CALCULUS AND ANALYTICAL GEOMETRY

EULER’S THEOREM
EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS
Statement: If u  f  x, y  is a homogeneous function of degree n, in x and y, then
u u
x y  nu 
x y
Proof: Since u  f x, y  is a homogeneous function of degree n in x and y, therefore it can be expressed as:
 y
u  xn f  
x
Differentiating partially with respect to x, we get
u  y  y   y  y  y  y 
 nx n 1  f    x n  f       nx n 1  f    x n  f      2 
x  
x  
x x  
x  x x  x 
u  y  y
 nx n 1  f    x n 2 y  f  
x x x
Multiplying both sides by x, we get
u  y  y u  y
x  nx n  f    x n 1 y  f    x  nu  x n 1 y  f   1
x x x x x
Now differentiating partially w.r.t y, we get
u  y  y   y  y 1  y
 0  f    x n  f       x n  f       x n 1  f  
y  
x  
x y  
x  
x  x  x
Multiplying both sides by y, we get
u  y
y  x n 1 y  f   2
y x
Adding 1  and 2  , we get
u u  y  y
x y  nu  x n 1  f    x n 1 y  f  
x y  
x x
u u
Therefore, x y  nu 
x y
Note: Euler’s Theorem can be extended to a homogeneous function of any number of variables. Thus, if
f x1 , x 2 , x 3 ,  , x n  be a homogeneous function of n variables x1 , x2 , x3 ,, xn of degree n, then
f f f f
x1  x2  x3    xn  nf 
x1 x 2 x 3 x n
 x3  y3 
EXAMPLE 01: If u  arctan  , show that x u  y u  sin2u 
 x y  x y
 
Solution: We have
 x3  y3  x3  y3
u  tan 1    tan u  
 x y  x y
 
 y3 
x 3 1  3 
x3  y3  x  1   y x 3

Let z  tan u    x2 
x y  y 1 y x
x1  
 x
Thus z is a homogeneous function of degree 2 in x and y. From Euler’s theorem we have

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 CALCULUS AND ANALYTICAL GEOMETRY

z z
y
x  2z 1
x y
Substituting the value of z in 1 , we get

x

x
   
 
tan u   y  tan u   2tan u   x sec 2 u  u   y sec 2 u  u   2 tan u
y 
 x  y 
 u u  u u sin u
sec 2 u x  y   2 tan u  x y 2  cos 2 u
 x y  x y cos u
u u u u
x y  2 sin u cos u  x y  sin 2u using sin 2  2 sin  cos  
x y x y
 x2  y2 
EXAMPLE 02: If u  ln  , prove that x u  y u  1 
 x y  x y
 
Proof: We have
 x2  y2 
  eu  x  y 
2 2
u  ln 
 x y  x y
 
 y2 
x 2 1  2 
x2  y2  x  1   y x 2

Let z  eu   x 
x y  y 1 y x
x 1  
 x
Thus z is a homogeneous function of degree 1 in x and y. From Euler’s theorem we have
z z
x y z 1
x y
Substituting the value of z in 1 , we get

x
 u
x
 
e y
 u
y
 
e  eu  x eu   ux   ye  uy   e
u u
   
 u u  u u
or e u  x y   e u  x y  1
 x y  x y
EXAMPLE 03: If U  f  x, y  is a homogeneous function of degree 10, prove that
x 2 f xx  2xyf xy  y 2 f yy  90f  x, y 
Solution: Since f is a homogeneous function of degree 10, we have
xf x  yf y  10 f 1
Differentiating 1 with respect to x, we get
xf xx  f x 1  yf yx  10 f x  xf xx  f x  yf yx  10 f x 2
Differentiating 1 with respect to y, we get
xf xy  yf yy  f y 1  10 f y  xf xy  yf yy  f y  10 f y 3
Multiplying 2  with x and 3  with y and then adding, assuming f xy  f yx , we get
   
x xf xx  f x  yf xy  y xf xy  yf yy  f y  10 xf x  10 yf y
 
x2 f xx  xf x  xyf xy  xyf xy  y 2 f yy  yf y  10 xf x  yf y 
x 2 f xx  2 xyf xy  y 2 f yy  10 f  1010 f   x 2 f xx  2 xyf xy  y 2 f yy  100 f  10 f
or x2 f xx  2 xyf xy  y 2 f yy  90 f 

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 CALCULUS AND ANALYTICAL GEOMETRY

TOTAL DIFFERENTIALS
EXAMPLE 01: A rectangular plate expands in such a way that its length changes from 10cm to 10.5 cm and
its breadth changes from 5 to 5.3 cm. Find the approximate change in its area.
Solution: Let x and y be the length and breadth
of the rectangular plate respectively. According
to the question, we have
x  10cm, dx  0.5cm, y  5cm and dy  0.3cm 
Since Area  A  x  y , using total differentials, we get
A A
dA  dx  dy  dA  ydx  xdy
x y
Substituting the values, we get
dA  50.5  100.3  dA  5.5cm 2 is the required change in the area.
EXAMPLE 02: A manufacturer of paper drinking cups decides to make its standard cup slightly smaller
1
than before. The cups are conical and hold volume V  πr 2 h, where r is the radius of the top and h is the
3
height (see figure). The radius will be changed from 1.5 inches to 1.4 inches, and the height will be changed
from 4 inches to 3.9 inches. Use differentials to approximate the reduction in volume that results from these
changes.
Solution: We have r r
r  1.5 in , dr  0.1 in , h  4 in , dh  0.1 in

and V
1 2
r h 1 h
3
Using total differentials, 1 becomes
V V
dh  dV  2rh dr   r 2 dh   dV  r 2hdr  rdh 
1 1 1
dV  dr 
r h 3 3 3
Substituting the given values, we get
dV  3.141.52  4  0.1  1.5  0.1  1.57 0.8  0.15  dV  1.4915 
1
3
3
Thus, negative sign indicates the reduction in volume. Thus, approximately 1.4915in volume will be reduced.
EXAMPLE 03: Approximate the change in the hypotenuse of a right triangle of legs 6 and 8 units, when the
shorter leg is increased by 1 2 units, the long leg is decreased by 1 4 units.
Solution: Let x and y be the length of shorter and long leg respectively. According to the conditions, we have

x  6, dx  0.5, y  8 and y  0.25  B

Since, z  x  y
2 2 2
Pythagoras theorem 1 B` z
z 2  62  82  z 2  100  z  10units 
Using total differentials 1 gives y=8units z`
2 z dz  2 x dx  2 y dy  z dz  x dx  y dy
Substituting the values, we get
10dz  60.5  8 0.25  10dz  3  2  dz 
1
 dz  0.1unit  A x = 6units C C`
10
Hence, the approximate change in the hypotenuse is 0.1unit.
[NOTE: ABC is the position of original right triangle and AB`C` is its position after the changes occur in its
sizes.]

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 CALCULUS AND ANALYTICAL GEOMETRY

Assignment Problems
2z 2z
1. Determine whether the following functions are solutions of Laplace equation   0:
x 2 y 2

a  z  e x cos y, b  z  1 e x  y , c  z  x2  y2 
2
 x3  y3 
2. If u  sec 1  , show that x u  y u  2 cot u 
 x y  x y
 
x y
3. Verify Euler’s theorem for the function u  sin 1  tan 1 
y x
4. If resistors of R1, R2 , and R 3 ohms are connected in parallel to make an R ohm resistor, the value of R can be
found from the equation:
1 1 1 1
   
R R1 R2 R3
R
Find the value of , where R1  20, R 2  35, and R3  50 
R3

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 CALCULUS AND ANALYTICAL GEOMETRY

INTEGRATION
INTRODUCTION
As we discussed that there are two branches of calculus, that are; Differential Calculus and Integral Calculus. In
differential calculus we begin with a function f(x) and obtain the derivative f   Interpreting f  as a rate of change
of f(x) led to a variety of applications. By contrast, there are situations in which we know the rate of change and seek
the function f(x). We need to be able to reverse the differentiation process in such cases. In other words, in
differential calculus we are given a function and we are required to find its derivative, while in integral calculus we
are required to find the function whose derivative is given. This process is depicted as under:
Differentiation process
f (x) f   x  = F(x)
Anti derivative/Integration process
f (x) f ‘(x) = F(x)
Thus integration is also known as anti-derivative. For example, if
f(x) = x3  f ` (x) = 3 x2, thus anti-derivative of (3 x2) = x3.

