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
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View Full DocumentAnother plug and chug leads us to plug in

1
,
0
,
1
,
and
,
2 in for
x
. Therefore the right hand
sum is
area
≈
Δ
x
(
f
(

1) +
f
(0) +
f
(1) +
f
(2))
⇒
area
≈
f
(

1) +
f
(0) +
f
(1) +
f
(2)
⇒
area
≈
(1
/
2 + 1 + 1
/
2 + 1
/
5)
2.
Section 5.2
1. Estimate the integral with
n
= 4 using the following techniques
Z
6
2
1 +
x
2
dx
(a) Lefthand sum
Solution:
First of all, Δ
x
for this problem is
Δ
x
= (
a

b
)
/n
= (6

2)
/
4 = 1
For the lefthand sum, the points that we plug in for
x
have the form
x
i
=
a
+
i
Δ
x
for
i
= 1
,
2
, ..., n
⇒
x
i
= 2 +
i
Δ
x
for
i
= 1
,
2
,
3
,
4
(b) Righthand sum
2. For question 1 above, how far oﬀ can your estimates be from the true value?
Solution:
The thing to notice (this can be seen from the graph of
f
(
x
) = 1 +
x
2
) is that the
righthand sum is an over estimate to the integral, while the lefthand sum is an under estimate to
the integral. So the integral value lies somewhere in the middle. Letting
R
(
f
) denote the right hand
sum and
L
denote the left hand sum, we have
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 Summer '10
 reed
 Math, Division, dx, Riemann sum, Riemann

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