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
Z
a
b
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 discusses a number of important applications 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
= x′ (t).
dt
(2)
From the Net Change property, which is simply a consequence of FTC II, we have that
Z
b
a
v(t) dt =
Z
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
259
means that we’ll have to change the integration variable, t, to something else, say, s. The result is
Z
t
v(s) ds =
Z
t
0
0
x′ (s) ds = x(t) − x(0).
(4)
The above equation may be rewritten as follows,
x(t) = x(0) +
Z
t
v(s) ds.
(5)
0
This result can be verified by differentiation. 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,
Z
d t
v(s) ds = v(t) = x′ (t).
dt 0
(6)
We’ll use this idea repeatedly in this lecture.
A return to the “near-earth” 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 = ma
becomes
f = ma,
(8)
a(t) = v ′ (t),
(9)
where
is the acceleration. Of course, in this simple case, f = −mg, so that Newton’s equation becomes
−mg = ma
⇒
a = −g.
In other words, the acceleration of the object is constant, as we know very well.
260
(10)
The “old method” of solving the equation of motion using antiderivatives
As you may recall, we proceeded to solve for the motion of the mass by expressing the above equation
in terms of the velocity function v(t),
dv
= v ′ (t) = −g.
dt
(11)
As such, we look for a v(t) of which the t-derivative is −g. In other words, we look for the antiderivatives of the constant function −g. The answer is
v(t) = −gt + C,
(12)
where C is an arbitrary constant. In light of our more recent discussion on antiderivatives, this solution
may also be written as follows,
v(t) =
Z
−g dt = −gt + C.
(13)
Here, we have introduced the indefinite integral, which represents the set of all antiderivatives of
the function in its integrand: Here, the integrand is the constant function −g.
A particular value of the arbitrary constant C can be extracted if we impose an initial condition
on the velocity function, e.g.,
v(0) = v0 .
(14)
If we impose this condition on the solution in (12), we obtain
v(0) = −g · 0 + C = v0
⇒
C = v0 .
(15)
Therefore, our solution to this initial value problem is
v(t) = v0 − gt.
(16)
Of course, this solution is well known to you.
A “new method” of obtaining solutions using definite integrals
We now consider another method to obtain the above solution for the velocity function v(t), starting
with Eq. (11). Instead of using antiderivatives, we’ll integrate both sides of Eq. (11) from time 0 to
general time t, as follows:
Z
t
′
v (s) ds =
0
Z
261
t
(−g) ds.
0
(17)
Note that we have to use an integration variable s instead of t, because we’re using t to denote the
upper limit. (Recall that we had the same situation when we were proving FTC I.) The LHS of the
above equation is a definite integral of a derivative. By the Net Change property, it becomes
Z
t
0
v ′ (s) ds = v(t) − v(0) = v(t) − v0 ,
(18)
where we have set v(0) to our prescribed initial velocity v0 .
The RHS of Eq. (17) may be evaluated from FTC II. An antiderivative of −g (with respect to
the variable s) is −gs, so that
Z
t
t
(−g) ds = −gs = −gt.
0
0
(19)
Equating the results of Eqs. (18) and (19), we obtain
v(t) − v0 = −gt
⇒
v(t) = v0 − gt,
(20)
in agreement with the result obtained by the “old” antiderivative method.
Note that in this method, we avoid the use of the arbitrary constant C. The method may seem
somewhat lengthy, but this is only because we have described each step in detail. In principle, we
could go from Eq. (17) to Eq. (20) directly.
And from Eq. (20), we may obtain the position function, x(t), with prescribed initial position,
x(0) = x0 ,
(21)
x′ (t) = v0 − gt,
(22)
We’ll first write Eq. (20) as follows,
and then perform a definite integration from s = 0 to s = t.
Z
t
x′ (s) ds =
Z
t
(v0 − gs) ds.
(23)
The integrals on both sides may be evaluated using FTC II, i.e.,
t
1 2 t
x(s) = v0 s − gs ,
2
0
0
(24)
0
0
which becomes
1
x(t) − x0 = v0 t − gt2 .
2
262
(25)
This leads to the well-known result,
1
x(t) = x0 + v0 t − gt2 .
2
(26)
In summary, we state that if the acceleration function a(t) of a particle moving on the x-axis is
known, then the velocity function v(t) corresponding to a given initial condition v(0) = v0 may be
found by definite integration from s = 0 to s = t as follows,
v(t) − v0 =
Z
t
a(s) ds
0
⇒
v(t) = v0 +
Z
t
a(s) ds.
(27)
0
From this velocity function v(t), we may find the position function x(t) in a similar fashion, by definite
integration from s = 0 to s = t:
x(t) − x0 =
Z
t
v(s) ds
0
⇒
x(t) = x0 +
Z
t
v(s) ds.
(28)
0
The work done by a nonconstant force
We start with a result that is well-known to you from high school physics. Suppose that a constant
force F = F i acts on a mass m, causing it to move along the x-axis from position x = a to x = b.
Then the total work done W by the force is given by the product of the magnitude of the force and
the displacement of the mass, i.e.,
W = F (b − a)
(29)
This is a special case of the more general result in which a constant force F moves the mass in a
straight line that is not necessarily parallel to F. If the displacement vector of the mass is d, then the
total work W done by F is
W = F · d.
(30)
In the discussion that follows, it will be sufficient to consider Eq. (29).
Now suppose that the force F is no longer constant, i.e., F = f (x)i, where the function f (x) is
not necessarily constant. If the mass m is moved from position x = a to x = b, what is the total work
263
W done by the force? You have most probably seen the answer in your first-year Physics course. It
is given by the definite integral,
W =
Z
b
f (x) dx.
(31)
a
We now derive this result mathematically in terms of our Riemann sum definition of the definite
integral. And our derivation will be done by employing what we have previously called the “Spirit
of Calculus.” Very briefly, we’ll subdivide the interval [a, b] into tiny subintervals Ik of length ∆x,
and then approximate the force function f (x) as a constant over each subinterval. We then use the
constant-force result from Eq. (29) over each subinterval Ik , to approximate the work ∆Wk done in
moving the mass over the subinterval Ik . Finally, we sum over the contributions from all subintervals.
As before, we first consider an n > 0 (with the idea of letting n → ∞) and define
∆x =
b−a
.
n
(32)
Then define the partition points,
xk = a + k∆x,
k = 0, 1, 2, · · · .
(33)
Note that x0 = a and xn = b. These partition points define a set of n subintervals Ik = [xk−1 , xk ],
k = 1, 2, · · · n, of equal length ∆x.
Now select a sample point x∗k ∈ [xk−1 , xk ] from each subinterval Ik . Then evaluate the force
function f at each sample point x∗k . We now consider each value f (x∗k ) as the approximation of f (x)
over the subinterval Ik . In other words, the function f (x) is approximated by a constant function
f (x∗k ). In this way, we may use Eq. (29) to approximate the work ∆Wk done by the function f (x) in
moving the mass over the subinterval Ik , i.e., from xk−1 to xk , as follows,
∆Wk ∼
= f (x∗k )∆x
(constant force strength × displacement).
