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∫ b

aF ′(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

dt= x′(t). (2)

From the Net Change property, which is simply a consequence of FTC II, we have that

∫ b

av(t) dt =

∫ b

ax′(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

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means that we’ll have to change the integration variable, t, to something else, say, s. The result is

∫ t

0v(s) ds =

∫ t

0x′(s) ds = x(t) − x(0). (4)

The above equation may be rewritten as follows,

x(t) = x(0) +

∫ t

0v(s) ds. (5)

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,

d

dt

∫ t

0v(s) ds = v(t) = x′(t). (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)

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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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

dt= v′(t) = −g. (11)

As such, we look for a v(t) of which the t-derivative is −g. In other words, we look for the antideriva-

tives 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) =

∫

−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:∫ t

0v′(s) ds =

∫ t

0(−g) ds. (17)

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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

∫ t

0v′(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∫ t

0(−g) ds = −gs

∣

∣

∣

t

0= −gt. (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)

We’ll first write Eq. (20) as follows,

x′(t) = v0 − gt, (22)

and then perform a definite integration from s = 0 to s = t.

∫ t

0x′(s) ds =

∫ t

0(v0 − gs) ds. (23)

The integrals on both sides may be evaluated using FTC II, i.e.,

x(s)∣

∣

∣

t

0= v0s −

1

2gs2

∣

∣

∣

∣

t

0

, (24)

which becomes

x(t) − x0 = v0t −1

2gt2. (25)

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This leads to the well-known result,

x(t) = x0 + v0t −1

2gt2. (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 =

∫ t

0a(s) ds ⇒ v(t) = v0 +

∫ t

0a(s) ds. (27)

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 =

∫ t

0v(s) ds ⇒ x(t) = x0 +

∫ t

0v(s) ds. (28)

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

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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 =

∫ b

af(x) dx. (31)

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∑

k=1

∆Wk∼=

n∑

k=1

f(x∗

k)∆x. (35)

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

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piecewise continuous), we have

W = limn→∞

n∑

k=1

f(x∗

k)∆x

=

∫ b

af(x) dx. (36)

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 MLT−2. (Think of F = ma and the dimensions of mass and acceleration.

Therefore the dimensionality of work is force times distance, or ML2T−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.

xy

z

P

Q

m F(r(t))r(t)

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

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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∫

CF · 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) or f(x) = −U ′(x). (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.,

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

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 ⇒ U(x) = mgx. (42)

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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

af(s) ds. (43)

There are two noteworthy points regarding this definition:

1. From FTC I, we have that

U ′(x) = − d

dx

∫ x

af(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) = −∫ a

af(s) ds = 0. (45)

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∫ x

af(s) ds (46)

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

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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

U(x) = −∫ x

0(−mg) ds =

∫ x

0mg ds = mg

∫ x

0ds = mgx, (47)

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) = − GMm

(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

U(x) = −∫ x

0f(s) ds

= GMm

∫ x

0

1

(R + s)2ds

= GMm

[

− 1

R + s

]x

0

(by FTC II)

= GMm

[

− 1

R + x+

1

R

]

=GMm

R− GMm

R + x. (49)

Note that U(0) = 0, as it should be.

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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

E(0) =1

2mv2

0 + U(0) =1

2+

GMm

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.,

E(x(t)) =1

2mv(t)2 +

GMm

R− GMm

R + x(t)= E(0) =

1

2mv2

0. (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

GMm

R− GMm

R + h=

1

2mv2

0 . (52)

After a little algebra, we may solve for h,

h =v20R

2GMR − v2

0

. (53)

As the initial speed v0 approaches the value

√

2GM

R, the height of the trajectory h → ∞. Therefore,

if the projectile were (at least theoretically) launched with initial velocity

vesc =

√

2GM

R, (54)

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

U(x) = −∫ x

0(−ks) ds = k

∫ x

0s ds =

1

2kx2, (56)

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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

af(s) ds, (57)

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 =

∫ x2

x1

f(s) ds. (58)

But from the additive property of definite integrals, we can write the above as follows,

W =

∫ x2

x1

f(s) ds

=

∫ 0

x1

f(s) ds +

∫ x2

0f(s) ds

= −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. (61)

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The work done by the gravitational force f(x) = −mg is

W =

∫ x2

x1

f(s) ds

=

∫ x2

x1

(−mg) ds

= −mg(x2 − x1)

= −∆U. (62)

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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]. Further-

more, 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.

