Integral Calculus: Two Major Approaches & Antiderivatives
BY
SUWARDI
2.1 INTRODUCTION
•
TWO MAJOR APPROACHES
–
ANTIDERIVATIVES – to mathematically generate
integrals
–
INTEGRATION – to assign a physical meaning to the
integral
•
The other approach:
–
To consider the integral as the sum of many similar,
infinitesimal elements
•
The process of taking integration, as the inverse
of differentiation
•
In this chapter we shall consider the reverse
process.
Knowing the effect of individual
changes, we wish to determine the overall
effect of adding together these changes such
that sum equals a finite change
•
Before considering the physical significance
and the application of integral calculus, let us
briefly
review the general and special methods
of integration
2.2 INTEGRAL AS
ANTIDERIVATIVE
•
Function y = f(x), differentiation of this
function, symbolized by the equation
Where
f’
(
x
) denotes the first derivative of
the function
f
(
x
) with respect to
x
•
In this chapter we shall pose the following
question:
What function f(x), when
differentiated, yields the function f’(x)?
•
For example,
what function f(x) when
differentiated, yields the function f’(x) = 2x?
•
Substituting f’(x) = 2x into equation gives
•
This function,
f(x)
, for which we are looking is
called the
integral
of the differential and
symbolized by the equation
:
the integral sign
; f’(x):
the integrand
•
If f’(x) = 2x, then f(x) = x
2
. f(x) = x
2
is not the
complete solution but f(x) = x
2
+ C is the
complete solution.
•
C is the constant of integration, and this
constant
always
is included as part of the
answer to any integration. Thus,
2.3 GENERAL METHODS OF INTEGRATION
•
Let us consider several general methods of
integration
Examples:
2.4 SPECIAL METHODS OF INTEGRATION
•
Many of the functions encountered in physical
chemistry are not in one of the general forms
given above
•
Special methods of integration:
–
Algebraic Substitution
–
Trigonometric Transformation
–
Partial Fractions
Algebraic Substitution
•
The mathematical functions can be
transformed into one of the general forms in
section 2.3 or into one of the forms found in
table of integral by some form of algebraic
substitution
•
Examples
•
Let us attempt to transform this integral into
the form
•
Let u = (1 – x
2
), Then du = -2x dx. Hence,
Let u = V – nb. Then du = dv. Hence,
Let u = sin x. Then du = cos x dx. Hence
Trigonometric Transformation
•
Many trigonometric integrals can be
transformed into a proper form for integration
by making some form of trigonometric
transformation using
trigonometric identities
.
•
For example, to evaluate the integral
we must make use of the identity
•
Thus,
•
Again, integration of integral of this type are
more parctically done by using the Table of
Integrals
•
Example:
Partial Fractions
•
Consider an integral of the type
•
This type of integral can be trasformed into simpler integral
by the following method. Let A = (a – x) and B = (b –x). Then
•
Therefore,
2.5 The Integral as a summation of
infinitesimally
•
In the previous sections we have considered
integration as the purely mathematical
operation of finding antiderivatives
.
•
Let us now turn to the more physical aspects
of integration in order to understand the
physical importance of the integral
•
Consider, as an example:
1.
The expansion of an ideal gas from a volume of
V
1
to a volume V
2
against a constant external
pressure
P
ext
,
•
The diagram shows what the external
pressure is doing on one axis and what the
volum is doing on the other axis
•
Both can vary independently
•
The work done by the gas does depend on
both the external prssure and the volume
change
•
For expansion, the work is
w
= - P
ext
(V
2
– V
1
)
•
W is just the negative of the area under the
P
ext
versus V diagram
2.
The
external prssure changes
, in some
fashion, as the volume changes
–
The external pressure is not changing as a
function of volume
–
To emphasize this point, let us assume that the
external prsessure actually goes up as the
volumes changes (see indicator diagram )
–
A gas could be expanding against atmospheric
pressure which perhaps is increasing over the
period of time that the expansion take place
Indicator diagram showing PV work done by
a gas expanding against a variable external
pressure
•
The approximate area under the curve, then,
is just
the sum of the four rectangles
•
Hence, we can write
•
Where the symbol is read “the integral
from V
1
to V
2
”, and V
1
and V
2
are called the
limits of integration
.