(34)
The total work done by the force in moving the mass from x0 = a to xn = b will then be approximated
as follows,
W =
n
X
k=1
∆Wk ∼
=
n
X
f (x∗k )∆x.
(35)
k=1
But by construction, the RHS of this equation is a Riemann sum for the definite integral of f (x) from
x = a to x = b. Assuming that the definite integral of f exists (which is ensured if f is continuous or
264
piecewise continuous), we have
W
=
=
lim
n→∞
Z
n
X
f (x∗k )∆x
k=1
b
f (x) dx.
(36)
a
This concludes our mathematical justification of the definite integral formula for work.
An important note regarding the dimensions of work and the integral formula
The dimensionality of force is M LT −2 . (Think of F = ma and the dimensions of mass and acceleration.
Therefore the dimensionality of work is force times distance, or M L2 T −2 . Note that the dimensionality
of the Riemann sum in Eq. (35) is also force times distance. Since the definite integral in Eq. (36)
is the limit of Riemann sums with this dimensionality, it follows that the definite integral has the
dimension of work. Basically, we can think of the integrand as having the dimensionality of force and
the infinitesimal dx as the dimensionality of length.
The extension of the work integral to several dimensions and motion along curves
In a future course on advanced calculus that includes the subject of “vector calculus” (e.g., AMATH
231 or MATH 227), you will consider the more general case of a nonconstant force F(r) in R3 acting
on a mass m as the mass moves along a curve C from a point P to point Q. The situation is sketched
in the diagram below.
z
Q
P
m
F(r(t))
r(t)
y
x
The goal is once again to compute the total amount of work W done by the force. Once again, in
the “Spirit of Calculus,” the idea is to break up the motion into tiny pieces over which we can use the
265
constant-force-straight-line formula “W = F (b − a)” to approximate the work over these pieces. We
then “sum up,” i.e., integrate over all contributions to obtain W . In this case, since the force vectors
F(r) will not, in general, be parallel to the motion of the mass, we’ll have to take scalar products
of these vectors F with the instantaneous direction of motion of the mass m – in other words, the
velocity vectors v = r′ along the curve. The net result is that we have an integral of the following
form
Z
C
F · dr,
(37)
which is known as the line integral of the vector field F over the curve C. You may already have
seen this integral in your Physics course.
Defining potential energy in terms of definite integrals
Recall the definition of a conservative force in one dimension, F(x) = f (x)i: There exists a potential
energy function U (x) such that
U ′ (x) = −f (x)
f (x) = −U ′ (x).
or
(38)
It follows that any force that is dependent only on the position coordinate x is a conservative force.
From Eq. (38), it follows that U (x) is the negative antiderivative of f (x), i.e.,
Z
U (x) = − f (x) dx.
(39)
As we’ll see below, this formulation introduces the need for an arbitrary constant. A definite integral
formulation will avoid the arbitrary constant.
Example: In the case of free fall near the earth, the force function is given by
f (x) = −mg.
(40)
From Eq. (39), the associated potential energy function is
Z
Z
U (x) = − (−mg) dx = mg dx = mgx + C,
(41)
a well-known result. If we now impose the condition that U (0) = 0, we have that
U (0) = mg0 + C = 0
⇒
C=0
266
⇒
U (x) = mgx.
(42)
In this case, all potential energy is measured with respect to the reference point x = 0, at which
U (0) = 0.
You may recall that the definition of the potential energy function according to Eq. (38) turned
out to be very convenient. From this definition, we were able to prove that the total mechanical
energy function E(t), which is a sum of kinetic and potential energies, is constant in time. But the
formulation of the potential energy function in terms of definite integrals turns out to be even more
useful, from both computational as well as physical viewpoints.
Given a force function f (x) and a convenient reference point x = a, we define the potential energy
U (x) associated with f (x) as follows,
U (x) = −
x
Z
f (s) ds.
(43)
a
There are two noteworthy points regarding this definition:
1. From FTC I, we have that
d
U (x) = −
dx
′
Z
x
a
f (s) ds = −f (x).
(44)
Therefore, Eq. (38) is satisfied.
2. Since the lower limit of the definite integral is a, we have that
U (a) = −
Z
a
f (s) ds = 0.
(45)
a
In other words, the definite integral isolates a particular negative antiderivative of the force
function f (x).
Let us now examine the physical interpretation of the definite integral definition in Eq. (38). The
integral
Z
x
f (s) ds
(46)
a
represents the work done by the force F = f (x)i in moving the mass m from the position a to
the position x. As such, the function U (x), which includes a negative sign in front of the integral,
represents the work done against the force F = f (x)i in moving the mass m from position a to
267
position x. If U (x) is positive, it means that we have had to exert a force against the force F in order
to displace the mass from a to x. This causes a “buildup” of potential energy in the system.
Example 1: We return to the free-fall problem examined earlier, in which f (x) = −mg. We’ll also
let a = 0 be the reference point. Then
Z
Z x
(−mg) ds =
U (x) = −
0
x
mg ds = mg
Z
x
ds = mgx,
(47)
0
0
with U (0) = 0, in agreement with our earlier treatment. But note that no arbitrary constant was
needed in this derivation. We simply “built” the initial condition into the function U (x) in terms of
the lower limit of integration. Note that U (x) increases with x. If the x-coordinate of the mass is
increased, i.e., it is elevated, then its potential energy increases.
Example 2: Let’s now return to the “far-from-earth” gravitational problem, in which the distance
from the mass m to the earth’s surface is sufficiently large that the approximation f (x) = −mg is no
longer valid. In this case, the force exerted by the earth on the mass at a height x ≥ 0 above the
surface of the earth is
f (x) = −
GM m
,
(R + x)2
(48)
where R is the radius of the earth, M its mass and G is the gravitational constant.
Recall that in an earlier assignment, you were required to find the potential U (x) associated with
this force, with the additional condition that U (0) = 0. At that time, you most probably worked
with the antiderivatives of f (x) and then adjusted the arbitrary constant. In our definite integral
formalism, we can simply state that the desired potential is given by
Z x
f (s) ds
U (x) = −
0
Z x
1
= GM m
ds
2
0 (R + s)
x
1
= GM m −
(by FTC II)
R+s 0
1
1
= GM m −
+
R+x R
GM m GM m
−
.
=
R
R+x
Note that U (0) = 0, as it should be.
268
(49)
Before moving on to another example, let us also recall the utility of this potential. If a projectile
is launched from the earth’s surface, x = 0, with speed v0 > 0, then its initial total mechanical energy
is given by
1
1 GM m
E(0) = mv02 + U (0) = +
.
2
2
R
(50)
But since the gravitational force is conservative (i.e., a potential energy function U (x) exists), the
total mechanical energy of the projectile will be constant in time, i.e.,
GM m
GM m
1
1
−
= E(0) = mv02 .
E(x(t)) = mv(t)2 +
2
R
R + x(t)
2
(51)
At the highest point of the projectile’s trajectory, call it x = h, its velocity v is zero, after which the
particle begins to return to earth. This implies that h satisfies the equation
GM m GM m
1
−
= mv02 .
R
R+h
2
(52)
After a little algebra, we may solve for h,
h=
v02 R
.