So, as before, let n > 0 and define ∆x =b − a

n. Then define the partition points,

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

f̄n =1

n

n∑

k=1

f(x∗

k). (64)

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,

f̄n =1

n

1

∆x

n∑

k=1

f(x∗

k)∆x

=1

n∆xSn. (65)

Here, Sn is a Riemann sum corresponding to the definite integral

∫ b

af(x) dx. There remains the

question about what to do about the factor1

n∆x. Recalling the definition of ∆x:

∆x =b − a

n⇒ n∆x = b − a. (66)

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Therefore, the average value in (65) becomes

f̄n =1

b − aSn. (67)

Assuming that f is continuous (or at least piecewise continuous), the limit of the Riemann sums Sn

exists, and we have

limn→∞

f̄n =1

b − alim

n→∞

Sn =1

b − a

∫ b

af(x) dx. (68)

This is the average value of f over the interval [a, b], which we shall denote as follows,

f̄[a,b] =1

b − a

∫ b

af(x) dx. (69)

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,

∫ b

af(x) dx = f̄[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 f̄[a,b]. The rectangular region is shaded.

a bx

y = f(x)

f̄[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.

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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

f̄[0,1] =1

1

∫ 1

01 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

f̄[0,1] =1

1

∫ 1

0x dx =

[

x2

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

f̄[0,1] =1

1

∫ 1

0x2 dx =

[

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

f̄[0,1] =1

1

∫ 1

0xn dx =

[

xn+1

n + 1

]1

0

=1

n + 1. (74)

Note that as n → ∞, the average value of the function xn behaves as follows,1

n + 1→ 0. Does

this make sense? For any x such that 0 ≤ x < 1, raising it to higher powers makes it smaller,

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

∆x=

m(x2) − m(x1)

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) = limh→0

m(x + h) − m(x)

h= ρ(x). (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), (78)

275

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) =

∫ x

0ρ(s) ds. (79)

Just to check this:

m(0) =

∫ 0

0ρ(s) ds = 0, (80)

and

m′(x) =d

dx

∫ x

0ρ(s) ds (FTC II). (81)

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) =

∫ L

0ρ(s) ds =

∫ L

0ρ(x) dx. (82)

(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.,

M = L · 1

L

∫ L

0ρ(x) dx. (83)

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). (85)

276

(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] =

∫ d

0ρ(s) ds −

∫ c

0ρ(s) ds

=

∫ d

0ρ(s) ds +

∫ 0

cρ(s) ds

=

∫ 0

cρ(s) ds +

∫ d

0ρ(s) ds

=

∫ d

cρ(s) ds. (86)

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) =

∫ x

0ρ0 ds = ρ0

∫ x

0ds = ρ0x, 0 ≤ x ≤ L. (87)

The graph of m(x) vs. x is a straight line that runs from (0, 0) to (L, ρ0L). The total mass of the

wire is M = ρ0L.

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

m1d1 = m2d2. (88)

277

m2

d1 d2

m1

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?

x̄

m1 m2

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 =m1x1 + m2x2

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∑

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

∑

k=1

mkxk − x̄n

∑

k=1

mk = 0. (93)

The second sum on the LHS is the total mass M of the system. Therefore, the center of mass is given

by

x̄ =1

M

n∑

k=1

mkxk. (94)

278

Once again, you are most probably familiar with this equation.

Before going on, let’s rewrite Eq. (94) as follows,

x̄ =

n∑

k=1

(mk

M

)

xk, (95)

or

x̄ =n

∑

k=1

pkxk, (96)

where

pk =mk

M⇒

n∑

k=1

pk = 1. (97)

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 → ∞.