Hence,
2.6 Line Integral
•
Integral of the general type
Are called
line integral
, because such integral
represent the area under the specific curve
(path) connecting x
1
to x
2
•
Such integral can be evaluated analytically
(i.e., by finding the antiderivative) only if an
equation, the path, y = f(x) is known, since
under these circumstances the integral
contains only one variable
•
If y is not a function of x (as in the case of
P
ext
versus V), or
•
if y is a function of x, but the equation
relating y and x is not known or can not be
integrated (as in the case ), or
•
if y is a function of more than one
variable that changes with x,
then the line integral
cannot
be
evaluated analytically, and one must
resort to
a graphical or numeric
method of integration
in order
to
evaluate the integral
.
1.
The integral in below can be evaluated
analytically if we assume that
P
ext
is a constant
,
hence, it can be brought out of the integral
work = -
A
= -
P
ext
(V
2
- V
1
)
2.
A second way to evaluate the above integral
is found in the
concept of reversibility
. If the
expansion of the gas is reversible, then for all
practical purposes the external is equal to
the gas pressure,
P
ext
=
P
gas.
This gives
It can be integrated if P is a function of V only
so T must be constant (isothermal condition)
We can write
Example
•
The change in enthalpy as a function of
temperature is given by the equation
Solution
•
We first recognize that the above integral is
a
line integral
. This integral cannot be evaluated
unless C
P
as a function of T is known
•
We could assume that
C
P
is a constant
and
evaluate the integral that way, but over such a
large temperature range the approximation
would be poor
•
Another approach would be to determine the
integral numerically
•
One analytical approach is to
expand C
P
as a
power series in temperature
C
P
= a + bT + cT
2
The constant a, b, and c are known for many
common gases. Substituting this into the
H
equation gives
2.7 Double and Triple Integrals
•
Functions could be differentited more than
once
•
How about the determination of multiple
integrals
Example
,
•
The volume of cylinder
is a function of both
the radius and the height of the cylinder. That
is, V =
f
(r, h)
•
Let us suppose that we allow the height of the
cylinder, h, to change while holding the radius,
r
, constant
•
The integral from h = 0 to h = h, then, could be
expressed as
•
But the value of this line integral depends on
the value of the radius,
r
, and hence the
integral could be considered to be a function
of
r
.
•
To evaluate the above double integral, we
integrate first while holding
r
a
constant, which gives us g(r).
•
Then we integrate next while holding
h
constant. Such a process is known as
successive partial integration
•
For example, let us evaluate
•
The above argument can be extended to the
triple integral. For example, let us evaluate the
triple
•
Problem. The differential volume element in
spherical polar coordinates is d
V
=
r
2
sin
d
d
dr. Given that
goes from 0 to 2
, and r goes
from 0 to
r
, evaluate the triple integral
Solution
PROBLEMS
1.
Evaluate the following integral (consider all
uppercase letters to be constant):
2.
Evaluate the following integrals using the
Table of Integral (consider all uppercase
letters to be constant):
3.
Evaluate the following definite integrals using
the Table of indefinite and definite interals,
as needed
4.
Consider the ideal gas law equation P = nRT/V,
where in the case n, R, and T are assumed to be
constant. Prepare a graph of P versus V,
choosing suitable coordinates, for n = 1 mole, R
= 0,0821 liter atm/mol K and T = 298 K from a
volume of V = 1.00 liters to a volume of V = 10.0
liters. Consider now the area under the P versus
V curve from V = 2.00 liters to V = 6.00 liters.
Determine the approximate area graphically by
breaking up the area into four rectangles of
equal width
V; compare your answer to that
found by analytically integrating the function
between th
ese limits of integration
5.
Evaluate the following multiple integrals
using the Table of Integrals, as needed
6.
The equation of a straight line passing
through the origin of a cartesian coordinate
system is y = mx, where m is the slope of the
line. Show that the area of a triangle made
up of this line and the x axis between x = 0
and x = a is A = ½ ay.
7.
The Kirchoff equation for a chemical reaction
relating the variation of
H of a reaction with
absolute temperature is where
C
P
as a power series in T,
∆C
P
= a + bT + cT
2
.
Derive an equation for
H as a function of
temperature. (
Hint
: Write the above
derivative in differential form)
8.
The Gibbs-Helmholtz equation for a chemical
reaction is
9.
Find the probability of finding a particle
confined to e field-free one-dimensional box
in the state n = 1 at x = L/2 in a range L/2 ±
0.05 L, where L is the width of box, given
10.
