Section 5.1 Homework Solutions
1. Since f is given only as a graph (no formula is given), we must look at
the graph (page 359 of Stewart) to determine the value of f (x) for each
value of x.
(a) An upper estimate for the area under the graph of f using …ve rectangles is
R5 = 2 ¢ 3 + 2 ¢ 4:3 + 2 ¢ 5:3 + 2 ¢ 6:2 + 2 ¢ 7 = 51:6.
A lower estimate for the area under the graph of f using …ve rectangles is
L5 = 2 ¢ 1 + 2 ¢ 3 + 2 ¢ 4:3 + 2 ¢ 5:3 + 2 ¢ 6:2 = 39:6.
(b) An upper estimate for the area under the graph of f using ten rectangles is
R10 = 1 ¢ 2 + 1 ¢ 3 + 1 ¢ 3:7 + 1 ¢ 4:3 + 1 ¢ 5 + 1 ¢ 5:3 + 1 ¢ 5:8 + 1 ¢ 6:2 + 1 ¢ 6:7 + 1 ¢ 7
= 49:0.
A lower estimate for the area under the graph of f using ten rectangles is
1 ¢ 1 + 1 ¢ 2 + 1 ¢ 3 + 1 ¢ 3:7 + 1 ¢ 4:3 + 1 ¢ 5 + 1 ¢ 5:3 + 1 ¢ 5:8 + 1 ¢ 6:2 + 1 ¢ 6:7
= 43:0
5. The graph of f (x) = sin x on the interval [0; ¼] is shown below.
1
0.8
0.6
0.4
0.2
0
0.5
1
1.5x
2
2.5
Graph of f (x) = sin x on [0; ¼].
1
3
For n = 10, we obtain
R10
³
³
³
¼
¼´
¼
¼´
¼
¼´
¢ sin 0 +
+
¢ sin 0 + 2 ¢
+ ¢¢¢ +
¢ sin 0 + 10 ¢
10
10
10
10
10
10
10
³
´
X
¼
¼
sin i ¢
.
=
10 i=1
10
=
To compute R10 on our TI-83 calculator, we enter
(¼=10) £ sum(seq(sin(I £ (¼=10); I; 1; 10))
to obtain R10 ¼ 1:9835.
Similarly, we obtain
³ ¼´
¼ X
sin i ¢
30 i=1
30
30
R30 =
= (¼=30) £ sum(seq(sin(I £ (¼=30); I; 1; 30))
¼: 1:9982
and
³ ¼´
¼ X
sin i ¢
50 i=1
50
50
R50 =
= (¼=50) £ sum(seq(sin(I £ (¼=50); I; 1; 50))
¼ 1:9993
Based on these calculations, it appears that the exact area in question
might be 2.
6. We want to approximate the area under the curve y = 1=x2 over the interval [1; 2] using n = 10; n = 30; and n = 50 subdivisions of [1; 2] and using
right endpoints of the subdivisions to compute heights of approximating
rectangles. Based on these calculations, we will try to guess the exact area
under the curve.
First, let us do the approximation with n = 10 subdivisions. When we
divide the interval [1; 2] into 10 subintervals (of equal length), each subinterval has length 1=10. The x coordinates of the endpoints of these subintervals are
10 11 12
20
; ; ;:::; .
10 10 10
10
Since the curve we are working with is y = 1=x2 , the corresponding y
2
coordinates are
102
1
¡ 10 ¢2 = 2
10
10
1
102
¡ 11 ¢2 = 2
11
10
102
1
¡ 12 ¢2 = 2
12
10
..
.
etc.
The Riemann sum R10 is thus
R10 =
1 102
1 102
1 102
¢ 2+
¢ 2 + ¢¢¢ +
¢
10 10
10 11
10 202
which can be expressed as
R10
µ
1
1
1
= 10
+
+ ¢¢¢ + 2
102 112
20
¶
or, by using the summation notation, as
R10 = 10
20
X
1
.
i2
i=10
We can use the TI-83 calculator to compute R10 by entering the following:
10*sum(seq(1/I^2,I,10,20))
and we obtain
R10 = 0:5639551275.
For n = 30, each subinterval has length 1=30, the x coordinates of endpoints of these subintervals are
60
30 31 32
; ; ;:::;
30 30 30
30
and the corresponding y coordinates are
302 302 302
302
;
;
;
:
:
:
;
.
302 312 322
602
3
This gives us
1 302
1 302
1 302
¢ 2+
¢ 2 + ¢¢¢ +
¢ 2
30 µ 30
30 31
¶30 60
1
1
1
= 30
+
+ ¢¢¢ + 2
302 312
60
60
X
1
= 30
2
i
i=30
R30 =
= 30 ¤ sum(seq(1=I^2; I; 30; 60))
= 0:5209953305
Similarly, using n = 50, we obtain
R50
100
X
1
= 50
= 0:5125583282.
2
i
i=50
Note that, in general, we have
Rn = n
2n
X
1
.
i2
i=n
Our calculations lead us to guess that the exact area under the curve is
0:5.
11. Using one second time intervals (and the graph of the velocity shown on
page 360 of Stewart), we estimate that the distance traveled by the car
during braking is
1 ¢ 45 + 1 ¢ 35 + 1 ¢ 22 + 1 ¢ 13 + 1 ¢ 8 + 1 ¢ 0 = 123 ft .
14. The area under the curve f (x) = x3 from x = 0 to x = 1 is
A = lim Rn
n!1
where
Rn =
We can write this as
µ ¶3
n
X
1
i
¢
.
n
n
i=1
n
1 X 3
Rn = 4
i .
n
i=1
Since
n
X
i=1
3
i =
µ
n (n + 1)
2
¶2
4
=
1 4 1 3 1 2
n + n + n ,
4
2
4
we obtain
1
Rn = 4
n
µ
1 4 1 3 1 2
n + n + n
4
2
4
¶
=
1
1
1
+
+
.
4 2n 4n2
From this, we see that
A = lim Rn
n!1
¶
µ
1
1
1
+
+ 2
= lim
n!1 4
2n 4n
1
= +0+0
4
1
= .
4
5