2GM
2
R − v0
(53)
r
2GM
, the height of the trajectory h → ∞. Therefore,
R
if the projectile were (at least theoretically) launched with initial velocity
r
2GM
vesc =
,
(54)
R
As the initial speed v0 approaches the value
it would never return, since h = ∞. This critical velocity, vesc as you may well know, is the escape
velocity of the earth. (Roughly, vesc ∼
= 11, 000 m/s.)
The existence of an escape velocity is the consequence of the fact that the magnitude of the
attractive gravitational force goes to zero as the distance x goes to infinity.
Example 3: Recall the linear mass-spring problem, where the restorative force was given as
f (x) = −kx,
k > 0.
(55)
The equilibrium point of this force, i.e., the point at which the net force is zero, is x = 0. We choose
this to be our reference point, i.e., a = 0. The associated potential energy function is then
Z x
Z x
1
s ds = kx2 ,
(−ks) ds = k
U (x) = −
2
0
0
269
(56)
a familiar result. Once again, there is no need for the arbitrary constant, which will eventually be
fixed according to a specified condition, e.g., U (a) = 0.
Note that U (x) > 0 for x 6= 0. If we move the mass from the equilibrium position x = 0, then
the spring will be either compressed (x < 0) or extended (x > 0) from its natural length. As a result,
potential energy will be stored in the spring.
A final note on the relationship between potential energy and work
We conclude by making a connection between the above discussion and something that you know from
physics regarding the relationship between potential energy and work.
Let’s return to the definition of potential energy in Eq. (43):
U (x) = −
x
Z
f (s) ds,
(57)
a
where a is a reference point, chosen so that U (a) = 0. Now consider any two points, x1 and x2 , on
the real line, with no assumptions on which is greater. Then the work done by a force f (x) in moving
a mass m from x1 to x2 is
W =
Z
x2
f (s) ds.
(58)
x1
But from the additive property of definite integrals, we can write the above as follows,
W
=
=
Z
x2
f (s) ds
x1
Z 0
f (s) ds +
x1
Z
x2
f (s) ds
0
= −U (x1 ) + U (x2 )
= −∆U,
(59)
where
∆U = U (x2 ) − U (x1 )
(60)
is the net change in potential energy of the mass m. In words, the work done by the force is the
negative of the change in the potential energy. This is something that you know from your
elementary course in Physics. Lifting up an object of mass m from height x1 to height x2 > x1 against
gravity – and near the surface of the earth – is an example of this idea. The change in potential energy
U (x) = mgx is
∆U = mg(x2 − x1 ) > 0.
270
(61)
The work done by the gravitational force f (x) = −mg is
W
=
=
Z
x2
Zx1x2
f (s) ds
(−mg) ds
x1
= −mg(x2 − x1 )
= −∆U.
271
(62)
Lecture 34
The definite integral and its applications (cont’d)
The average value of a function
Suppose that a thin, straight wire is located on the x-axis, specifically on the interval [a, b]. Furthermore, suppose that the function f (x) represents the temperature of the wire at a point x ∈ [a, b]. The
question is, “What is the average temperature of the wire?” This is a specific example of the more
general question: What is the average value of a function f over the interval [a, b]?
We’ll address this problem in the usual way, i.e., by means of the “Spirit of Calculus.” We’ll divide
up the interval [a, b] into n subintervals Ik , take samples of the function f (x) on these subintervals,
and then compute the average of these sample values. The average value of f over the interval [a, b]
will be the limit n → ∞ of these average values, provided that the limit exists.
b−a
. Then define the partition points,
So, as before, let n > 0 and define ∆x =
n
xk = a + k∆x,
k = 0, 1, · · · , n,
(63)
so that x0 = a and xn = b. These points define the n subintervals Ik = [xk−1 , xk ], k = 1, 2, · · · , n.
From each subinterval Ik , choose a sample point x∗k ∈ Ik . Then evaluate the function at this sample
point. The result is a set of n function values f (x∗k ). These may be viewed as samples of the function
f (x) over the interval [a, b].
It seems reasonable to take the average of these n function values – we’ll denote this average as
n
1X
f¯n =
f (x∗k ).
n
(64)
k=1
Now the sum on the RHS looks almost like a Riemann sum to the definite integral of f . However, a
∆x is missing. So let’s multiply and divide by ∆x as follows,
n
f¯n =
1 1 X
f (x∗k )∆x
n ∆x
k=1
=
1
Sn .
n∆x
Here, Sn is a Riemann sum corresponding to the definite integral
(65)
Z
b
f (x) dx. There remains the
a
1
. Recalling the definition of ∆x:
n∆x
b−a
⇒
n∆x = b − a.
∆x =
n
question about what to do about the factor
272
(66)
Therefore, the average value in (65) becomes
f¯n =
1
Sn .
b−a
(67)
Assuming that f is continuous (or at least piecewise continuous), the limit of the Riemann sums Sn
exists, and we have
1
1
lim Sn =
b − a n→∞
b−a
lim f¯n =
n→∞
Z
b
f (x) dx.
(68)
a
This is the average value of f over the interval [a, b], which we shall denote as follows,
f¯[a,b] =
1
b−a
Z
b
f (x) dx.
(69)
a
In other words, we compute the definite integral of f over the interval [a, b] and then divide by the
length of the interval, b − a.
Let’s now rewrite Eq. (69) as follows,
Z
b
a
f (x) dx = f¯[a,b] (b − a).
(70)
If we assume, for the moment - for the sake of simplicity - that f (x) > 0 on [a, b], then Eq. (70) is
stating that the area enclosed by the graph of f (x), the lines x = a and x = b and the x − axis is given
by the average value of f on [a, b] multiplied by the length of the interval (b − a). In other words, as
sketched in the figure below, we have replaced the area enclosed by the graph, etc., by a rectangle of
height f¯[a,b] . The rectangular region is shaded.
y = f (x)
f¯[a,b]
x
a
b
That being said, we may now relax the restriction that f (x) be strictly positive on [a, b]. In this case,
the average value of f on [a, b] times the length (b − a) will be a signed area.
273
Some simple, yet illuminating, examples:
1. The function f (x) = 1 over the interval [a, b] = [0, 1]. Since f (x) assumes only one value over
the entire interval, namely, the value 1, we expect that its average value is 1. Let’s check this.
Since a = 0, b = 1 and f (x) = 1, we have
1
f¯[0,1] =
1
Z
1
0
1 dx = [x]10 = 1,
(71)
as expected.
2. The function f (x) = x over the interval [a, b] = [0, 1]. From a look at the graph of f over
[0, 1], we might guess that the average value is its average value is 1/2. Since a = 0, b = 0 and
f (x) = 1, we have
1
f¯[0,1] =
1
Z
1
0
x2
x dx =
2
1
0
1
= .
2
(72)
Our intuition was correct.
3. The function f (x) = x2 over the interval [a, b] = [0, 1]. A look at the graph of f (x) = x2 shows
that there are many more x-values for which f (x) < 1/2 than in the previous case, f (x) = x.