Definition: If F  x  is a differentiable function such that F  x   f  x  , then F  x  is called an integral or

d
dx
anti – derivative of f  x  and is written as  f  x  dx  F  x   The symbol dx indicates that integration is
performed with respect to x.
Integration: The process of finding the integral of a function is called integration.
Integrand: The function to be integrated is called the integrand. For example, if  f  x  dx  F  x  , then f(x) is the
integrand.
Integral sign: The symbol ‘  ’ is used to denote the ‘sign of integration or integral sign’ and is used to represent
the process of integration. The mathematician Leibniz introduced this symbol.
Constant of Integration and Indefinite Integral

F  x   f  x  , then
d
Let
dx


F  x  c  
F  x  c  F '  x  0  f  x 
d d d
dx dx dx

Therefore,  f  x  dx  F  x   c 
The arbitrary constant ‘c’ is called the constant of integration. Since  f  x  dx  F  x   c involves an arbitrary
constant ‘c’ that is why it is called the indefinite integral of f  x  because the value of ‘c’ is not definite.
EXAMPLE 01: Find the anti – derivative of f  x   3x 2 
Solution: We need to determine the function F  x  that yields f  x   3x 2 when differentiated. It does not take
long to realize that the derivative of x 3 is 3x 2  So,
 3x dx  x
2 3

But it is also true that

d 3
dx
 x  1  3x 2 ,  x 3  4   3x 2 ,  x 3  3  3x 2 
d
dx
d
dx

In fact,

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 CALCULUS AND ANALYTICAL GEOMETRY

d 3
dx
 x  c   3x 2 for any constant ‘c’.
In view of this, we say that the  3x 2 dx  x 3  c 
PROPERTIES OF INDEFINITE INTEGRALS
If f(x) and g(x) are any functions of the variable x, then following properties are always true.
(i)   f  x   g  x  dx   f  x  dx   g  x  dx 
(ii)   f  x   g  x  dx   f  x  dx   g  x  dx 
(iii)  kf  x  dx  k  f  x  dx, k  constant 
provided the anti-derivatives of f(x) and g(x) exist.
Integral of a constant:  kdx  kx  c, k being a real constant .
METHODS OF INTEGRATION
There is no uniform technique to find integral of a given function. Several methods have been developed to find the
integrals. You learn these methods gradually. To start with there are four major techniques.
1. Integration by formulas.
2. Integration by substitution.
3. Integration by parts.
4. Integration by partial fraction.
1. INTEGRATION BY FORMULAS
The integral of a given function may be found through direct formulas.
 Power Rule of integration
x n1
 x dx 
n
 c,  n  1
n 1
To integrate a power of x, increase the exponent by 1 and then divide by the new exponent.
x3  d  x3  
EXAMPLE 01:  x 2 dx   c  Note:   c   x 2  original function 
3  dx  3  
EXAMPLE 02:   x 2  4 x  5 dx   x 2 dx  4 xdx  5 1dx
3 2
x x
  x  4 x  5  dx 
 4  5 x  c  since  1dx   x dx  x 
2 0

3 2
3
x
  x  4 x  5  dx   2 x  5 x  c 
2 2

3
Note 1: In the power rule, if x is replaced by x  a (a being a constant), then the result also holds. So, we have
 x  a  x  2
n1 4

  x  a  dx   c, n  1 For instance,   x  2  dx 

 c
n 3

n 1 4
Note 2: If x be replaced by ax  b (a and b being constants) on both sides of the power rule, the standard form
remains true, provided the result on the right side is divided by ‘a’ the coefficient of x. That is,
1  ax  b 
n1

  ax  b  dx   c, n  1, a  0 
n

a n 1
1  2 x  3
32
1
For instance,   2 x  3 dx   c   2 x  3  c 
12 32

2 32 3
Integrals of the form   f  x   f   x  dx 
n


d   f  x   
n 1

  c    f  x   f   x  , n  1
n
Since
dx  n  1 
 

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 CALCULUS AND ANALYTICAL GEOMETRY

 f  x  
n 1

  f  x   f   x  dx   c, n  1 
n
Therefore,
n 1
This is known as the general power rule of integration.
EXAMPLE 03: Evaluate   x 2  x   2 x  1 dx 
4

d 2
Solution: Here the function x 2  x having power 4, together with derivative ( x  x)  2 x  1 which is present in
dx
multiplication, so using above formula, we get
x  x
2 5

x  x   2 x  1 dx  C
2 4

5
EXAMPLE 04: Evaluate   x 2  2 x   x  1 dx 
3

d 2
Solution: Here the derivative of x  2 x is not present in multiplication, but
2
( x  2 x)  2 x  2  2( x  1) , we
dx
observe that the constant 2 is missing, so we can write
1 1
  x  2 x   x  1 dx    x  2 x  2  x  1 dx    x  2 x   2 x  2  dx
3 2 3 2 3
2

2 2
  2x  2 4
1 x 1
   x 2  2 x   x  1 dx   c   x  2x   c 
3 4
2

2 4 8
f  x
 Integrals of the form  dx  ln f  x   c 
f  x
If the derivative of the denominator is present in numerator, then integral will be ln (denominator)  c 
You should keep in mind that
ln f  x   c, if n  1
f  x 
 dx    f  x     n 1

 f  x     c, if n  1
n

 n  1
2x  1
EXAMPLE 05: Evaluate  2 dx
x x
Solution: Here derivative of x2  x that is; 2 x  1 is present in the numerator, hence
2x  1
 x2  xdx  ln  x  x   c
2

x 1
EXAMPLE 06: Evaluate  2 dx
x  2x
Solution: Here the derivative of x2  2x , that is; 2 x  2 is not present in numerator, but if we multiply by 2, then its
derivative will be present, so
x 1 1 2  x  1 1
 2 dx   2 dx  ln  x 2  2 x   c 
x  2x 2 x  2x 2
 Integrals of exponential functions
d  ax 
We know that 
d x
e  c   e x and 
dx  ln a
 c   a x , therefore
dx 
x
a
 e dx  e  c and  a dx  c
x x x

ln a
d a   
 
f x
d f  x
e  c  e    f   x  and   c   a    f   x  , therefore
f x f x
Since,
dx dx  ln a 
a f  x
 e  f   x  dx  e  c and  a  f   x  dx 
f  x f  x f  x
c
ln a

7
 CALCULUS AND ANALYTICAL GEOMETRY

e dx (ii)  esin x cos xdx

5x
EXAMPLE 07: Evaluate (i)
5x
e
Solution: (i)  e dx 5x
c
5
(ii) Here we observe that function is of the type  e f  x   f   x  dx , that is; if derivative of power of exponential is
present in multiplication, then integral will be e  c, hence
f  x

 e cos xdx  e C

sin x sin x

tan 1 x
e
EXAMPLE 08: Evaluate  dx 
1 x 2

tan 1 x
e 1 d 1
Solution:  dx   e  tan 1 x
dx, here derivative of tan 1 x that is;  tan x   is present in 1

1 x 2
1 x 2
dx 1 x 2

tan 1 x
e
multiplication, hence using  e    f   x  dx  e    c , we get 
f x
dx  e
f x
c tan 1 x