Find the probability of finding an electron in
the 1s-state of the hydrogen atom at r = a
0
in
a range a
0
± 0.005a
0
, where a
0
is the Bohr
radius, given
11.
Find the expectation value <x> for an
electron in the 1s-state of the hydrogen
atom, given that
In this chapter, we delve into the fundamental concepts of integral calculus, focusing on two major approaches to mathematically generate integrals and assigning physical meanings to them. We explore antiderivatives, differentiation, integration, and the process of taking integration as the inverse of differentiation. Through examples and explanations, we aim to provide a clear understanding of integrals and their applications.
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Presentation Transcript
INTEGRAL CALCULUS BY SUWARDI
2.1 INTRODUCTION TWO MAJOR APPROACHES ANTIDERIVATIVES to mathematically generate integrals INTEGRATION to assign a physical meaning to the integral The other approach: To consider the integral as the sum of many similar, infinitesimal elements The process of taking integration, as the inverse of differentiation
In this chapter we shall consider the reverse process. Knowing the effect of individual changes, we wish to determine the overall effect of adding together these changes such that sum equals a finite change Before considering the physical significance and the application of integral calculus, let us briefly review the general and special methods of integration
2.2 INTEGRAL AS ANTIDERIVATIVE Function y = f(x), differentiation of this function, symbolized by the equation ( ) dy df x = = = ( ' ) dy f x dx ( ' ) f x dx dx Where f (x) denotes the first derivative of the function f(x) with respect to x
In this chapter we shall pose the following question: What function f(x), when differentiated, yields the function f (x)? For example, what function f(x) when differentiated, yields the function f (x) = 2x? Substituting f (x) = 2x into equation gives dy = = 2 or 2 dy x dx x dx
This function, f(x), for which we are looking is called the integral of the differential and symbolized by the equation = ( ) ( ' ) f x f x dx : the integral sign; f (x): the integrand
If f(x) = 2x, then f(x) = x2. f(x) = x2 is not the complete solution but f(x) = x2 + C is the complete solution. C is the constant of integration, and this constant always is included as part of the answer to any integration. Thus, = = + 2 2 y x dx x C
2.3 GENERAL METHODS OF INTEGRATION Let us consider several general methods of integration = + . 1 ( ) ( ) du x u x C = = + . 2 a du a du au C + 1 n u = + n . 3 u du C + 1 n
Examples: 4 x 4 = + 3 ( ) a x dx C RT RT 1 H H H H H = = = + = + 2 1 ( ) b dT dT T dT T C C 2 2 R T R R
du = = + . 4 ln ln d u u C u + = + . 5 ( ) ( ) ( ) ( ) f x g x dx f x dx g x dx 1 = + mx mx . 6 e dx e C m 1 = + . 7 sin cos kx dx kx C k 1 = + . 8 cos sin x dx x C k
2.4 SPECIAL METHODS OF INTEGRATION Many of the functions encountered in physical chemistry are not in one of the general forms given above Special methods of integration: Algebraic Substitution Trigonometric Transformation Partial Fractions
Algebraic Substitution The mathematical functions can be transformed into one of the general forms in section 2.3 or into one of the forms found in table of integral by some form of algebraic substitution Examples ( 1 2 Evaluate ) ( ) 5 2 . a x x dx
Let us attempt to transform this integral into the form un du Let u = (1 x2), Then du = -2x dx. Hence, ( 1 ) ( 1 ) 1 1 5 6 = = + = + 2 5 6 2 2 x x dx u du u C x C 6 6
kT E / E k T ( ) Evaluate b e dT 2 E E = = Let u Then . Hence du 2dT kT kT kT E = = + = + / / E k T u u E k T e dT e du e C e C 2
dV ( ) Evaluate . c V nb du u Let us attempt to transform the integral into the form . Let u = V nb. Then du = dv. Hence, dV du = = + = + ln ln( ) u C V nb C V nb u
2 ( ) Evaluate sin cos d x x dx Let u = sin x. Then du = cos x dx. Hence 1 1 = = + = + 2 2 3 3 sin cos sin x x dx u du u C x C 3 3
Trigonometric Transformation Many trigonometric integrals can be transformed into a proper form for integration by making some form of trigonometric transformation using trigonometric identities. For example, to evaluate the integral we must make use of the identity 2 sin x dx 1 ( ) x 2 = sin2 1 cos x 2