Therefore, we would expect the average value to be less than 1/2. Since a = 0, b = 0 and
f (x) = 1, we have
1
f¯[0,1] =
1
Z
1
x2 dx =
0
x3
3
1
=
0
1
.
3
(73)
4. In general, the function f (x) = xn over the interval [a, b] = [0, 1], where n > 0. Since a = 0,
b = 0 and f (x) = 1, we have
1
f¯[0,1] =
1
Z
1
0
xn+1
x dx =
n+1
n
1
0
=
1
.
n+1
(74)
1
→ 0. Does
n+1
this make sense? For any x such that 0 ≤ x < 1, raising it to higher powers makes it smaller,
Note that as n → ∞, the average value of the function xn behaves as follows,
i.e., xn → 0 as n → ∞. (Think of x = 1/2.) That means that the graph of f (x) = xn gets flatter
and flatter as n increases, except at x = 1, since 1n = 1 always. This is illustrated below.
Since all values of xn for x ∈ [0, 1) – note that we exclude the case x = 1 – approach zero as
n → ∞, we expect the average value of xn to approach zero in the limit n → ∞.
274
An important piece of advice: When you encounter a new concept in mathematics, it is often
most helpful to apply that concept to a set of cases, perhaps a one-parameter family of functions, and
to observe the behaviour of the results as you vary the parameter. We have done this with the concept
of the average value of a function, applying it to the one-parameter family of functions xn on [0, 1].
One-dimensional continuous distributions of mass: The “thin wire”
We return to the idea of a thin wire that is represented by the interval [0, L] on the x-axis. (The
cross-sectional area of the wire, assumed to be small in comparison to the length of the wire, b − a,
is essentially factored out, so we may view the wire as a one-dimensional object.)
Note: In class, we considered the wire to be located on an interval [a, b]. Without loss of generality,
we can “shift” this interval to [0, L]: it is convenient to let the left endpoint of the interval be 0.
Recall that in the section on derivatives, we introduced the mass function m(x) associated with
the wire:
For x ∈ [0, L], m(x) is the mass of the wire over the interval [0, x].
This implies that
m(0) = 0 and
m(L) = M, the total mass of the wire.
(75)
Also recall that we examined the idea of the average rate of change of the mass over any interval
[x1 , x2 ] ⊂ [0, L]:
m(x2 ) − m(x1 )
∆m
=
.
∆x
x2 − x1
(76)
We then considered the case in which ∆x → 0, to define an instantaneous rate of change of mass, i.e.,
the derivative of m(x),
m′ (x) = lim
h→0
m(x + h) − m(x)
= ρ(x).
h
(77)
We referred to the function ρ(x) as the linear density function. Once again, it represents the rate
of change of the mass m(x) at an x ∈ [0, L].
Let us now view m(x) and ρ(x) in terms of integrals and, in particular, definite integrals. Since
m′ (x) = ρ(x),
275
(78)
the function m(x) is an antiderivative of ρ(x). Moreover, m(x) is the particular antiderivative for
which m(0) = 0. This means that we can write m(x) in terms of ρ(x) as follows,
m(x) =
Z
Just to check this:
Z
m(0) =
x
ρ(s) ds.
(79)
ρ(s) ds = 0,
(80)
0
0
0
and
d
m (x) =
dx
′
Z
x
ρ(s) ds
(FTC II).
(81)
0
Therefore, Eq. (79) is correct.
A consequence of the above result is that the total mass M of the wire is given by
M = m(L) =
Z
L
ρ(s) ds =
Z
L
ρ(x) dx.
(82)
0
0
(In a definite integral with constant endpoints, it doesn’t matter what variable we use as the integration
variable.)
Before going on with other things, let’s multiply and divide the RHS by L, the length of the wire,
i.e.,
1
M =L·
L
Z
L
ρ(x) dx.
(83)
0
But this result may be rewritten as
M = Lρ̄[0,L] ,
(84)
where ρ̄[0,L] is the average or mean linear density of the wire, i.e., the average value of the density
function ρ(x) over the interval [0, L] representing the wire. In other words, the total mass M is the
average linear density of the wire times the length of the wire. This seems to make sense. In fact, it
is the mass/density analogy of Eq. (70) which relates the area enclosed by the graph of f (x) to the
area of a rectangle.
Let us now carry our analysis a little further. Suppose that we wish to characterize the mass of
the wire found in a subinterval [c, d] ⊂ [0, L]: We’ll call this mass M[c,d] . Since c < d, it follows that
M[c,d] = m(d) − m(c).
276
(85)
(The wire on [c, d] may be considered as taking the wire on [0, d] and then removing the wire on [0, c].)
From the definition of m(x) in (79),
M[c,d] =
Z
=
Z
=
=
Z
Z
d
0
ρ(s) ds −
Z
ρ(s) ds +
Z
d
0
0
ρ(s) ds +
c
Z
c
ρ(s) ds
0
0
ρ(s) ds
c
d
ρ(s) ds
0
d
ρ(s) ds.
(86)
c
In the special case that [c, d] = [0, L], then we have the total mass M of the wire on [0, L].
Returning to Eq. (79), consider the special case that ρ(x) = ρ0 , a constant. Assuming that the
cross-sectional area of the wire is constant, this implies that the wire is homogeneous, i.e., composed
of the same material throughout the wire. In this case, the mass function m(x) in (79) becomes
m(x) =
Z
x
ρ0 ds = ρ0
0
Z
x
ds = ρ0 x,
0
0 ≤ x ≤ L.
(87)
The graph of m(x) vs. x is a straight line that runs from (0, 0) to (L, ρ0 L). The total mass of the
wire is M = ρ0 L.
Center of mass of a thin wire
We now address the problem of finding the center of mass of a wire on [0, L], with linear density
function ρ(x). Recall that the center of mass is the point at which the wire can be balanced. In order
to solve this problem, it is helpful to return to the case of a finite number of masses.
The simplest problem is the two-mass case: the “teeter-totter,” sketched below. Given two masses
m1 and m2 located on opposite sides of the pivot point, and at distances of d1 and d2 , respectively,
from the pivot, balance is achieved when
m1 d1 = m2 d2 .
277
(88)
d1
d2
m1
m2
We now reformulate this problem as follows: Suppose that masses m1 and m2 are located at coordinate
positions x1 and x2 , with x1 < x2 . Where is the center of mass x̄?
m1
m2
x̄
x1
x2
Once again, x̄ is the location of the pivot point for perfect balance. Because x̄ lies between x1 and x2 ,
Eq. (88) translates to
m1 (x̄ − x1 ) = m2 (x2 − x̄).
(89)
We’ll rewrite this equation as follows,
m1 (x1 − x̄) + m2 (x2 − x̄) = 0.
(90)
We can solve for x̄:
x̄ =
m1 x1 + m2 x2
,
m1 + m2
(91)
a formula with which you are no doubt familiar as the center of mass of a two-body system.
We can now generalize this result to the case of n masses on the line: For k = 1, 2, · · · , n, a mass
of mk is situated at position xk . Eq. (90) generalizes to
n
X
k=1
mk (xk − x̄) = 0.
(92)
The LHS of this equation is known as the first moment of the masses about x̄.