1 x 2

 Integration with the use of trigonometric identities

The trigonometric identities are sometimes helpful for integrating certain functions. First you may remember the
following formulas:
 i   sin xdx   cos x  c ii   cos xdx  sin x  c
 iii   tan x dx   ln  cos x   c  ln sec x   c iv   cot x dx  ln sin x   c
 v   cos ec x dx   ln  cos ec x  cot x   c  vi   sec x dx  ln sec x  tan x   c
1 1
It may be noted that  i   sin mx dx  cos mx  c  ii   cos mx dx  sin mx  c
m m
Thus if any trigonometric function appears with multiple angle, then above six rules are applicable with division by
the multiple factor.
EXAMPLE 09: Integrate the following functions.
 i  sin2 x  ii  cos 2 x  iii  sec 2 xcosec 2 x 
1  cos 2 x 1 1 1
Solution: (i)  sin xdx  2
dx   1  cos 2 x  dx   1dx   cos 2 xdx
2 2 2 2
1 1 sin 2 x 1 1
 sin xdx  x   c   sin 2 x  c 
2

2 2 2 2 4
1  cos 2 x 1 1 1
(ii)  cos 2 xdx   dx   1  cos 2 x  dx   1dx   cos 2 xdx
2 2 2 2
1 1 sin 2 x 1 1
 sin xdx  x   c   sin 2 x  c 
2

2 2 2 2 4
1 1 1
(iii)  sec2 x cos ec 2 xdx   dx  4 dx  4 dx
 2cos x sin x 
2 2 2 2 2
cos x sin x 4cos x sin x
1
 4 dx  4 cos ec 2 xdx  4   cos ec 2 x  dx
2 2

 sin 2 x 
2

 cot 2 x 
Thus  sec2 x cos ec 2 xdx  4    c  2cot 2 x  c 
 2 
2. INTEGRATION BY SUBSTITUTION
Almost this technique is to be used when we observe the function is the product or quotient of two functions and the
derivative of one function is present either in multiplication or in quotient form.
Note: There are no hard and fast rules for making suitable substitutions. Experience is the best guide in this matter.
If after a substitution the new integrand becomes more complicated, then either some other substitution should be
tried or other methods need to be employed.
EXAMPLE 10: Evaluate   ax 2  2bx  c   ax  b  dx
n

8
 CALCULUS AND ANALYTICAL GEOMETRY

 
Solution: Here we observe that derivative of ax2  2bx  c  2  ax  b  is present in multiplication form without
constant 2, so let
dz
z  ax 2  2bx  c  2ax  2b  2  ax  b 
dx
dz
dz  2ax  2b  2  ax  b  dx    ax  b  dx
2
1  ax  2bx  c 
n1
1 z n1
2
dz 1
So,   ax  2bx  c   ax  b  dx   z    z n dz  c   c
2 n n

2 2 2 n 1 2 n 1
EXAMPLE 11: Evaluate  1  cos 2 xsin2x dx
Solution: Here we see that derivative of 1  cos2 x is 2sin x cos x =  sin 2x is present, so let
dz
z  1  cos2 x . Therefore,  2 cos x   sin x   dz  2sin x cos xdx   sin 2xdx
dx
z3 2 2
 1  cos x sin 2 xdx   z  dz    z dz   3 2  c   3 1  cos x   c 
2 12 2 32

1
EXAMPLE 12: Integrate 
x lnx
dz 1 dx
Solution: Put z  ln x    dz  
dx x x
dx 1
Now,  x ln x   z dz  ln z  c  ln  ln x   c 
Some well – known substitutions
The following substitutions are generally helpful to transform the integrand to an easier from.
If the integrand contains: Make the substituion:
a x 2 2
x  a sin  or a cos 
a x
2 2
x  a tan  or a sinh 
x a
2 2
x  a sec  or a cosh 
ax  b ax  b  z 2

 ax  b  ax  b  z
r s s

Note: If the integrand contains some linear expression of the variable, put that linear factor  z . If the integrand is a
rational function of e x , put e x  z  If the integrand contains a logarithm of f  x  or an exponential function in
which the index is f  x  , put f  x   z 

 1
dx
EXAMPLE 13: Evaluate 
x1
Solution: Let x  1  z so that z  x  1  2 zdz  dx and on substitution, we have
2

dx 2z z  1   1 
 1  x  1   1  z dz  2 1  z dz  2 1  1  z  dz  2  1dz   1  z dz 
dx

 1  x  1  2  z  ln 1  z   2  x  1  ln 1  x  1   c  
dx
EXAMPLE 14: Evaluate  x2  4


dx dx x
Solution: Let I     sinh 1  c.
x 42
x 22 2 2

9
 CALCULUS AND ANALYTICAL GEOMETRY

3. INTEGRATION BY PARTS
Considering the product rule of differentiation from integration point of view leads to another method of integration
called integration by parts. Integration performed by using following formula is called integration by parts.
d 
 u  v dx  u   vdx    dx u . vdx  dx
In this integral consists two parts – a function u and the differential of some other function v. An example should
provide some feeling for the process.

EXAMPLE 15: Evaluate x  e x dx


Solution: This integral cannot be evaluated by any of the formulas we have studied so far. So, it will be evaluated
using the formula of integration by parts. One part of the integral must be chosen as u, and what remain must be dv.
In this example, we will select:
u  x  du  dx and dv  e x dx  v  e x
Using the above given formula, we get
 x  e dx  x  e   e  dx  xe  e  c
x x x x x

The choices of u  x and dv  e x dx were correct. Suppose

1
u  e x  du  e x dx and dv  xdx  v  x 2 .Then,
2

 
1 1
x  e x dx  x 2  e x  x 2 e x dx .
2 2
Clearly, this result is not as good as the integral we began with because the first integrand is containing the power of
x as unity whereas the resulting integrand contains the power of x two. This shows that if we take a wrong choice of
the functions u and v, the resulting integral will never come to an end. The experience suggests the following
guidelines that are often useful in using the “Integration by Parts” method.
Guidelines:
(i) If one factor of the product is a power of x, take it as the first function provided the integral of the
second function is easy to evaluate.
(ii) If, the integral of the second function is not readily available (in case of inverse circular function or a
logarithmic function) in that case, take that factor as the first function.
(iii) If the integral is a single inverse circular or logarithmic function, take that function as the first and
unity (1) as the second function.
EXAMPLE 16: Evaluate the following:
1  x ln x 
 i  x 2 tan 1 xdx  ii  ln xdx  iii  e x 
    dx 
 x 
dx 1
Solution: (i) Let u  tan 1 x  du  and dv  x 2 dx  v  x 3
1  x2 3
Using the formula, we get
1  x  
  
1 1 x3 1
x 2 tan 1 xdx  x3 tan 1 x  dx  x 3 tan 1 x    x   dx 
3 3 1  x2 3 
3   1  x 2  


1

x 2 tan 1 xdx   x 3 tan 1 x  xdx 
3  
1 2x
2 1  x 2  3 
 1 1
2   1 
dx   x 3 tan 1 x  x 2  ln 1  x 2   c 
2 
1
(ii) Let u  ln x  du  dx and dv  1dx  v  x
x
Using the formula, we get
1
 ln xdx  x ln x   x  x  dx  x ln x  1dx  x ln x  x  c 

10
 CALCULUS AND ANALYTICAL GEOMETRY

 1  x ln x  x  1

 e  
 dx  e   dx  e ln xdx  1
x x
(iii)
x   x
1
Let u  e x and dv  dx  Then , du  e x dx and v  ln x 
x
Therefore 1 becomes
 1  x ln x 
 e   
 dx  e ln x  e ln xdx  e ln xdx  e ln x 
x x x x x

x 
EXAMPLE 17: Show that
x n1 x n1
 
1
x n tan1 xdx  tan1 x  dx 
n1 n  1 1  x2

 x tan
1
Hence evaluate 3
x dx 

Solution: Let u  tan 1 x and dv  x n dx, then du  dx /(1  x 2 ) and v  x n 1 /(n  1) 

