Lecture 33
The definite integral and its applications (cont’d)
Using definite integrals instead of indefinite integrals (antiderivatives) in solving
problems
There is no relevant section in the textbook by Stewart for the material presented in today’s lecture.
In the previous lecture, we discussed an important application of the Fundamental Theorem of
Calculus II, namely, that
integraldisplay
b
a
F
′
(
x
)
dx
=
F
(
b
)
−
F
(
a
)
.
(1)
This is true for any differentiable function
F
(
x
) since, by definition,
F
(
x
) is an antiderivative of
F
′
(
x
).
The quantity on the left hand side of the equation represents an integration of the
rate of change
of
F
(
x
), namely,
F
′
(
x
), from
x
=
a
to
x
=
b
. The quantity on the right hand side represents the
net
change
of
F
(
x
) from
x
=
a
to
x
=
b
.
Section 5.4 of the textbook by Stewart (Sixth Edition) discusses a number of important applica
tions for this “Net Change” property of definite integrals. And, indeed, in the previous lecture, we
examined an important application to classical mechanics: Suppose that
v
(
t
) represents the velocity
of a particle travelling on a straight line (represented by the
x
axis). The relation of the position
x
(
t
)
and the velocity function
v
(
t
) is, of course,
v
(
t
) =
dx
dt
=
x
′
(
t
)
.
(2)
From the Net Change property, which is simply a consequence of FTC II, we have that
integraldisplay
b
a
v
(
t
)
dt
=
integraldisplay
b
a
x
′
(
t
)
dt
=
x
(
b
)
−
x
(
a
)
.
(3)
In other words, the definite integral of
v
(
t
) over the time interval [
a, b
] is the
net displacement
of
the particle over this time interval.
We now consider a slightly different form of the above Net Change property. We’ll integrate the
velocity function from a starting time, say
t
= 0 to a future time
t >
0. But we’ll keep this upper
limit as a variable. With regard to the previous equation, we’ll replace
a
with 0 and
b
with
t
. This
219
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means that we’ll have to change the integration variable,
t
, to something else, say,
s
. The result is
integraldisplay
t
0
v
(
s
)
ds
=
integraldisplay
t
0
x
′
(
s
)
ds
=
x
(
t
)
−
x
(0)
.
(4)
We can rewrite the above equation as follows,
x
(
t
) =
x
(0) +
integraldisplay
t
0
v
(
s
)
ds.
(5)
We can check this result by differentiating. The derivative of the LHS is simply
x
′
(
t
). Since
x
(0) is a
constant, the derivative of the RHS is, by the FTC I,
d
dt
integraldisplay
t
0
v
(
s
)
ds
=
v
(
t
) =
x
′
(
t
)
.
(6)
We’ll use this idea repeatedly in this lecture.
A return to the “nearearth” falling body problem
We consider the motion of an object of mass
m
near the surface of the earth, where the force acting
on it is assumed to be
F
(
x
) =
f
(
x
)
i
, where
f
(
x
) =
−
mg.
(7)
As before, the positive
x
axis points upward, with
x
= 0 denoting the surface of the earth.
If we let
x
(
t
) denote the position of the mass at time
t
, then Newton’s equation of motion,
F
=
m
a
becomes
f
=
ma,
(8)
where
a
(
t
) =
v
′
(
t
)
,
(9)
is the acceleration. Of course, in this simple case,
f
=
−
mg
, so that Newton’s equation becomes
−
mg
=
ma
⇒
a
=
−
g.
(10)
In other words, the acceleration of the object is constant, as we know very well.
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 Fall '08
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 Calculus, Antiderivatives, Derivative, Integrals, Mass, dx, RHS

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