(116) = 348
(b) For the same reasons given in part (a), a lower estimate for the distance is obtained by using left
sums. In this case
distance
≈
3(
f
(3) +
f
(6) +
f
(9) +
f
(12)) = 3(96) = 278
Remark
: You can also approximate distance traveled, or equivalently area under curves, by
using rectangles of diﬀerent widths. For example, a left handed Riemann sum with nonuniform
rectangles would have the following form
L
(
f
) =
n
X
i
=0
f
(
a
+
i
Δ
x
i
)Δ
x
i
The subscript
i
with the Δ
x
indicates that the width of each rectangle varies.
2. Consider the function
f
(
x
) =
1
1 +
x
2
for

2
≤
x
≤
2
Use Δ
x
= 1 to ﬁnd the:
(a) Left sum
(b) Right sum
Solution:
(a) First note that since we are insisting on Δ
x
= 1, the number of rectangles we will be using is
n
= (
b

a
)
/
(Δ
x
) = (2

(

2))
/
1 = 4. Furthermore, the points at which we need to evaluate
f
(
x
) have the form
x
i
=
a
+
i
Δ
x
for
i
= 0
, .., n

1
⇒
x
i
=

2 +
i
for
i
= 0
,
1
,
2
,
3
After
plugging and chugging
, we see that
f
(
x
) must be evaluated at

2
,

1
,
0
,
and 1. This
generates the left hand sum
area
≈
Δ
x
(
f
(

2) +
f
(

1) +
f
(0) +
f
(1))
⇒
area
≈
f
(

2) +
f
(

1) +
f
(0) +
f
(1)
⇒
area
≈
(1
/
5 + 1
/
2 + 1 + 1
/
2)
(b) The reasoning is exactly the same as that given for part (a), the only diﬀerence being that
f
(
x
)
must be evaluated at points of the form
x
i
=
a
+
i
Δ
x
for
i
= 1
,
2
, ..., n
⇒
x
i
=

2 +
i
for
i
= 1
,
2
,
3
,
4
1