Using the formula of integration by parts, we have
xn 1 xn 1
 
1
x n arctan xdx  arctan x  dx 
n 1 n  1 1  x2
Replacing n by 3, we have
 2 
  
x4 1 x4 1 1 1
x3 arctan xdx  arctan x  dx  x 4 arctan x   x  1   dx
4 4 1  x2 4 4  1  x2 
x4 1 1 1
 x arctan xdx  arctan x  x3  x  arctan x  c 
3

4 12 4 4
4 INTEGRATION OF RATIONAL ALGEBRAIC FUNCTIONS
An expression of the type f  x  g  x  , where f  x  and g  x  are polynomials with real coefficients, is called a
rational function. In this section we perform the integration of such functions by the method of PARTIAL
FRACTIONS and COMPLETING THE SQUARES.
In the fraction f  x  / g  x  , if the degree of the numerator is less than that of the denominator, the fraction is called
a proper fraction, otherwise improper fraction.
NOTE: Every improper fraction can be reduced to a sum of a polynomial and a proper fraction.
Thus,
N  x R  x
 Q  x  
D  x D  x
Here arise four cases of partial fractions.
CASE 1: When all factors of the denominator D  x  are linear and distinct
2x  3
For example, . In this case, we write
 x  1 x  1 2 x  3
2x  3 A B C
   
  
x  1 x  1 2 x  3  x  1 x  1 2 x 3
CASE 2: When all the factors of the denominator D  x  are linear but some are repeated
x2
For example, . In this case
 x  1  x  1
3

x2 A B C D
    
 x  1  x  1 x  1  x  1  x  1 x  1
3 2 3

CASE 3: When D  x  have non – repeated irreducible quadratic factors

11
 CALCULUS AND ANALYTICAL GEOMETRY

1
For example,
 x  1  x 2
 4
1 A Bx  C
In this case,   2 
 x  1  x  4 
2
x 1 x  4
CASE 4: When D  x  has repeated irreducible quadratic factors
8x2 8x2 8x2
 
1  x 1  x  1  x 1  x 1  x  1  x 1  x  1  x 
4 2 2 2 2 2
2

In this case,
8x2 A B Cx  D Ex  F
    
1  x 1  x 
4 2
1  x 1  x 1  x 2 1  x 2 2
NOTE: It may be noted that before we use the techniques of partial fraction, it has to be checked out whether or not
the degree of a polynomial in the numerator is less than the degree of polynomial in the denominator. If yes, then we
directly start the partial fraction technique. In case it isn’t, we divide the numerator by the denominator till the
degree of the polynomial in the numerator becomes less than that of the polynomial in the denominator. After this
process is completed, we resolve the resultant fractions in to partial fractions.
All these cases are studied by means of few examples.
EXAMPLE 01: Evaluate the following integrals
x1 x
i   dx  ii  
 
dx
 x  2  x  3   x  1 x 2  4
Solution: (i) Here degree of a polynomial in the numerator is less than the degree of polynomial in the denominator,
so by using partial fractions, we have
x 1 A B
  (1)
 x  2  
x  3 x  2 x 3
Multiplying both sides of (1) by  x  2  x  3 , we get
x  1  A  x  3  B  x  2  (2)
Now put x  2  0  x  2. Substitute this in(2), we have
3  A2  3  A  3 
Again if x  3  0  x  3, put into (2), we have
4  B 3  2  B  4 
Substituting these values in 1 , we have
x 1 3 4
  (3)
 x  2 x  3 x  2 x  3
Now integrating both sides of (3), we have
x 1
  
dx dx
dx  3 4  3ln  x  2   4ln  x  3  c 
 x  2  
x  3 x  2 x 3
(ii) Here we observe that degree of the numerator is less than that of the denominator, but in the denominator one
factor is linear and other one is quadratic, so by partial factions we have
x A Bx  C
  2 (1)
 x  1  x  4  x  1 x  4
2

Multiplying both sides of 1 by  x  1 x2  4 , we have 

x  A  x2  4    Bx  C  x  1 (2)
Substituting x  1, in (2) we have: 1  A 1  4   A  1/ 5 

12
 CALCULUS AND ANALYTICAL GEOMETRY

Now to find B and C, we simplify (2) that is;

x  Ax 2  4 A  Bx 2  Bx  Cx  C  x   A  B  x 2   C  B  x  4 A  C (3)
2
Now comparing the coefficients of x , x and constants on both sides of (3), we get
A  B  0, C  B  1, 4 A  C  0
B  1/ 5  0  B  1/ 5  C   1/ 5  1  C  1  1/ 5  C  4 / 5 
Substituting the values of A, B and C in (1), we have
x 1  1 5  x  4 5 1 1 1 x  4
     
 x  1  x  4  5  x  1
2
x2  4 5 x  1 5 x2  4
Now integrating, we have
x4 1 1 dx 
  x  x  1  5  2  x 
x 1 dx 1 1 dx 2x
dx   dx  dx  4 2
 x  1  x  4 
2
5 x 1 5 2
4 5 2
4 x  4 
1 1 1 1 x
 ln  x  1   ln  x 2  4   4  tan 1 
5 5 2 2 2
1 1 1 x
 ln  x  1   ln  x 2  4   2 tan 1   c 
5 5 2 2
EXAMPLE 02: Evaluate the following integrals:
x2
 
cosx
i  dx  ii  dx 
 x  1  x  1
3
 1  sinx  2  sinx  3  sinx 
x2 x2 A B C D
Solution: (i)      (1)
 x  1  x  1
3
 x  1 x  1 x  1 x  1 x  1  x  12  x  13 x  1
Multiplying both sides of (1) with  x  1  x  1 , we have
3

x2  A  x  1  x  1  B  x  1 x  1  C  x  1  D  x  1
2 3
(2)
Substituting x  1  0  x  1 into 2 , we have
1  C  2  C  1 2 
2

Substituting x  1  0  x  1 into (2), we have

 1  D  2   D  1 8 
2 3

Now to find the values of A and B, we rewrite the equation 2 in the following way:
x 2
 A  x  1  x  1  B  x  1  C  x  1  D  x  3x  3x  1
2 2 3 2

x3  A  x3  x  x 2  1  Bx 2  B 
1 1 1 3 3 1
x   x3  x 2  x  
2 2 8 8 8 8
 1 3  3 2  1 3  1 1 0
0 x  1x  0 x  0 x   A   x    A  B   x    A    x   A  B    x
3 2 0

 8  8  2 8  2 8
Comparing the coefficients from both sides, we have
A 1/ 8  0  A  1/ 8 and  A  B  3 / 8  1  B  3 / 4 
Hence, A  1/ 8, B  3 / 4, C  1/ 2 and D  1/ 8 
Substituting all these v alues into equation (1) we have
x2 1 3 1 1
    
 x  1  x  1 8  x  1 4  x  1 2  x  1 8  x  1
3 2 3

Integrating both sides with respect to x, we have

x2
     x  1  x 1
1 dx 3 dx 1 dx 1 dx
dx    
 x  1  x  1 8 x  1 4  x  1
3 2 3
2 8

13
 CALCULUS AND ANALYTICAL GEOMETRY

3  x  1 1  x  1
1 2
1 1
 ln  x  1    ln  x  1
8 4 1 2 2 8
1 x 1 3 1
 ln   2  c
8 x  1 4  x  1 4  x  1
(ii) Let z  sin x  dz  cos x dx  Therefore,