So, as before, for an n > 0, let ∆x =L

n(our interval [a, b] is now [0, L]) and define the partition

points

xk = k∆x =k

nL, 0 ≤ k ≤ n. (98)

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

∑

k=1

∆mk∼=

n∑

k=1

ρ(x∗

k)∆x (100)

279

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

0ρ(x) dx, (101)

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∑

k=1

∆mkx∗

k =

n∑

k=1

x∗

kρ(x∗

k)∆x. (102)

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∫ L

0xρ(x) dx. (103)

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

x̄ =1

M

∫ L

0xρ(x) dx, where M =

∫ L

0ρ(x) dx. (104)

We’ll examine some examples in the next lecture.

280

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 =

∫ 1

0ρ(x) dx =

∫ 1

0dx = 1. (105)

The first moment of the wire with respect to the origin is given by

Mx =

∫ 1

0xρ(x) dx =

∫ 1

0x dx =

[

1

2x2

]1

0

=1

2. (106)

The center of mass of this wire is therefore

x̄ =Mx

M=

1/2

1= 1, (107)

as expected.

2. Example 2: We now consider a perturbation of the above mass distribution, the density function

ρ(x) = 1 +1

2x. (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

M =

∫ 1

0ρ(x) dx =

∫ 1

0

(

1 +1

2x

)

dx =

[

x +1

4x2

]1

0

= 1 +1

4=

5

4. (109)

281

The first moment of the wire with respect to the origin is

Mx =

∫ 1

0xρ(x) dx =

∫ 1

0

(

x +1

2x2

)

dx =

[

1

2x2 +

1

6x3

]1

0

=1

2+

1

6=

2

3. (110)

The center of mass of the wire is therefore

x̄ =Mx

M=

2/3

5/4=

8

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

Consider a wire over the interval

[

−L

2,L

2

]

with mass density function ρ(x). Now let a point mass

m be situated outside the wire, at position coordinate a >L

2, as sketched below. (By symmetry,

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.

x

L/2−L/2

m

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 =Gm1m2

d2. (112)

Once again, in the “Spirit of Calculus,” we divide up the interval

[

−L

2,L

2

]

into n tiny subinter-

vals Ik = [xk−1, xk] of width ∆x =L

n, and approximate the mass of wire ∆mk in each subinterval

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

∆Fk∼= Gm∆mk

(a − x∗

k)2≈ Gmρ(x∗

k)∆x

(a − x∗

k)2

. (113)

We now add up the magnitudes of the forces from all n subintervals,

F =

n∑

k=1

∆Fk∼= Gm

∑

k=1

ρ(x)

(a − x∗

k)2∆x. (114)

In the limit n → ∞, the Riemann sum on the right converges to the integral,

F = Gm

∫ L/2

−L/2

ρ(x)

(a − x)2dx. (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.

d = a − x

L/2−L/2 a

m

0 x

dx

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

dx= ρ(x) ⇒ dm =

dm

dxdx = ρ(x) 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 =Gm dm

(a − x)2=

Gmρ(x) dx

(a − x)2= Gm

ρ(x)

(a − x)2dx. (118)

283

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

∫ L/2

−L/2

ρ(x)

(a − x)2dx, (119)

the result obtained earlier.

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

M =

∫ L/2

−L/2ρ(x) dx =

∫ L/2

−L/2ρ0 dx = ρ0L. (120)

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

F = Gmρ0

∫ L/2

−L/2

1

(a − x)2dx. (121)

The integral is not difficult to compute, since the antiderivative of the integrand is relatively

straightforward,∫ L/2

−L/2

1

(a − x)2dx =

[

1

a − x

]L/2

−L/2

=1

a − L/2− 1

a + L/2

=L

a2 − L2/4. (122)

Therefore,

F =Gmρ0L

a2 − L2/4. (123)

Recalling that the total mass of the wire is M = ρ0L, we have the final result,

F =GMm

a2 − L2/4. (124)

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 =GMm

a2, (125)

284

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 =GMm

a2

1

1 −(

L2a

)2 . (126)

For any given L > 0, we see that the ratioL

2amust be sufficiently small.

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 =q1q2

4πǫ0d2. (127)

Here, ǫ0 denotes the permittivity of the vacuum.

285

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

F =q

4πǫ0

∫ L/2

−L/2

ρ(x)

(a − x)2dx. (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 =qρ0

4πǫ0· 1

a2 − 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.

