Transcript INTEGRAL CALCULUS - Yogyakarta State University

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)

 f ' ( x)
dx dx
dy  f ' ( x) 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 x dx or
dy
 2x
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,
y   2 x dx  x  C
2
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
n 1
u
3.  u du 
C
n 1
n
Examples:
4
x
(a)  x 3 dx 
C
4
H
H 1
H
H 1
H
2
(b) 
dT 
dT 
T dT  
T  C 
C
2
2


RT
R T
R
R
RT
du
4.    d ln u  ln u  C
u
5.   f ( x)  g ( x)dx   f ( x) dx   g ( x) dx
1 mx
6.  e dx  e  C
m
mx
1
7.  sin kx dx   cos kx  C
k
1
8.  cos x dx  sin 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

(a) Evaluate 2 x 1  x
 dx.
2 5
• Let us attempt to transform this integral into
the form
n
 u du
• Let u = (1 – x2), Then du = -2x dx. Hence,
 2 x1  x 
2 5

1 6
1
dx    u du   u  C   1  x 2
6
6
5

6
C
(b) Evaluate e
 E / k T
E
Let u  
kT
e
 E / k T
 E 
 2  dT
 kT 
E
Then du  2 dT . Hence
kT
 E 
u
u
 E / k T
C
 2  dT   e du  e  C  e
 kT 
dV
(c) Evaluate 
.
V  nb
du
Let us attempt to transform the integral into the form  .
u
Let u = V – nb. Then du = dv. Hence,
dV
du
 V  nb   u  ln u  C  ln(V  nb)  C
(d ) Evaluate
2
sin
 x cos x dx
Let u = sin x. Then du = cos x dx. Hence
1 3
1 3
 sin x cos x dx   u du  3 u  C  3 sin x  C
2
2
Trigonometric Transformation
• Many trigonometric integrals can be
transformed into a proper form for integration
by making some form of trigonometric
transformation using trigonometric identities.
2
sin
x dx
• For example, to evaluate the integral 
we must make use of the identity
1
sin x  1  cos 2 x 
2
2
• Thus,
1
1
1
 2 1  cos 2 x   2  dx  2  cos 2 x dx
• Integrating each term separatly gives,
x 1
 sin x dx  2  4 sin 2 x  C
2
• Again, integration of integral of this type are
more parctically done by using the Table of
Integrals
3
Evaluate  cos 2 x dx
• Example:
• Integration of this function can be
accomplished using integral from the Table of
Integral. Here, a = 2 and b = 0
1
1
3




ax  b   C
cos
ax

b
dx

sin
ax

b

sin

a
3a
3
1
1 3
 cos (2 x) dx  2 sin( 2 x)  6 sin (2 x)  C
3
Partial Fractions
• Consider an integral of the type
dx
 (a  x) (b  x) , wherea andb areconstants
• 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) 
dx
1  dx
dx 
 (a  x)(b  x)  (b  a)  (a  x)   (b  x) 
• Which can be integrated to give,
(b  x)
dx
1
1
 (a  x)(b  x)  (b  a)  ln(a  x)  ln(b  x)  C  (b  a) ln (a  x)  C
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 V2 against 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
Aapprox  P1V  P2 V  P3V  P4 V   Pi V
i 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
A  lim
N
 P V
N   i 1
i
• However, since as N approaches infinity, V
approaches zero, we also can write
A  lim
N
 P V
V  0 i 1
i
• But, by definition
V2
N
lim  P V   P dV
V  0 i  1
i
i
V1
V2
• Where the symbol

is read “the integral
V1
from V1 to V2”, and V1 and V2 are called the
limits of integration. Hence,
V2
A   Pext dV
V1
This integral is
called a definite
integral
2.6 Line Integral
• Integral of the general type
A
x2
 y dx
x1
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
x2
 f ( x) dx
x1
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
integrated (as in the case
y e
 ax 2
), 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
V2
V2
V1
V1
A   Pext dV  Pext  dV  Pext (V2  V1 )
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
V2
A   Pext dV 
V1
V2
P
gas
(T , V ) dV
V1
It can be integrated if P is a function of V only
so T must be constant (isothermal condition)
We can write
V2
A   Pext dV 
V1
V2
P
gas
V1
V2
V2
nRT
dV  
dV
V
V1
dV
 nRT 
 nRT ln V2  ln V1 
V
V1
V2
 nRT ln
V1
Example
• The change in enthalpy as a function of
temperature is given by the equation
T2
H   C P dT
T1
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
T2
H   (a  bT  cT 2 ) dT
T1
b 2
c 3
2
3
H  a(T2  T1 )  (T2  T1 )  (T2  T1 )
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
g ( r )   f ( r , h) dh
0
• 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
h
integrate
 f (r , h) dh first while holding r a
0
constant, which gives us g(r).
r
• Then we integrate  g (r ) dr next while holding
0
h constant. Such a process is known as
successive partial integration
• For example, let us evaluate
r
h
0
0
  2 r dh dr
h
First,
g (r )   2 r dh  2 r h
0
Next, we integrate
r
r
0
0
2
g
(
r
)
dr