 1  sin x  2  sin x 3  sin x  dx   1  z  2  z 3  z 

cos x dz
I

1 A B C
Now,    (1)
1  z  2  z  3  z  1  z 2  z 3  z
Multiplying both sides of (1) with 1  z  2  z  3  z  , we get
1  A  2  z  3  z   B 1  z  3  z   C 1  z  2  z  (2)
Substituting 1  z  0  z  1 into (2), we get
1  A 1 2   A  1 2 
Substituting 2  z  0  z  2 into (2), we get
1  B  11  B  1
Substituting 3  z  0  z  3 into (2), we get: 1  C  2  1  C  1 2 
Hence, A  1/ 2, B  1 and C  1/ 2 
Now, equation (1) becomes
1 1 1 1
   (3)
1  z  2  z  3  z  2 1  z  2  z 2 3  z 
Integrating both sides of (3) with respect to z, we get

   
dz 1 dz dz 1 dz 1 1
    ln 1  z   ln  2  z   ln  3  z 
 
1  z 2  z  
3  z 2 1  z 2  z 2 3  z 2 2
Replacing z by sin x, we get


cos x 1 1
dx  ln 1  sin x   ln  2  sin x    3  sin x   c 
1  sin x  2  sin x 3  sin x  2 2
Integration by Completing the Squares Method
If the rational algebraic function f(x)/g(x) is such that g(x) is quadratic function that do not have real factors, the
method of completing the squares is used to evaluate the integrations of f(x)/g(x). The method is well presented by
the following two examples.
EXAMPLE 03: Evaluate the following by “Completing the Squares Method”.
x 1
 
x
(i) dx (ii) dx
x  4x  5
2
x  6x  7
2

Solution: (i)
1 2x  4  4 2x  4
    
x 1 2x 1 1
Let I  dx  dx  dx  dx  2 2 dx
x  4x  5
2
2 x  4x  5
2
2 x  4x  5
2
2 x  4x  5
2
x  4x  5


1 1
 ln( x 2  4 x  5)  2 2 dx
2 x  4x  5

 
1 1
 ln ( x 2  4 x  5)  2 2 dx  ln ( x 2  4 x  5)  2 dx
x  4x  4  1 ( x  2) 2  5
Let x + 2 = z  dx = dz. Thus,

14
 CALCULUS AND ANALYTICAL GEOMETRY

 z   5  dz  ln
1 2 z
I  ln ( x 2  4 x  5)  2 2
( x 2  4 x  5)  tan 1 c
2
5 5

2 ( x  2)
 ln ( x 2  4 x  5)  tan 1 c
5 5
(ii)
x 1 2x  2 2x  6  4 2x  6
 x  6x  7 dx  2  x  6x  7 dx  2  x  6 x  7 dx  2  x  6 x  7 dx  2 x  6 x  7 dx
1 1 1 1
Let I  2 2 2 2 2

 
1 1 1
 ln( x  6 x  7)  2
2
dx  ln x  6 x  7  2 dx 2

2 ( x  6 x  9)  9  72
( x  3)  16 2

Let x - 3 = z  dx = dz. Thus,

1 z4 1 x 34

1
I  ln x 2  6 x  7  2 2 dx  ln x 2  6 x  7  2 ln  c  ln x 2  6 x  7  ln c
z 4 2
4 z4 2 x 3 4
1 x7
 ln x 2  6 x  7  ln c
2 x 1


1
EXAMPLE 04: Evaluate dx
x 1
4

1 x2  1  x2  1 1 x2  1 1 x2 1
    
1 1 2
Solution: Let I  dx  dx  dx  dx  dx (1)
x4  1 2 x4  1 2 x4  1 2 x4  1 2 x4  1
x2  1 x 2 (1  1/ x 2 ) (1  1/ x 2 )
Now consider, I1 

x4  1
dx 

x 2 ( x 2  1/ x 2 )
dx 
( x 2  1/ x 2 )
dx

Let z = x – 1/x  dz = (1 + 1/x2) dx. Also z2 = x2 – 2 + 1/x2  x2 + 1/x2 = z2 + 2. Thus,
1 ( x  1/ x) 1 ( x  1)

   
2
1 1 1 z 1 1
I1  dz  dz  tan 1
 tan  tan
z2  2
2
z2  2 2 2 2 2 2 2x

x 1 x (1  1/ x ) (1  1/ x )
 x  1 dx  x ( x  1/ x ) dx  ( x  1/ x ) dx
2 2 2 2

Now consider, I  2 4 2 2 2 2 2

Let z = x + 1/x  dz = (1 - 1/x2) dx. Also z2 = x2 + 2 + 1/x2  x2 + 1/x2 = z2 - 2. Thus,

z 2 ( x  1/ x  2) ( x 2  2 x  1)
   
1 1 1 1 1
I2  dz  dz  ln  ln  ln
z 2 z 2 ( x  1/ x  2) ( x 2  2 x  1)
2 2
z2  2 2 2 2 2 2 2
Thus equation (1) becomes:
1 1 ( x 2  1) ( x 2  2 x  1) 

1 1 1
dx  [ I1  I 2 ]   tan 1  ln 2 c
x 1
4
2 2 2 2x 2 2 ( x  2 x  1) 
NOTE: The above example does not belong to “Completing the Squares Method”, nevertheless it is an
important type of integral. Students are advised to see the steps taken in solving this integral. The following
formulae have been used here.
xa
 
1 1 1 1 x
dx  ln and dx  tan 1
x2  a2 2a x  a x2  a2 a a
Integration by Completing the Squares Method
If the integrals are of the form

 
dx
OR Quadratic dx
Quadratic
the following tips can be used to evaluate such integrals. This technique is known as “Completing the Squares
Method”.
2
 Make the coefficient of x unity by taking its numerical coefficient outside the square root sign.
 Complete the square in terms containing x by adding and subtracting the square of half the coefficient of x.
 Use proper standard form.

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 CALCULUS AND ANALYTICAL GEOMETRY

EXAMPLE 02: Evaluate

 3x 2  4x  1dx 
Solution:

 
4 1
I 3x 2  4 x  1dx  3 x 2  x  dx
3 3
Completing the squares, we get
 2 4 2  2
2 2
 2 1
2

I 3
  x  x         1dx  3
 3 3  3   x    dx
 3 9
2 1 1
Substituting z  x   dz  dx and a 2   a  , we get
3 9 3
z 2 a2 z
I 3

z 2  a 2 dz  3 
 2
z  a 2  cosh 1   c
2 a 
Replacing z by x  2 3, we get
1  2 
2
2 1 1 x  2 3
I  3   x    x     cosh 1  c
2  3  3  9 18 3 
 
8.5.2 Five Standard Cases
In this section, we shall study the five important forms of integrations.


1
CASE 1: Integral of the form dx
Linear Linear
Rule to evaluate: Put Linear  z 

 Quadratic
1
CASE 2: Integral of the form dx
Linear
Rule to evaluate: Put Linear  z 

 Linear
1 1
CASE 3: Integral of the form dx , dx,
Linear Quadratic n
Quadratic
Rule to evaluate: Put Linear  1/ z 

  Pure Quadratic 
x
CASE 4: Integral of the form dx
Pure Quadratic
Rule to evaluate: Put Pure Quadratic  z 

  Pure Quadratic 
1
CASE 5: Integral of the form dx
Pure Quadratic
Rule to evaluate: Put x  1 / y and in the resulting integral put Pure Quadratic  z 
EXAMPLE 03: Evaluate the following integrals:

   
1 1 dx x dx
i  dx  ii  dx  iii   iv  
 2x  3  x  5  x  1  x 2
 1 x 2
x 2
 1  x 2
 2x  2 x  1
 
  2 x  3  Linear
1dx 1dx
Solution: (i)Let I form 
x5  Linear 
Put z  x  5  z 2  x  5  2 z dz  dx  Then

 2  z  2z z  z 
2 zdz 2dz 2 dz dz
I   
 5   3  z   10  3 2 7 2

2 2 2 2
2
72

16
 CALCULUS AND ANALYTICAL GEOMETRY

1 z 7 2 1 x5  7 2
I ln  ln  c
2 72 z 7 2 14 x5  7 2
 
  x  1  Linear
1 dx 1 dx
(ii) Let I  form 
x  1 
2
Quadratic 
1 1 1
Put x  1   x   1  dx   2 dz  Then
z z z
 1   1 z  dz2