From this equation, we may easily solve for x̄. First rewrite it as follows,
n
X
k=1
mk xk − x̄
n
X
mk = 0.
(93)
k=1
The second sum on the LHS is the total mass M of the system. Therefore, the center of mass is given
by
n
1 X
m k xk .
x̄ =
M
k=1
278
(94)
Once again, you are most probably familiar with this equation.
Before going on, let’s rewrite Eq. (94) as follows,
x̄ =
n X
mk k=1
or
x̄ =
n
X
M
xk ,
pk xk ,
(95)
(96)
k=1
where
pk =
mk
M
⇒
n
X
pk = 1.
(97)
k=1
From Eq. (96), the center of mass x̄ may be viewed as a weighted average of the xk values. The
greater the mass mk , the greater the weighting factor pk . In fact, because the weighting factors pk are
nonnegative and sum to 1, the weighted sum in Eq. (96) has a special name: it is called a convex
combination of the xk .
The next step is to carry this idea over to continuous distributions of mass. Our goal: to find the
continuous version of Eq. (94) for the center of mass corresponding to a density function ρ(x). The
way to do this is to use – guess what? – the “Spirit of Calculus.” Once again, we divide up the mass
into a finite number n of tiny pieces, consider each piece as a point mass mk , then compute the center
of mass of this ensemble, and then let n → ∞.
L
So, as before, for an n > 0, let ∆x =
(our interval [a, b] is now [0, L]) and define the partition
n
points
k
(98)
xk = k∆x = L, 0 ≤ k ≤ n.
n
These partition points define a set of n subintervals Ik = [xk−1 , xk ]. We now let ∆mk denote the mass
of wire over each subinterval Ik .
Once again, we choose sample points x∗k ∈ Ik from each subinterval. From our definition of the
density function, the mass ∆mk of the wire on Ik is approximated as follows,
∆mk ∼
= ρ(x∗k )∆x.
(99)
The total mass of the wire, M , is then approximated as follows,
M=
n
X
k=1
∆mk ∼
=
279
n
X
k=1
ρ(x∗k )∆x
(100)
But the sum on the RHS is the Riemann sum for the function ρ(x) over the interval [0, L]. As such,
in the limit n → ∞, we have
M=
L
Z
ρ(x) dx,
(101)
0
which is consistent with our earlier definition in (82).
But we haven’t finished! We still have to compute the continuous version of the sum in Eq. (94).
Returning to the n masses ∆mk produced by our partition above, we shall consider them as point
masses situated at the sample points x∗k . Of course, this is an approximation, but as n → ∞, this
approximation gets better and better. The moment of these masses is then approximated by
n
X
∆mk x∗k
=
k=1
n
X
x∗k ρ(x∗k )∆x.
(102)
k=1
Note that the sum on the RHS has the form of a Riemann sum over the interval [0, L], but it is the
Riemann sum corresponding to the function f (x) = xρ(x). In the limit, this Riemann sum converges
to the definite integral
Z
L
xρ(x) dx.
(103)
0
As a result, the center of mass x̄ of the continuous distribution of mass corresponding to the density
function ρ(x), 0 ≤ x ≤ L is given by
1
x̄ =
M
Z
L
xρ(x) dx,
where M =
Z
0
0
We’ll examine some examples in the next lecture.
280
L
ρ(x) dx.
(104)
Lecture 35
The definite integral and its applications (cont’d)
One-dimensional continuous distributions of mass (cont’d)
Center of mass (cont’d)
Some of the following examples were discussed in the Thursday tutorial. As such, this subsection was
skipped in the lecture.
We now compute the centers of mass of some simple continuous mass distributions. In all cases, the
wire is located on the interval [0, 1].
1. Example 1: The mass distribution ρ(x) = 1. Since the mass density function is constant, the
wire may be considered homogeneous, i.e., identical composition throughout the wire. In this
case, we would expect the center of mass to be located at its center point, i.e., x̄ = 1/2. The
total mass of the wire is
M=
Z
1
ρ(x) dx =
0
Z
1
dx = 1.
(105)
0
The first moment of the wire with respect to the origin is given by
Z 1
Z 1
1 2 1 1
x
= .
x dx =
xρ(x) dx =
Mx =
2
2
0
0
0
(106)
The center of mass of this wire is therefore
x̄ =
Mx
1/2
=
= 1,
M
1
(107)
as expected.
2. Example 2: We now consider a perturbation of the above mass distribution, the density function
1
ρ(x) = 1 + x.
2
(108)
The density function ρ(x) increases as x increases from 0 to 1. As such, the wire is heavier on the
right side than on the left, and we expect the center of mass to lie to the right of the geometric
center x = 1/2. The total mass of the wire is
Z 1
Z 1
1 2 1
5
1
1
=1+ = .
ρ(x) dx =
M=
1 + x dx = x + x
2
4
4
4
0
0
0
281
(109)
The first moment of the wire with respect to the origin is
Z 1
Z 1
2
1 2 1 3 1 1 1
1 2
x + x
= + = .
xρ(x) dx =
Mx =
x + x dx =
2
2
6
2 6
3
0
0
0
(110)
The center of mass of the wire is therefore
x̄ =
Mx
2/3
8
=
= .
M
5/4
15
(111)
As expected, the center of mass lies to the right of the geometrical center point x = 1/2 (although
not that far away from it).
From an understanding of integration and the “Spirit of Calculus,” we are now in a position
to consider a wide variety of applications in physics that involve continuous mass distributions.
The following is an example of such an application.
The total gravitational force exerted by a one-dimensional rod
L L
Consider a wire over the interval − ,
with mass density function ρ(x). Now let a point mass
2 2
L
m be situated outside the wire, at position coordinate a > , as sketched below. (By symmetry,
2
the case a < −L/2 will yield the same magnitude.) We wish to find the total gravitational force
exerted by the wire on the point mass.
m
x
−L/2
L/2
0
We’ll be using the following basic fact from Physics: The magnitude of the force of gravitational
attraction between two point masses m1 and m2 situated a distance d > 0 apart is
F =
Gm1 m2
.
d2
(112)
L L
into n tiny subinterOnce again, in the “Spirit of Calculus,” we divide up the interval − ,
2 2
L
vals Ik = [xk−1 , xk ] of width ∆x = , and approximate the mass of wire ∆mk in each subinterval
n
282
by ρ(x∗k )∆x, where x∗k is a sample point in that interval. The magnitude of the gravitational
force between this mass element and the point mass at x = a is
Gmρ(x∗k )∆x
Gm∆mk
≈
.
∆Fk ∼
=
(a − x∗k )2
(a − x∗k )2
(113)
We now add up the magnitudes of the forces from all n subintervals,
F =
n
X
k=1
∆Fk ∼
= Gm
X
k=1
ρ(x)
∆x.
(a − x∗k )2
(114)
In the limit n → ∞, the Riemann sum on the right converges to the integral,
F = Gm
Z
L/2
−L/2
ρ(x)
dx.