We consider the same thin wire of the previous example - it is positioned over the interval

[

−L

2,L

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

2(dm) v2, (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. (131)

As for the velocity, it is given by v = ω|x|, implying that v2 = ω x2.

286

L/2−L/2 0

dm = ρ(x) dx

ω

x

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

2πcycles per unit time. (132)

This means that it takes1

ν=

2π

ωunits of time to complete one cycle. The circumference of a circular

orbit of radius r is C = 2πr units of length. This is one cycle of revoluton. Therefore the speed of a

particle on this circular orbit is

v =distance

time=

2πr

2π/ω= ωr units of length per unit time. (133)

In the above discussion, r = |x|.

Putting all of this together, the kinetic energy dK of the mass element is

dK =1

2ρ(x) dx (ω|x|)2 =

1

2ω2 ρ(x)x2 dx. (134)

We now integrate over the rod to obtain the total kinetric energy,

K =

∫ L/2

−L/2dK =

1

2ω2

∫ L/2

−L/2x2 ρ(x) dx. (135)

Note that the above result may be written in the form,

K =1

2I ω2, where I =

∫ L/2

−L/2x2 ρ(x) dx. (136)

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 = ρ0L. We first compute the moment of inertia:

I =

∫ L/2

L/2x2ρ0 dx =

1

3ρ0 x3

∣

∣

L/2

−L/2=

1

3ρ0 (2)

L3

8=

1

12ρ0L

3 =1

12ML2. (137)

Substitution into Eq. (136) yields

K =1

24ρ0ω

2L3 =1

24ML2ω2. (138)

You are probably familiar with this result from your first-year Physics course.

288

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

∫ √5x + 4 dx.

We’ll let u = 5x + 4 so that du =du

dxdx = 5 dx. This means that dx =

1

5du. Making these

replacements in the integral yields

∫ √5x + 4 dx =

∫ √u

1

5du =

1

5

∫ √u du. (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

∫ √u du =

1

5

2

3u3/2 + C =

2

15u3/2 + C. (140)

We would now like to return to our original problem, namely, finding the antiderivative of√

5x + 4.

This means that we now replace u with 5x + 4, according to our initial change of variable:

2

15u3/2 + C =

2

15(5x + 4)3/2 + C. (141)

In summary, we have∫ √

5x + 4 dx =2

15(5x + 4)3/2 + C. (142)

289

It never hurts to check the result:

d

dx

[

2

15(5x + 4)3/2 + C

]

=2

15

3

2(5x + 4)1/2 (5) (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.,

∫

√

5x2 + 4 dx. (144)

In this case, we could let u = 5x2 + 4, implying that du =du

dxdx = 10x dx. This implies that

dx =1

10xdu. If we simply wrote

∫

√

5x2 + 4 dx =

∫ √u

1

10xdu, (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 ⇒ x =1√5

√u − 4. (146)

Substituting this result into the earlier integral yields

1

10√

5

∫√

u

u − 4du. (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.,∫

x√

5x2 + 4 dx, (148)

290

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

dxdx = 10x dx ⇒ x dx =

1

10du. (149)

In other words, the term x dx of the original integrand gets absorbed into the du. As a result,

∫

x√

5x2 + 4 dx =1

10

∫ √u du

=1

10

2

3u3/2 + C

=1

15(5x2 + 4)3/2 + C. (150)

Let’s once again check this result (we can ignore the constant C),

d

dx

1

15(5x2 + 4)3/2 =

1

15

3

2(5x2 + 4)1/2 (10x)

= 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),∫

f(g(x)) g′(x) dx (152)

If we make the change of variable,

u = g(x) ⇒ du =du

dxdx = g′(x) dx, (153)

and then substitute these results into our original integral, we obtain

∫

f(g(x)) g′(x) dx =

∫

f(u) du. (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∫

f(u) du = F (u) + C. (156)

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Substitution of this result into (154) yields

∫

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

dxF (g(x)) = F ′(g(x))g′(x)

= 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.,

∫

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.,∫

sinx

cos xdx. (160)

Now the derivative of sin x is cos x and the derivative of cos x is − sinx. 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 = − sinx dx ⇒ sin x dx = −du. (161)

We substitute these results into our original integral and proceed, i.e.,

∫

sinx

cos xdx =

∫

1

u(−1)du

= −∫

1

udu

= − ln |u| + C

(162)

292

Now resubstitute for x to obtain the final result,

∫

sinx

cos xdx = − ln |u| + C

= − ln | cos x| + C

= ln | cos x|−1 + C

= ln | sec x| + C. (163)

Is this result correct? Let’s check it by differentiation:

d

dxln | sec x| =

1

sec x· d

dxsec x

=1

sec x· sec x tan x

= tan x. (164)

It is correct.