 
1
I   2  dz 
1 
2
 z  1  2z  z 2  z 2
1 z    1  1 1 z 
z  z2

 1 z  dz   dz 2

 1 z  1  2 z  1  2z   1  2z 
1 2
I 2 12
 dz

1 1  2 z 
12
2 x 1 2 x 1
I  1  2z  1    c
2 12 x 1 x 1 x 1
 
x 
1dx dx
(iii) Let I  form 
x  1 
2
Quadratic Quadratic 
2

1 1
Put x   dx   2 dy  Then
y y
 1 y  dy 2
 1 y  dy2

 1 y    
ydy 1 2 ydy
I   
1 y  1 1 y 1 y 2 1 y2
2 2

1 y 
2 2
2

y2

1  1  y 
 2 1 2 1   f   x  dx  f  x 
 n 1

I 
2  1 2  1 



 using


  f  x  n

n  1



x2  1 x2  1
2
1
I   1 y2   1       c
x
2
x x
 
  x  2x  2  Quadratic
x dx x dx
(iv) Let I  form 
2
x 1  Linear 
Put z  x  1  z 2  x  1  2 zdz  dx  Then
 z  1  2 z  dz 2
1  z  dz
2

I
  z  1  2  z  1  2  z   2 z
2 2 2
4
 2z 2  1  2z 2  2  2

1 z2 11 z2
I 2
 1 z4
dz  2
1 z2  z2  (1)

1  1
Put z   t  1  2  dz  dt into (1), we get
z  z 
1  t  1  z  1 z  1  z  1 

   
2
dt dt 2
I 2 2 2  tan    2 tan    2 tan  
t 2
2
t2  2 2  2  2   2z 

 x 1 1  1  x  2   c 
I  2 tan 1    2 tan  
 2 x  1   2x  2 
THE DEFINITE INTEGRAL AS AN AREA UNDER A CURVE

17
 CALCULUS AND ANALYTICAL GEOMETRY

The following integral is so important for the study of calculus that it is known as the Fundamental Theorem of
Calculus. It provides a key link between differential calculus and integral calculus.
b

 f  x  dx   F  x   F b   F  a  ,
b
a
a

If f is continuous on the interval a,b, then in above integral F is any anti-derivatived of f., the
numbers a and b are called limits of integration, b is the upper limit and a is the lower limit.

18
 CALCULUS AND ANALYTICAL GEOMETRY

 x dx using the Fundamental Theorem of Calculus.

2
EXAMPLE 03: Evaluate
1
2 2
1 
x 2 dx   x3    x3    23  13   

1 2 1 7
Solution:
1
 3 1 3 1 3 3
Note: Because the Fundamental Theorem of Calculus says we can use any anti-derivative, we will always use the
anti-derivative with C  0  However, if you did use some other value of C, it would be eliminated in the process
anyway. If x3 3  C (rather than just x 3 3 ) had been used in the above example, we would have had
2 2
1  8  1  7

7
x 2 dx   x3  C     C     C    C  C  
1
3 1  3  3  3 3
EXAMPLE 04: Determine the (exact) area under the curve y  x from x  1 to x  4 
Solution: We have f  x   x , a  1 and b  4  Thus,
4 4 4
 x3 2 
4
2 3 2 
     x    4  1   7   
2 32 2 14
Area  xdx  x1 2 dx  
1 1
 3 2 1  3 1 3 3 3
The area between the graph of y  x and the x – axis on the interval 1, 4 has been shown to be 14 3 square
units. The area is shown shaded in the following figure:

Fundamental Properties of Definite Integrals

b b

Property 1:
 f  x  dx   f u  du 
a a

Proof: Let
b

 f ( x) dx  F ( x) 
 f ( x) dx  F (b)  F (a). Similarly
a
b b b

 f (u ) du  F (u ) 

a
f (u ) du  F (b)  F (a ) 

a
f ( x ) dx 
 f (u) du
a

This property states that definite integral is independent of the variable used. For example,
 /2  /2

 cos x dx   cos u du  1
0 0

b a
Property 2: If f(x) is an integrable function, then  f x dx   f x dx 
a b
Proof: Let

19
 CALCULUS AND ANALYTICAL GEOMETRY

b a b

 f ( x) dx  F ( x)   f ( x) dx  F (b)  F (a) and  f (x) dx  F (a)  F (b)  [F (b)  F (a)]   f (x) dx
a b a

This property states that interchange of the limits of integration changes the sign of the integral. For example,
 /2 0


0
cos x dx  1 whereas
 cos x dx  1
 /2

b c b

Property 3:
 a
f  x  dx 
 a
f  x  dx 
 f  x  dx, where a  c  b 
c
Proof: Let
b c b

 f ( x) dx  F ( x) 
 a
f ( x) dx  F (b)  F (a),

a
f ( x) dx  F (c)  F (a) and
 f ( x) dx  F (b)  F (c)
c

c b b

Now
 f ( x) dx   f ( x) dx  F (c)  F (a)  F (b)  F (c)  F (b)  F (a)   f (x) dx
a c a

This property states that if c is any real number that lies in the interval [a, b] then integral from a to b is equal to sum
of the integrals taken from a to c and then from c to b. For example, consider


 sin x dx   cos x


0
 (cos   cos 0)  (1  1)  2
0

Now consider,
  /2 

 sin x dx   sin x dx   sin x dx   cos x 

  cos x  / 2    cos( / 2)  cos 0  cos   cos( / 2) 
 /2 

0
0 0  /2

 [0  1  1  0]  2. We see that two results are equal.

Generalization of Property 3
b a1 a2 a3 an b

 f x dx   f x dx   f x dx   f x dx     f x dx   f x dx,

a a a1 a2 an1 an
where a  a1  a2  a3   an1  an  b 
This means that if the interval of integration [a, b] is divided into any finite number of subintervals, the integration
taken over the interval [a, b] is equal to sum of the integrals taken over the subintervals.
a a

Property 4:
 f  x  dx   f  a  x  dx 
0 0

Proof: Let u = a – x  dz =- dx. Also if x = 0 the u = a, and if x = a then u = 0. Thus

a 0 a

 f  a  x  dx   f (u)du   f ( x)dx. (Using the property 1.)

0 a 0

Let us try an example now. Consider

 /2

 cos x dx  sin x
 /2

0
 sin( / 2)  sin 0  1  0  1
0

Now consider,
 /2  /2  /2

   sin x dx   cos x
 /2
cos x dx  [cos ( / 2)  x] dx  0
 [cos( / 2)  cos 0]  [0  1]  1 .
0 0 0

We observe that a result of both integrals is same. It may be noted that cos (90 o – x) = sin x.
This property states that if the lower limit in any definite integral is zero, the value of the integral remains
unchanged on replacing x by “upper limit minus x “in the integrand.

20
 CALCULUS AND ANALYTICAL GEOMETRY

Before we provide an example, let us consider the geometrical meaning of this property. Geometrically this property
states that “area under the curve y = f(x) from 0 to any real number x = a is same as the area under the curve
between 0 to a when the function f(x) is shifted “a units|” on the left of origin. This is depicted in the following
figure.