(a − x)2
(115)
Before proceeding, let us now present a slightly abbreviated version of the above derivation, of the
kind that you will probably encounter in your Physics courses. (It was also the version presented
in this lecture.) Instead of considering the partition of the interval into n tiny subintervals Ik , we
simply go to the “infinitesimal limit” and consider an infinitesimal interval of width dx situated
at x ∈ [−L/2, L/2], as sketched below.
dx
−L/2
0
m
x
L/2
a
d=a−x
The infinitesimal mass dm of the element of wire situated in this interval is
dm = ρ(x) dx.
(116)
This comes from the definition of the mass density function,
dm
= ρ(x)
dx
⇒
dm =
dm
dx = ρ(x) dx.
dx
(117)
The magnitude dF of the force between this infinitesimal mass element at x and the point mass
at a is given by
dF =
Gmρ(x) dx
ρ(x)
Gm dm
=
= Gm
dx.
2
2
(a − x)
(a − x)
(a − x)2
283
(118)
The magnitude of the total force exerted by the rod on the point mass is obtained by integrating
over all mass elements on [−L/2, L/2]:
F = Gm
Z
L/2
−L/2
the result obtained earlier.
ρ(x)
dx,
(a − x)2
(119)
Once we specify the mass density function ρ(x), we may, at least in principle, compute the
magnitude of the total force F . In what follows, we consider the particular case ρ(x) = ρ0 ,
constant, the case of a homogeneous wire. In this case, the total mass of the wire is
Z L/2
Z L/2
ρ0 dx = ρ0 L.
ρ(x) dx =
M=
(120)
−L/2
−L/2
And since the wire is homogeneous, the center of mass is located at x̄ = 0.
The total force exerted by this homogeneous wire is then given by the integral
Z L/2
1
F = Gmρ0
dx.
2
−L/2 (a − x)
(121)
The integral is not difficult to compute, since the antiderivative of the integrand is relatively
straightforward,
Z
L/2
−L/2
1
dx =
(a − x)2
=
=
Therefore,
F =
L/2
1
a − x −L/2
1
1
−
a − L/2 a + L/2
L
.
2
a − L2 /4
(122)
Gmρ0 L
.
a2 − L2 /4
(123)
GM m
.
− L2 /4
(124)
Recalling that the total mass of the wire is M = ρ0 L, we have the final result,
F =
a2
This is a very interesting result, and worthy of some comment and analysis. First of all, the
most obvious observation is that the force is not given by
F =
284
GM m
,
a2
(125)
the case if the rod were replaced by a point mass M at its center of mass x = 0. Many of
you may be aware of the result that the gravitational force exerted by a three-dimensional
spherical and homogeneous mass M is the same as the force due to a point mass M located
at the center of the sphere. But this is not the case in one-dimension. Nor is it the case in
two-dimensions. And even in three dimensions, the mass must be spherical and homogeneous
(or at least have a spherically symmetric mass density function ρ) for the ability to replace it by
a point mass at its center.
Note that for a very large, the term L2 /4 in the denominator of Eq. (124) is negligible, in which
case the magnitude of the force is well approximated by Eq. (125). Perhaps it is helpful to
characterize how large a would have to be for the approximation to be valid. We can do this by
rewriting Eq. (124) as follows,
F =
1
GM m
.
a2 1 − L 2
2a
For any given L > 0, we see that the ratio
(126)
L
must be sufficiently small.
2a
One-dimensional charge distributions
Because the classical electrostatic force between two charges is also inversely proportional to the square
of their separation, the gravitational example examined above has an electrostatic counterpart. The
rod now supports a one-dimensional distribution of charge with linear density ρ(x), x ∈ [−L/2, L/2].
And the mass m at x = a is now a test charge q. There is one important difference, however – the
electrostatic force can be either (i) repulsive or (ii) attractive, depending on whether the charge q has
the (i) same or (ii) opposite sign to that of the rod. For simplicity, we’ll assume that the charges have
the same sign so that the force is repulsive.
We start with the electrostatic analogy to Eq. (112), namely, “Coulomb’s Law”, in which the
electrostatic force between two charges q1 and q2 a distance d apart is given by
F =
q1 q2
.
4πǫ0 d2
Here, ǫ0 denotes the permittivity of the vacuum.
285
(127)
We proceed as before, considering the electrostatic force between an infinitesimal element of charge
ρ(x) dx situated at x ∈ [−L/2, L/2] and the test charge q at x = a. Integration over the entire rod
yields the net force
q
F =
4πǫ0
Z
L/2
−L/2
ρ(x)
dx.
(a − x)2
(128)
This result may be compared with its gravitational counterpart in Eq. (119).
In the special case that the ρ(x) = ρ0 , a constant, we have the result,
F =
1
qρ0
· 2
,
4πǫ0 a − L2 /4
(129)
which may be compared to its gravitational counterpart in Eq. (124).
The total kinetic energy of a rotating rod
You have probably encountered this idea in your first-year Physics course, but we include it here as
another example of an integration problem over a one-dimensional mass distribution.
L L
We consider the same thin wire of the previous example - it is positioned over the interval − ,
2 2
and has linear mass density ρ(x). It is assumed to rotate about the point x = 0 with angular frequency
ω > 0 (radians/unit time).
Very quickly, in the “Spirit of Calculus,” we once again consider an infinitesimal element of mass
dm of thickness dx and situated at x. During the rotation, this mass element dm will travel over
a circle of radius |x|, as sketched below. (Recall that x can assume negative values in the way that
this problem was formulated. This wasn’t mentioned in the lecture.) The kinetic energy of this mass
element is given by
dK =
1
(dm) v 2 ,
2
(130)
where v is the speed of the revolving mass. Note that we have written the kinetic energy of this element
as dK since it is an an infinitesimal amount of energy – we’ll integrate over all of these infinitesimal
elements in order to compute the total kinetic energy K of the rotating rod.
It remains to express the quantities dm and v in terms of x. The dm part is easy – as before
dm = ρ(x) dx.
As for the velocity, it is given by v = ω|x|, implying that v 2 = ω x2 .
286
(131)
ω
dm = ρ(x) dx
−L/2
0
x
L/2
Aside: If you are still uncertain about the above result v = ω|x|, think of it this way: The unit of
angular frequency ω is radians per unit time (not degrees per unit time). This means that its cyclical
frequency is
ν=
ω
cycles per unit time.
2π
(132)
2π
1
=
units of time to complete one cycle. The circumference of a circular
ν
ω
orbit of radius r is C = 2πr units of length. This is one cycle of revoluton. Therefore the speed of a
This means that it takes
particle on this circular orbit is
v=
2πr
distance
=
= ωr units of length per unit time.
time
2π/ω
(133)
In the above discussion, r = |x|.
Putting all of this together, the kinetic energy dK of the mass element is
dK =
1
1
ρ(x) dx (ω|x|)2 = ω 2 ρ(x) x2 dx.
2
2
(134)
We now integrate over the rod to obtain the total kinetric energy,
L/2
1
dK = ω 2
K=
2
−L/2
Z
Z
L/2
x2 ρ(x) dx.
(135)
−L/2
Note that the above result may be written in the form,
1
K = I ω2,
2
where I =
Z
L/2
x2 ρ(x) dx.