Thus, 1 1 1 ( ) = 1 cos 2 cos 2 x dx x dx 2 2 2 Integrating each term separatly gives, 1 x 2 = + sin2 sin 2 x dx x C 4
Again, integration of integral of this type are more parctically done by using the Table of Integrals Example: Evaluate 3 cos 2 x dx Integration of this function can be accomplished using integral from the Table of Integral. Here, a = 2 and b = 0 1 1 ( ) ( ) ( ) + = + + + 3 3 cos sin sin ax b dx ax b ax b C 3 a a 1 1 = + 3 3 cos 2 ( ) sin( 2 ) sin 2 ( ) x dx x x C 2 6
Partial Fractions Consider an integral of the type dx , where and are constants a b ( ( ) x ) a b x This type of integral can be trasformed into simpler integral by the following method. Let A = (a x) and B = (b x). Then 1 1 B A B A = = A B AB AB AB
Therefore, 1 1 1 1 = ( ) AB B A A B 1 1 1 1 = ( )( ) ( ) ( ) ( ) a x b x b a a x b x 1 dx dx dx = ( )( ) ( ) ( ) ( ) a x b x b a a x b x Which can be integrated to give, ( ) 1 1 b x dx )( = + + = + ln( ) ln( ) ln a x b x C C ( ) ( ) ( ) ( ) a x b x b a b a a x
2.5 The Integral as a summation of infinitesimally In the previous sections we have considered integration as the purely mathematical operation of finding antiderivatives. Let us now turn to the more physical aspects of integration in order to understand the physical importance of the integral
Consider, as an example: 1. The expansion of an ideal gas from a volume of V1 to a volume V2against a constant external pressure Pext, The plot seen in the indicator diagram is not a graph of Pext as a function of V. There is no functional dependence between the external pressure on the gas and the volume of the gas
The diagram shows what the external pressure is doing on one axis and what the volum is doing on the other axis Both can vary independently The work done by the gas does depend on both the external prssure and the volume change For expansion, the work is w = - Pext(V2 V1) W is just the negative of the area under the Pext versus V diagram
2. The external prssure changes, in some fashion, as the volume changes The external pressure is not changing as a function of volume To emphasize this point, let us assume that the external prsessure actually goes up as the volumes changes (see indicator diagram ) A gas could be expanding against atmospheric pressure which perhaps is increasing over the period of time that the expansion take place
Indicator diagram showing PV work done by a gas expanding against a variable external pressure
The approximate area under the curve, then, is just the sum of the four rectangles 4 = i = + + + = approx A P V P V P V P V i V P 1 2 3 4 1 If we extend this process-that is if we divide the area under the curve into more and more rectangles of smaller and smaller V-the sum approaches a fixed value as N approaches infinity
Hence, we can write N = i lim = A P V i N 1 However, since as N approaches infinity, V approaches zero, we also can write N = i lim = A P V i 0 V 1 But, by definition V N 2 = i V lim V = P V P dV i i 0 1 1
V 2 Where the symbol is read the integral V 1 from V1 to V2 , and V1 and V2 are called the limits of integration. Hence, V 2 This integral is called a definite integral = A P dV ext V 1
2.6 Line Integral Integral of the general type x 2 x = A y dx 1 Are called line integral, because such integral represent the area under the specific curve (path) connecting x1 to x2
Such integral can be evaluated analytically (i.e., by finding the antiderivative) only if an equation, the path, y = f(x) is known, since under these circumstances the integral x 2 x ( ) f x dx 1 contains only one variable
If y is not a function of x (as in the case of Pext versus V), or if y is a function of x, but the equation relating y and x is not known or can not be 2 = ax y e integrated (as in the case ), or
if y is a function of more than one variable that changes with x, then the line integral cannot be evaluated analytically, and one must resort to a graphical or numeric method of integration in order to evaluate the integral.