0 a -a 0
We observe that the shape of the curve is same but it is shifted a units to the left of the origin. Thus area will be
same. It may be noted that in the second figure, the lower limit is –a, nevertheless the area is always positive, hence
the integral an be taken from 0 to a as the numerical values of both integrals will be same.
Property 5:
2a a a

 f  x  dx   f  x  dx   f  2a  x  dx 
0 0 0
2a a 2a

Proof: By property 3,
 f  x  dx   f  x  dx   f  x  dx 
0 0 a

2a a

But
 f  x  dx   f  2a  x  dx. [Students may verify this by letting u  2a  x] . Thus
a 0

2a a a


0
f  x  dx 

0
f  x  dx 
 f  2a  x  dx 
0

Property 6:
2a a

i.
 f  x  dx  2 f  x  dx,
0 0
if f  2a  x   f  x  

2a

ii.
 f  x  dx  0,
0
if f  2a  x    f  x  

2a a 2a a a

Proof: (i)
 f  x  dx   f  x  dx  f  x  dx   f  x  dx  f  2a  x  dx
0 0 a 0 0
[By property 4]

a a a



0

f  x  dx  f  x  dx  2 f  x  dx
0
 0
[ f (2a  x)  f ( x)]

2a a 2a a a

(ii)

0
f  x  dx 
 0

f  x  dx  f  x  dx 
a

0

f  x  dx  f  2a  x  dx
0
[By property 4]

a a



0

f  x  dx  f  x  dx  0
0
[ f (2a  x)   f ( x)]

EXAMPLE 05: (i) We know that cos (2π- x) = cos x, hence

21
 CALCULUS AND ANALYTICAL GEOMETRY

2


LHS  cos x dx  sin x 0  sin 2  sin 0  0  0  0
2

0
   

   
RHS  cos x dx  cos  2  x  dx  cos x dx  cos x dx  sin x 0  sin x 0  2 sin x 0  2(sin   sin 0)  0 (ii)
  

0 0 0 0

We se that LHS = RHS

We know that sin (2π- x) =- sin x, hence

2

 sin x dx   cos x
2

0
 (cos 2  cos 0)  (1  1)  0 . Hence the result.
0

na a

Property 7: If f x   f a  x , then  f  x  dx  n f  x  dx 
0 0
na a 2a 3a na

Proof:
 f  x  dx   f  x  dx   f  x  dx   f  x  dx  ...  
0 0 a 2a ( n1) a
f  x  dx (1)

In the 2nd integral on the right hand side substitute, x = a + u  dx = du. Now if x = a then u = 0 and if x=
2a then u = a. Thus,
2a a a a


a
f  x  dx 

0
f  a  u  du 

0
f  a  x  dx 
 f  x  dx
0

Similarly, in the 3rd integral on the right side of (1) if we make the same substitution, we get

3a a

 f  x  dx   f  x  dx . Proceeding in this manner we shall see that each integral on the right side of (1) is equal to
2a 0

 f  x  dx . Thus,
0

na a


0

f  x  dx  n f  x  dx provided f ( x )  f (a  x ).
0

2


EXAMPLE 06: RHS  n cos x dx  n sin x 0  n(sin 2  sin 0)  n(0  0)  0 [ cos(2  x)  cos x]
2

2 n 2

 cos x dx  n cos x dx  n sin x

2
LHS  0
 n(sin 2  sin 0)  n(0  0)  0. Hence the result.
0 0

Property 8:
 If f  x  is an even function of x, then
a a

 f  x  dx  2 f  x  dx 
a 0

 If f  x  is an odd function of x, then

a

 f  x  dx  0 
a

 Even function: A function f  x  is said to be an even function of x if f   x   f  x  

For example, f  x   x 2 and f  x   cos x are even functions.

22
 CALCULUS AND ANALYTICAL GEOMETRY

 Odd function: A function f  x  is said to be an odd function of x if f   x    f  x  

For example, f  x   x 3 and f  x   sin x are odd functions.
If these conditions are not satisfied, then the given function is neither even nor odd
Proof: Let f(x) be an even function, that is, f(-x) = f(x). Then
a 0 a

 f  x  dx   f   x  dx   f  x  dx.
a a 0

Put  x  u  dx  du , in the first integral. If x  a then u  a and if x  0 then u  0.Thus

a 0 a a a a a a

 f  x  dx   f u  du  f  x  dx   f u  du   f  x  dx   f  x  dx   f  x  dx  2 f  x  dx
a a 0 0 0 0 0 0

Now let f(x) be an odd function, that is, f(-x) = - f(x). Then
a 0 a


a
f  x  dx 

a
f   x  dx   f  x  dx.

0

Put  x  u  dx  du, in the first integral. If x  a then u  a and if x  0 then u  0.Thus EX

a 0 a a a a a

 f  x  dx   f u  du  f  x  dx   f u  du   f  x  dx   f  x  dx   f  x  dx  0. This proves the result.

a a 0 0 0 0 0

AMPLE 06: We know that f(x) = x2 is an even function and f(x) = x3 is an odd function. Hence we may see that
4 4 4 4
 x3   x3 
 
1 128 2 128
LHS  x 2 dx     [64  (64)]  , RHS  2 x 2 dx  2    (64  0)   LHS = RHS
4
 3  4 3 3 0
 3 0 3 3
4 4
 x4 

1
Also, x dx     (256  256)  0
3

4
 4  4 3
With the help of these properties we are now solving some important definite integrals.

EXAMPLE 07: Prove the following:

π 2 π 2
π
  ln  tanθ  cotθ  dθ  πln2 
sinx
i  dx   ii 
0
sinx  cosx 4
0
Solution: (i) Let
 2


sin x
I dx 1
0
sin x  cos x

 
 2 sin   x   a a

2 
I
    
dx since
  f  x  dx 
 f  a  x  dx 

0 sin   x   cos   x  0 0
2  2 
 2


cos x
I dx  2
0
cos x  sin x
Adding 1 and  2  , we have
 2  2
sin x  cos x   
 
dx  1dx   x 0    0  
 2
2I 
0
cos x  sin x 0
 2  2
 2


sin x
I  dx  
0
cos x  sin x 4
(ii) Let

23
 CALCULUS AND ANALYTICAL GEOMETRY

 2  2  2
 sin  cos   sin   cos  
  
2 2

I ln  tan   cot   d  ln    d  ln   d
0 0
 cos sin   0
 sin  cos 
 2  2  2
 
   ln sin  ln cos  d
1
I ln   d  ln 1  ln  sin  cos   d  
0
 sin  cos  0 0

 2  2  2

 
I   ln sin  d  ln cos d  2 ln sin  d ,
0 0
0

 
 2  2  2
 
 since
 
0
ln cos d 
0
ln cos     d 
2   ln sin d 
0

  
I  2   ln 2    ln 2 
 2 

24
 CALCULUS AND ANALYTICAL GEOMETRY

IMPROPER INTEGRAL
EXAMPLE 01: Evaluate the following improper integrals
 

 x
1
i  xdx  ii  2
dx 
1 1

Solution:
 b
 x3 2  2
b

 
2
xdx  Lim  x  dx  Lim   3 Lim b3 2  13 2   Lim b3 2  1    Hence the given integral does not
12

b 
(i)
 1
b b b
1 1
3 2 3
exist.
 b b
 x 1  1 
 
1
   1    0  1  1
2
(ii) dx  Lim x dx  Lim
b     Lim
b 
b
 1 1 b 
2
1
x 1

Note: The integral evaluated in part (i) is said to be divergent, since it has no finite numerical value. By contrast, the
integral in part (ii) is said to be convergent, since it has a finite value. There is indeed a finite area under the curve.
b

Improper integrals of the type

 f  x  dx are defined and evaluated in similar manner. That is;

b b


 f  x  dx  Lim  f  x  dx,
m
m

provided that f is continuous on the interval  ,b  

2

 4  x
dx
EXAMPLE 02: Evaluate 2



Solution:
2
2 2
  4  x 1   1 
2
1 1  1
 4  x  4  x
dx 2
 Lim dx  Lim     Lim
a     Lim
a  
   
      a 6
2 a  a 
 a  1  a 4 x a 6 4

Improper integrals of the type

 f x dx are defined and evaluated in a way that involves both types of improper
integrals you have seen already.
 c 


 f  x  dx   f  x  dx   f  x  dx,
 c

provided that f is continuous on the interval  ,    The number c used in the two integrals can be any real
number, although 0 or 1 is often a good choice. If either (or both) of the two integrals diverges, then the original
integral also diverges. Otherwise, the original integral converges.