(136)
−L/2
Here, the integral I is known as the moment of inertia of the rod or the second moment of the
density function ρ(x) (with respect to x = 0). Different density functions, ρ(x), i.e., different mass
distributions will yield different moments of inertia.
287
Special case: In the special case that ρ(x) = ρ0 , a constant, the rod is homogeneous. In this
case, its total mass is M = ρ0 L. We first compute the moment of inertia:
I=
Z
L/2
x2 ρ0 dx =
L/2
L/2
L3
1
1
1
1
ρ0 x3 −L/2 = ρ0 (2)
=
ρ0 L3 = M L2 .
3
3
8
12
12
(137)
Substitution into Eq. (136) yields
K=
1
1
ρ0 ω 2 L 3 = M L 2 ω 2 .
24
24
You are probably familiar with this result from your first-year Physics course.
288
(138)
Lecture 36
Substitution rules for integrals
(Relevant section from Stewart: Section 5.5)
This material was closely based on Section 5.5 of the textbook of Stewart. As such, its presentation
here will be abbreviated. You are encouraged to read this section of the textbook and study the
examples carefully.
We’ll review the idea of the substitution rule with an example, and then summarize why the
method works.
Example 1: Find the antiderivative
Z
We’ll let u = 5x + 4 so that du =
√
5x + 4 dx.
1
du
dx = 5 dx. This means that dx = du. Making these
dx
5
replacements in the integral yields
Z
Z
Z
√
√ 1
1 √
5x + 4 dx =
u du =
u du.
5
5
(139)
Note that we have rewritten the x-integral entirely as a u-integral – there are no x-terms appearing on
the right. We may now consider u as the new independent variable and antidifferentiate with respect
to it, forgetting x:
1
5
√
Z
u du =
2 3/2
1 2 3/2
u +C =
u + C.
53
15
We would now like to return to our original problem, namely, finding the antiderivative of
(140)
√
5x + 4.
This means that we now replace u with 5x + 4, according to our initial change of variable:
2 3/2
2
u +C =
(5x + 4)3/2 + C.
15
15
(141)
In summary, we have
Z
√
5x + 4 dx =
2
(5x + 4)3/2 + C.
15
289
(142)
It never hurts to check the result:
d 2
3/2
(5x + 4) + C
=
dx 15
2 3
(5x + 4)1/2 (5)
15 2
(Chain Rule)
= (5x + 4)1/2 .
(143)
The result is correct. You can see that the checking procedure – involving the Chain Rule – is roughly
the reverse of the antidifferentiation process. As such, the method of substitution may be viewed as
a kind of “reverse” Chain Rule.
Example 2: Suppose that integrand of the above example were slightly different, i.e.,
Z p
5x2 + 4 dx.
(144)
du
dx = 10x dx. This implies that
In this case, we could let u = 5x2 + 4, implying that du =
dx
1
dx =
du. If we simply wrote
10x
Z p
5x2 + 4 dx =
Z
√
u
1
du,
10x
(145)
we would not be finished, because of the appearance of the x. The above integral is incorrect as it
stands. OK, we could use the definition of u to substitute for x:
u = 5x2 + 4
⇒
1 √
x = √ u − 4.
5
Substituting this result into the earlier integral yields
Z r
1
u
√
du.
u−4
10 5
(146)
(147)
At this time, we don’t have the tools to handle such a complicated integral. This tells us that the
method of substitution, as attempted above, might not be the best way to proceed. (Just in case you
are wondering: The method of trigonometric substitution is quite suitable for this integral. You’ll
learn this method in MATH 138.)
Example 3: On the other hand, if the integrand from the above example had an additional term,
i.e.,
Z
p
x 5x2 + 4 dx,
290
(148)
then our current method of substitution will work. This is because the factor “x” outside the square
root sign is, up to a constant, the derivative of the expression inside the square root sign. To see this,
let
u = 5x2 + 4
⇒
du =
du
dx = 10x dx
dx
⇒
x dx =
1
du.
10
(149)
In other words, the term x dx of the original integrand gets absorbed into the du. As a result,
Z p
Z
√
1
u du
x 5x2 + 4 dx =
10
1 2 3/2
u +C
=
10 3
1
=
(5x2 + 4)3/2 + C.
(150)
15
Let’s once again check this result (we can ignore the constant C),
1 3
d 1
(5x2 + 4)3/2 =
(5x2 + 4)1/2 (10x)
dx 15
15 2
= x(5x2 + 4)1/2 .
(151)
The result is therefore correct.
The mathematical basis of the Method of Substitution is the following: Our integrals have the form
(up to a constant),
Z
f (g(x)) g′ (x) dx
(152)
If we make the change of variable,
u = g(x)
⇒
du =
du
dx = g′ (x) dx,
dx
and then substitute these results into our original integral, we obtain
Z
Z
f (g(x)) g′ (x) dx = f (u) du.
(153)
(154)
We now simply proceed by finding the antiderivative of f . Suppose that F is an antiderivative of f ,
i.e.,
F ′ (u) = f (u).
(155)
Then
Z
f (u) du = F (u) + C.
291
(156)
Substitution of this result into (154) yields
Z
f (g(x) g′ (x) dx = F (u) + C
= F (g(x)) + C,
(157)
where we have resubstituted u = g(x), to obtain a final result in terms of x. Let’s check this result
(once again, ignoring the C because it vanishes with differentiation). Using the Chain Rule,
d
F (g(x)) = F ′ (g(x))g′ (x)
dx
= f (g(x))g′ (x)
(since F ′ (x) = f (x)).
(158)
Therefore, we have found the required antiderivative.
Example 4: Here is an interesting example. Find, if possible, the antiderivative of tan x, i.e.,
Z
tan x dx.
(159)
This problem looks formidable. Where is the g(x)? And where is the g′ (x)? Let’s start by expressing
tan x in terms of sin x and cos x, i.e.,
Z
sin x
dx.
cos x
(160)
Now the derivative of sin x is cos x and the derivative of cos x is − sin x. So which is our g(x) and which
is our g′ (x). A look at Eq. (152) shows that g′ (x) should be “upstairs” and not in the denominator.
So let’s choose u = cos x, which implies that
du = − sin x dx
⇒
sin x dx = −du.
(161)
We substitute these results into our original integral and proceed, i.e.,
Z
sin x
dx =
cos x
1
(−1)du
u
Z
1
= −
du
u
= − ln |u| + C
Z
(162)
292
Now resubstitute for x to obtain the final result,
sin x
dx = − ln |u| + C
cos x
= − ln | cos x| + C
Z
= ln | cos x|−1 + C
= ln | sec x| + C.
(163)
Is this result correct? Let’s check it by differentiation:
d
ln | sec x| =
dx
1
d
·
sec x
sec x dx
1
=
· sec x tan x
sec x
= tan x.
(164)
It is correct.
Example 5: We know that
Z
1
dx = arctan(x) + C,
1 + x2
(165)
but what about the indefinite integral,
Z
1
dx,
a2 + x2
(166)
where a is a constant?
The first step is to try to convert the integral in (166) into the “standard form” of (165), first by
dividing out the a2 :
Z
1
1
dx = 2
a2 + x2
a
Z
1
1+
x2
a2
dx.
