7
Integration
Up to now we have been concerned with extracting information about how a function
changes from the function itself. Given knowledge about an object’s position, for example,
we want to know the object’s speed.
Given information about the height of a curve we
want to know its slope. We now consider problems that are, whether obviously or not, the
reverse of such problems.
EXAMPLE 7.1
An object moves in a straight line so that its speed at time
t
is given
by
v
(
t
) = 3
t
in, say, cm/sec. If the object is at position 10 on the straight line when
t
= 0,
where is the object at any time
t
?
There are two reasonable ways to approach this problem. If
s
(
t
) is the position of the
object at time
t
, we know that
s
′
(
t
) =
v
(
t
). Because of our knowledge of derivatives, we
know therefore that
s
(
t
) = 3
t
2
/
2+
k
, and because
s
(0) = 10 we easily discover that
k
= 10,
so
s
(
t
) = 3
t
2
/
2 + 10. For example, at
t
= 1 the object is at position 3
/
2 + 10 = 11
.
5. This
is certainly the easiest way to deal with this problem. Not all similar problems are so easy,
as we will see; the second approach to the problem is more difficult but also more general.
We start by considering how we might approximate a solution. We know that at
t
= 0
the object is at position 10. How might we approximate its position at, say,
t
= 1? We
know that the speed of the object at time
t
= 0 is 0; if its speed were constant then in the
first second the object would not move and its position would still be 10 when
t
= 1. In
fact, the object will not be too far from 10 at
t
= 1, but certainly we can do better. Let’s
look at the times 0
.
1, 0
.
2, 0
.
3,
...
, 1
.
0, and try approximating the location of the object
133
134
Chapter 7 Integration
at each, by supposing that during each tenth of a second the object is going at a constant
speed. Since the object initially has speed 0, we again suppose it maintains this speed, but
only for a tenth of second; during that time the object would not move. During the tenth
of a second from
t
= 0
.
1 to
t
= 0
.
2, we suppose that the object is traveling at 0
.
3 cm/sec,
namely, its actual speed at
t
= 0
.
1. In this case the object would travel (0
.
3)(0
.
1) = 0
.
03
centimeters: 0
.
3 cm/sec times 0
.
1 seconds.
Similarly, between
t
= 0
.
2 and
t
= 0
.
3 the
object would travel (0
.
6)(0
.
1) = 0
.
06 centimeters. Continuing, we get as an approximation
that the object travels
(0
.
0)(0
.
1) + (0
.
3)(0
.
1) + (0
.
6)(0
.
1) +
· · ·
+ (2
.
7)(0
.
1) = 1
.
35
centimeters, ending up at position 11.35. This is a better approximation than 10, certainly,
but is still just an approximation. (We know in fact that the object ends up at position
11
.
5, because we’ve already done the problem using the first approach.)
Presumably,
we will get a better approximation if we divide the time into one hundred intervals of a
hundredth of a second each, and repeat the process:
(0
.
0)(0
.
01) + (0
.
03)(0
.
01) + (0
.
06)(0
.
01) +
· · ·
+ (2
.
97)(0
.
01) = 1
.
485
.