1. The integral in below can be evaluated analytically if we assume that Pext is a constant, hence, it can be brought out of the integral V V 2 2 = = = ( ) A P dV P dV P V V 2 1 ext ext ext V V 1 1 work = -A = - Pext(V2- V1)
2. A second way to evaluate the above integral is found in the concept of reversibility. If the expansion of the gas is reversible, then for all practical purposes the external is equal to the gas pressure, Pext = Pgas. This gives V V 2 2 V V = = ( , ) A P dV P T V dV ext gas 1 1 It can be integrated if P is a function of V only so T must be constant (isothermal condition)
We can write V V V nRT 2 2 2 = = = A P dV P dV dV ext gas V V V V 1 1 1 V dV 2 ( ) = = ln ln nRT nRT V V 2 1 V V 1 V = ln 2 nRT V 1
Example The change in enthalpy as a function of temperature is given by the equation T 2 = H C dT P T 1 Find the change in enthalpy for one mole of real gas when the temperature of the gas is increased from, say, 298.2 K to 500.0 K
Solution We first recognize that the above integral is a line integral. This integral cannot be evaluated unless CP as a function of T is known We could assume that CP is a constant and evaluate the integral that way, but over such a large temperature range the approximation would be poor Another approach would be to determine the integral numerically
One analytical approach is to expand CP as a power series in temperature CP = a + bT + cT2 The constant a, b, and c are known for many common gases. Substituting this into the H equation gives T 2 1 T = + + 2 ( ) H a bT cT dT b c = + + 2 2 3 3 ( ) ( ) ( ) H a T T T T T T 2 1 2 1 2 1 2 3
2.7 Double and Triple Integrals Functions could be differentited more than once How about the determination of multiple integrals Example, The volume of cylinder is a function of both the radius and the height of the cylinder. That is, V = f(r, h)
Let us suppose that we allow the height of the cylinder, h, to change while holding the radius, r, constant The integral from h = 0 to h = h, then, could be expressed as h 0 ( , ) f r h dh
But the value of this line integral depends on the value of the radius, r, and hence the integral could be considered to be a function of r. h 0 = ( ) ( , ) g r f r h dh If we allow r to vary from r = 0 to r = r and integrate over the change, we can write r r h 0 0 0 = ( ) ( , ) g r dr f r h dh dr
To evaluate the above double integral, we 0 h integrate first while holding r a dh h r f ) , ( constant, which gives us g(r). r Then we integrate next while holding 0 ( ) g r dr h constant. Such a process is known as successive partial integration
r h For example, let us evaluate 0 2 r dh dr 0 h First, 0 = = ( ) 2 2 g r r dh r h Next, we integrate r r 0 0 = = 2 ( ) 2 g r dr rh dr r h That is the volume of a cylinder
The above argument can be extended to the triple integral. For example, let us evaluate the triple 0 0 0 y z x dx dy dz x 0 First, evaluate = dx x Substituting this back into the above equation gives 0 0 y z x dy dz
Next, evaluate y 0 = x dy xy Substituting this back into the above equation gives xy 0 z dz Integrating this gives z 0 = xy dz xyz Which is the volume of a rectangular box x by y by z
Problem. The differential volume element in spherical polar coordinates is dV = r2sin d d dr. Given that goes from 0 to 2 , and r goes from 0 to r, evaluate the triple integral 2 r 0 0 0 = d d 2 sin V r dr Solution 2 0 0 2 d = 4 d = 2 2 2 2 2 sin sin sin r r r r r 4 0 = = 2 3 4 r dr r V 3
PROBLEMS 1. Evaluate the following integral (consider all uppercase letters to be constant): RT 2 ( ) 4 a x dx + 2) 5 ( g) dp ( ) 3 ( d x x dx p 1 ( ) b dx d M x ( ) 4 e e dx ( h) 2 x d ( f ) P ( ) cos 2 ( ) i W t dt ( ) sin 3 c x dx
2. Evaluate the following integrals using the Table of Integral (consider all uppercase letters to be constant): 2 2 / 1 ) 2 4 x ( ) ( e x A dx ( ) a e dx d 3 ( ) cos sin i N x 2 ( ) sin f x dx d + 4 4 2 3 ( ) cos k ( ) ( 2 ) 4 b x x x dx A 2 2 xcos ( ) ( ) ( ) c x A dx g e x dx + 6 ( ) sin 3 ( ) 4 l x dx dx N x ( ) m 2 2 ( ) sin d dx ( ) sin 2 ( ) h Wt dt 4 ( )( 3 ) x x A
3. Evaluate the following definite integrals using the Table of indefinite and definite interals, as needed T d T 2 T H 2 a a bT + + + 2 n x a . a cT dT . d dT 2 2 . sin g x dx 2 RT T 1 T 0 1 P V RT P 2 nRT V n a V 2 2 . b dP . e dV 2 2 ax 2 . h x e dx nb P V 1 1 0 /2 2 2 . sin cos f d . c d 0 0