(167)
Comparing this result with (165), we now set
u=
x
a
⇒
du =
1
dx
a
⇒
dx = a du.
(168)
The integral on the RHS of (167) becomes
1
a2
Z
1
a du =
1 + u2
=
=
293
Z
1
1
du
a
1 + u2
1
arctan(u) + C
a
x
1
+ C.
arctan
a
a
(169)
In summary, we have found that
Z
x
1
1
+ C.
dx
=
arctan
a2 + x2
a
a
(170)
This is the more general form of antiderivative that you will find in standard tables.
As always, it doesn’t hurt to check this result:
x d 1
=
arctan
dx a
a
=
1
1
1
2
x
a 1+
a
a
1
1
2
a 1+ x 2
a
=
1
.
1 + x2
(171)
The result is therefore correct.
You are advised to read Section 5.5 of Stewart’s text, “The Substitution Rule,” for more discussion
on this topic along with a good number of illustrative examples.
Method of Substitution and Definite Integrals
Let’s now return to our integrand of Example 1, but now appearing in the following definite integral,
Z 10
√
5x + 4 dx.
(172)
5
√
2
(5x + 4)3/2 + C.
5x + 4 to be
15
Here, we omit all of the work that went into finding this antiderivative in terms of x and simply use
In Example 1, we found the general antiderivative of the integrand
the result. From the FTC II, we may evaluate the above definite integral as follows,
10
Z 10
√
2
3/2
(5x + 4)
5x + 4 dx =
15
5
0
i
2 h 3/2
3/2
.
54 − 39
=
15
(173)
We now propose a slightly alternate method of evaluating this definite integral that can save some
work – instead of returning from “u-space” to “x-space” and working with the x values, as was done
above, we simply remain in “u-space”, making all evaluations in terms of u. This is done as follows.
294
The first step, as before, is to make the change of variable,
u = 5x + 4
⇒
du
dx = 5 du.
dx
du =
(174)
We know what happens to the indefinite integral – see Example 1. The question is what happens to
the definite integral:
Z
10 √
5
1
5x + 4 dx =
5
Z
??
u1/2 du.
(175)
??
The answer is that if we change variables, i.e., from x to u, and we change the integral to a u-integral,
and decide that we wish to work with the u-integral, then we must change the limits of integration
as well, consistent with our change of variable. Our change of variable u = 5x + 4 implies that
x=5
implies
u = 29
x = 10
implies
u = 54.
(176)
We use these values of u as limits of integration in the u-definite integral and proceed as follows,
Z 10
Z
√
1 54 1/2
5x + 4 dx =
u du
5 29
5
1 2 h 3/2 i54
=
u
53
29
i
2 h 3/2
(177)
54 − 293/2 ,
=
15
which agrees with the result obtained earlier.
Once again, the point is that once we make the change of variable u = g(x) to produce an integral
in u, we may stay there and evaluate the definite integral in terms of u. But one must change the limits
of integration, in accordance with the change of variable u = g(x). The procedure can be summarized
as follows,
Z
b
′
f (g(x)) g (x) dx =
Z
g(b)
f (u) du.
(178)
g(a)
a
An illustrative example that is most relevant to Physics
We conclude this section – and this course! – with a very nice example of the change of variables
method that yields a fundamental result from Physics. Most of you are familiar with this result from
your first-year Physics course, but it may not have been proved mathematically:
295
Suppose that a force F(x) = f (x)i acts on a mass m moving along a straight line (i.e., the
x-axis) according to Newton’s Second Law, i.e., F = ma. Show that the work W done by
the force on the mass during the time that the mass moves from x = x1 to x = x2 is the
change in kinetic energy of the mass between these two points, i.e.,
W = ∆K = K(x1 ) − K(x2 ).
(179)
Before proceeding with the proof of this result, we emphasize that the force F is not assumed to be
conservative. This result will hold for nonconservative forces, e.g., friction, as well. (In such cases, f
could also be a function of the velocity v of the mass. But we’ll simply keep the notation to f (x) –
the most important point is that f (x) is defined at each point x of the trajectory of the mass.)
Proof: The work W done by the force is, by definition,
W =
Z
x2
f (x) dx.
(180)
x1
By assumption, Newton’s Second Law, f = ma is obeyed, so we replace the integrand as follows,
W =
Z
x2
ma dx.
(181)
x1
In this formulation, a is considered as a function of the position x. But let us now change variables,
so that the integration variable is time t. In this case we let
x = x(t)
⇒
dx =
dx
dt = v dt.
dt
(182)
This means that the limits of integration must be changed to times – we’ll call t1 the time that mass
m is at x1 and t2 the time that it is at x2 . The above integral now becomes
W =m
Z
t2
av dt
(183)
t1
But the integrand is, up to a constant, the derivative of v(t)2 . Recall, from our earlier discussion of
conservation of energy, that
d
dv
v(t)2 = 2v(t)
= 2v(t)a(t).
dt
dt
(184)
Therefore
a(t)v(t) =
1 d
v(t)2 ,
2 dt
296
(185)
so the above integral becomes
1
W = m
2
Z
t2
t1
d v(t)2 dt.
dt
(186)
The integrand is a derivative with respect to t. Since the antiderivative of the derivative of a function
is simply the function itself, we have, by the FTC II,
W
t
1 m v(t)2 t21
2
1
1
=
mv(t2 )2 − mv(t1 )2
2
2
1
1
mv 2 − mv 2
=
2 2 2 1
= ∆K.
=
(187)
Comment: In the special case that F is conservative, the above result can be proved quite easily
as follows. Recall, from the definition of the potential function U (x) associated with f (x), that the
work done by the force is given by∗
W = U (x1 ) − U (x2 ) = −∆U.
(188)
But conservation of total mechanical energy implies that
K(x1 ) + U (x1 ) = K(x2 ) + U (x2 ).
(189)
W = U (x1 ) − U (x2 ) = K(x2 ) − K(x1 ) = ∆K.
(190)
A simple rearrangement yields
In the case that F is not conservative, a potential function U (x) does not exist, and the previous proof
must be used.
∗ Just
in case you don’t recall this important result, we may compute the work done by the
conservative force as follows,
Z x2
f (x) dx
W =
x1
Z x2
U ′ (x) dx
= −
x1
= −[U (x2 ) − U (x1 )]
by FTC II since U (x) is an antiderivative of U ′ (x)
= −∆U.
(191)
297
A final note: In Physics books, you may see the following derivation of the earlier result W = ∆K,
starting at Eq. (183), which we’ll write as
W =m
Z
t2
a(t)v(t) dt.
(192)
t1
Note that
v ′ (t) = a(t)
⇒
dv
= a(t)
dt
⇒ dv = a(t)dt.
(193)
Then the integral in (192) may be rewritten as follows,
W
= m
= m
Z
t2
Zt1v2
v(t)a(t) dt
v dv,
(194)
v1
where we have made a change of variable so that the integration is performed velocity variable. The
above integral easily becomes
W
1 2 v2
= m· v 2 v1
1
1
=
mv(x2 )2 − mv(x1 )2
2
2
= K(x2 ) − K(x1 ),
in agreement with our earlier result.
298
(195)