Example 5: We know that∫

1

1 + x2dx = arctan(x) + C, (165)

but what about the indefinite integral,∫

1

a2 + x2dx, (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:∫

1

a2 + x2dx =

1

a2

∫

1

1 + x2

a2

dx. (167)

Comparing this result with (165), we now set

u =x

a⇒ du =

1

adx ⇒ dx = a du. (168)

The integral on the RHS of (167) becomes

1

a2

∫

1

1 + u2a du =

1

a

∫

1

1 + u2du

=1

aarctan(u) + C

=1

aarctan

(x

a

)

+ C. (169)

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In summary, we have found that

∫

1

a2 + x2dx =

1

aarctan

(x

a

)

+ C. (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:

d

dx

[

1

aarctan

(x

a

)

]

=1

a

1

1 +(

xa

)2

1

a

=1

a2

1

1 +(

xa

)2

=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,

∫ 10

5

√5x + 4 dx. (172)

In Example 1, we found the general antiderivative of the integrand√

5x + 4 to be2

15(5x + 4)3/2 + C.

Here, we omit all of the work that went into finding this antiderivative in terms of x and simply use

the result. From the FTC II, we may evaluate the above definite integral as follows,

∫ 10

5

√5x + 4 dx =

[

2

15(5x + 4)3/2

]10

0

=2

15

[

543/2 − 393/2]

. (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.

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The first step, as before, is to make the change of variable,

u = 5x + 4 ⇒ du =du

dxdx = 5 du. (174)

We know what happens to the indefinite integral – see Example 1. The question is what happens to

the definite integral:∫ 10

5

√5x + 4 dx =

1

5

∫ ??

??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,

∫ 10

5

√5x + 4 dx =

1

5

∫ 54

29u1/2 du

=1

5

2

3

[

u3/2]54

29

=2

15

[

543/2 − 293/2]

, (177)

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,∫ b

af(g(x)) g′(x) dx =

∫ g(b)

g(a)f(u) du. (178)

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:

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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 =

∫ x2

x1

f(x) dx. (180)

By assumption, Newton’s Second Law, f = ma is obeyed, so we replace the integrand as follows,

W =

∫ x2

x1

madx. (181)

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

dtdt = v 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

∫ t2

t1

av dt (183)

But the integrand is, up to a constant, the derivative of v(t)2. Recall, from our earlier discussion of

conservation of energy, thatd

dtv(t)2 = 2v(t)

dv

dt= 2v(t)a(t). (184)

Therefore

a(t)v(t) =1

2

d

dtv(t)2, (185)

296

so the above integral becomes

W =1

2m

∫ t2

t1

d

dt

[

v(t)2]

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 =1

2m

[

v(t)2]t2t1

=1

2mv(t2)

2 − 1

2mv(t1)

2

=1

2mv2

2 − 1

2mv2

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)

A simple rearrangement yields

W = U(x1) − U(x2) = K(x2) − K(x1) = ∆K. (190)

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,

W =

∫ x2

x1

f(x) dx

= −∫ x2

x1

U ′(x) dx

= −[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

∫ t2

t1

a(t)v(t) dt. (192)

Note that

v′(t) = a(t) ⇒ dv

dt= a(t) ⇒ dv = a(t)dt. (193)

Then the integral in (192) may be rewritten as follows,

W = m

∫ t2

t1

v(t)a(t) dt

= m

∫ v2

v1

v dv, (194)

where we have made a change of variable so that the integration is performed velocity variable. The

above integral easily becomes

W = m · 1

2v2

∣

∣

∣

∣

v2

v1

=1

2mv(x2)

2 − 1

2mv(x1)

2

= K(x2) − K(x1), (195)

in agreement with our earlier result.

298