2

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
z y x
triple
   dx dy dz
0
0
0
• First, evaluate
x
 dx  x
0
• Substituting this back into the above equation
gives
z y
  x dy dz
0
0
• Next, evaluate
y
 x dy  xy
0
• Substituting this back into the above equation
z
gives
xy dz

0
• Integrating this gives
z
 xy dz  xyz
0
• 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
Solution
r

2
0
0
0
V 

2
r
 sin  d d dr
2

0
0
2
2
r
sin

d


2

r
sin 

r
2
2
2

r
sin

d


4

r

4 3
0 4 r dr  3  r  V
2
PROBLEMS
1. Evaluate the following integral (consider all
uppercase letters to be constant):
(a )  4 x 2 dx
( d )  (3 x  5) x dx
RT
( g) 
dp
p
1
(b)  2 dx
x
(e)  4e x dx
( h)  M d
(c )  sin 3 x dx
( f )  P d
2
(i )  cos ( 2 W t ) dt
2. Evaluate the following integrals using the
Table of Integral (consider all uppercase
letters to be constant):
(a)  e
4 x
(e)  ( x 2  A2 )1/ 2 dx
dx
(i )  cos 3  sin  d
(b)  ( x  2 x  4) x dx
 N x
( f )  sin 2 
 x dx
A


( k )  cos 4  d
(c)  ( x 2  A2 ) dx
( g )  e x cos x dx
(l )  sin 6 (3 x  4) dx
 N x
(d )  sin 
 dx
 A 
(h)  sin (2Wt ) dt
4
2
2
3
2
(m) 
dx
(4  x)(3  x)
3. Evaluate the following definite integrals using
the Table of indefinite and definite interals,
as needed
T2
d

a.   a  bT  cT 2 
T
T1 
P2
RT
b. 
dP
P
P1
2
c.  d
0

 dT

H
d. 
dT
2
RT
T1
T2
 nRT n 2 a 
e.  
 2  dV
V  nb V 
V1 
V2
 /2
f.

0
n x
g.  x sin
dx
a
0
a
2

h.  x e
0
sin 2  cos  d
2
2  ax 2
dx

i.  e
2 r / a0
r dr
0

j.  e
 m 2 / 2 kT
 d
3
0

k .  (2 J  1) e
0
a( J 2  J )
dJ
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 these limits of integration
5. Evaluate the following multiple integrals
using the Table of Integrals, as needed
a.  yx dx dy
2
b.  ( x  y ) dx dy
2
2
c.  y ln x dx dy
d .  x 2 ln y e 2 x dx dy dz
e.
 /2
2
0
0
2


2
r
 sin  dr d d
0
0
0
  r cos  dr d
f.

g. 
0

r

 e
0
0
2
h2  nx2 n y nz2
  

8 mkT  a 2 b 2 c 2





dnx dny dnz
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
 (H ) 
 T   CP
P
where
CP as a power series in T, ∆CP = a + bT + cT2.
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
H
 (G / T ) 
 T
   T 2
P
Where G is the Gibbs free energy change
attending the reaction, and T is absolute
temperature. Expressing H in a power series
in T, ∆H = a + bT + cT2. where a, b, and c are
experimentally determined constants, derive an
expression for G as a function of temperature
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
L 2  0.05 L
2
2  x
Probabilit y 
sin
dx

L L 2  0.05 L
L
10. Find the probability of finding an electron in
the 1s-state of the hydrogen atom at r = a0 in
a range a0 ± 0.005a0, where a0 is the Bohr
radius, given
a0  0.005 a0
 2 r a0
1
Pr obability  4   e
 a0  a0  0.005 a0
r dr
11. Find the expectation value <x> for an
electron in the 1s-state of the hydrogen
atom, given that
 1
 x  4  
 a0 
3
e
2 r / a0
3
r dr
0
12.The differential volume element in cylindrical
coordinates is dV = r d dr dz. Show that if r
goes from 0 to r,  from 0 to 2, and z from 0
to h, the volume of a cylinder is V =  r2 h