EXAMPLE 03: Evaluate the following improper integrals

 

  xe
x
i  dx  ii   x2
dx 
x 1
4
 

 x  1 dx  Substituting x  z  2xdx  dz  xdx  2 dz  Then

x 2 1
Solution: (i) Consider 4

 x  1 dx   z  1  2  z  1 dz  2 tan z  2 tan  x  
x 1 2 dz 1 1 1 1 1 1 2
4 2 2

Now,
0  0 b

   
x x x x
 dx  dx  Lim dx  Lim dx

x 1
4
0
x 1
4 a
a
x 1
4 b
0
x 1
4

25
 CALCULUS AND ANALYTICAL GEOMETRY

0 b
 1 1 2   1 1 2  1 1
 Lim
a 
tan x   Lim
b 
tan x   Lim  tan 1 0  tan 1 a 2   Lim  tan 1 b 2  tan 1 0 
 2 a  2 0 2 a 
2 b


1  1 
 x  1 dx   2 Lim tan
x 1 1
 1
a 2  Lim tan 1 b 2         0 

4 a 
2 b 2 2  2 2 

 xe
 x2 1
(ii) Consider dx  Substituting x 2  z  2 xdx  dz  xdx  dz  Then
2
1 2
 xe 
1 z 1
 x2
dx  e dz  e  z  1  e  x 
2 2 2
Now let
 0  0 b

I
 xe x dx 
  
xe  x dx  xe  x dx  Lim xe  x dx  Lim xe  x dx

2 2 2 2 2

a  b 
  0 a 0

 1    x2  0  1    x2  b 1
 Lim
a 
 2 
  e  a
 b 
Lim
 2
 e  0 
2
Lim
a
1
e0  e  a  Lim
2

2 b
e b  e0
2
   
1 1 1 1
 I   1  0    0  1     0 
2 2 2 2

APPLICATIONS OF INTEGRATION
EXAMPLE 01: A toy rocket is shot vertically upward from the ground with an initial velocity of 300feet per
second. The acceleration due to gravity is -32feet per second per second-negative because it is downward. (No
other acceleration is applied to the rocket.)
(a) Find a formula for the rocket’s velocity t seconds after the launch.
(b) Find a formula for the rocket’s distance above the ground at any time t.
Solution: (a) As we know that the acceleration is the derivative of velocity, so, a  dv / dt
The acceleration is given as -32, so we have
dv / dt  32
Integrating both sides with respect to t, we get
v
  32 dt  v  32t  C (1)
To determine C, use the fact that the initial velocity is 300 feet per second. This means v  300 when t = 0.
Substituting these two numbers into the equation v  32t  C yields
300  32  0   C  C  300.
Thus, we have
v  32t  300. using 1
(b) We also know that the velocity is the derivative of distance. That is;
v  ds / dt  32t  300.
Integrating both sides with respect to t, we get

s    32t  300dt  
32 2
t  300t  C  s  16t 2  300t  C (2)
2
To find C, note that at the beginning (when t = 0) the rocket’s distance s above the ground is zero, because it is shot
upward from the ground. Substituting 0 for t and 0 for s into (2), we get
0  16  0   300  0   C  C  0.
2

Thus, (2) becomes: s  16t 2  300t.

EXAMPLE 02: Let R  x  be the revenue a company receives from the sale of x units of its product. If its
marginal revenue R  x  is 100  0.2x dollars, determine:
(a) R  x  (b) the revenue from the sale of 20 units. Assume there is no revenue when zero units are sold.
Solution: (a) Since R  x   100  0.2 x.

26
 CALCULUS AND ANALYTICAL GEOMETRY

Integrating both sides with respect to x, we get


0.2 2
R  x   100  0.2 x  dx  100 x  x  C R  x   100 x  0.1x 2  C (1)
2
It is assumed that there is no revenue when zero units are sold, that is, when x  0 , R0  0. Therefore, from (1),
we get
0  100  0   0.1 0   C  0  0  0  C  C  0.
2

Thus, (1) becomes,

R  x   100 x  0.1x 2 .
(b) The revenue from the sale of 20 units will be
R  20   100  20   0.1 20   2000  40  R  20   $1960.
2

EXAMPLE 03: A ball is shot vertically upward from the edge of a building with initial velocity 352feet per
second. The building is 768feet tall. Acceleration due to gravity is -32feet per second per second.
(a) Determine the equations that describe the velocity of the ball and its distance from the ground.
(b) How far above the ground is the ball after 6 seconds, and how fast is it going then?
Solution: (a) We know that the acceleration is the derivative of velocity. Therefore,
a  dv / dt = -32 [since a  32feet/sec2 . ]
Integrating both sides, we get


v  32 1dt  v  32t  C (1)
To find C, use the fact that the initial velocity is 352 feet/sec. This means v  352 when t = 0. Thus, (1)
becomes, 352  32  0   C  C  352.
Thus, we have v  32t  352 (2)
Also, we know that velocity is the derivative of distance, that is;
ds ds
v or  32t  352, [from (2)]
dt dt
Integrating both sides, we get
t2

s   32t  352 dt  32  352t  C  s  16t 2  352t  C
2
(3)
To find C, note that at the beginning (when t=0) the ball’s distance s from the ground is zero, because it is
shot upward from the ground. Substituting 0 for t and 0 for s into  3 , we get
0  16  0   352  0   C  C  0.
2

Thus, s  16t 2  352t (4)

Hence the equation that describes the velocity of the ball is v  32t  352 and its distance from
the ground is s  16t 2  352t.

(b) At t = 6 sec, we have from (4), s  16  6   352  6   576  2112  s  1536 feet.
2

Also from (2), we have v  32  6   352  192  352  v  544 feet/sec.
EXAMPLE 04: A tourist accidentally drops his camera from the top of a cliff that is 576feet above the water
below. Assume the acceleration due to gravity to be -32feet per second per second.
(a) Determine the velocity v t  of the camera at any time t during its fall.
(b) Determine s(t), the height of the camera above the water at any time t during its fall.
(c) How fast is the camera falling 4 seconds after it is dropped?
(d) How long will it take the camera to hit the water? (Hint: What is the value of s when the camera hits
the water?)
Solution: (a) As we know that the acceleration is the derivative of velocity. That is;
a  dv / dt = - 32 [ since a  32feet/sec2 . ]
Integrating both sides, we get

27
 CALCULUS AND ANALYTICAL GEOMETRY


v  32 1dt  v  32t  C (1)

Initially, the velocity was zero, that is; when t = 0, v = 0, therefore 1 becomes,
0  32  0   C  C  0.
Again (1) becomes, v  32t or, v  t   32t (2)
(b) Also, we know that velocity is the derivative of distance, that is;
v  ds / dt   32 t [ from (2)]
t2
Integrating both sides, we get s  32  C  s  16t 2  C (3)
2
Since it is given that s = 576 feet, so at t = 0, (3) becomes, 576  16  0   C  C  576.
2

Again, (3) becomes , s  16t 2  576 (4)

(c) At t = 4 sec, (2) becomes, v  4   32  4   128.
Thus, the camera is being fallen 128 feet/sec fast, 4 seconds after it is dropped.
(d) Since when the camera hits the water, its distance will vanish, so from (4) we have
0  16t 2  576  t 2  36  t  6.
Thus, the camera will take 6seconds to hit the water.
EXAMPLE 05: A woman gets into her car and then drives it with a constant acceleration of 22 feet per
second per second.
(a) Determine the velocity function.
(b) Determine the distance function.
(c) How far does the car go in 6 seconds?
Solution: (a) We know that the acceleration is the derivative of velocity. That is;
a  dv / dt  22 [ since a  22feet/sec2 . ]
Integrating both sides with respect to t, we get
v
  22 dt  22t  C (1)

Initial velocity of the car was zero when t = 0, so (1) becomes, 0  22  0   C  C  0

Substituting C = 0 into(1), we get v  22t , which is the required velocity function.
(b) Also, we know that velocity is the derivative of distance, that is;
v  ds / dt  22 t .
Integrating both sides with respect to t, we get


s  22 tdt  11t 2  C  s  11t 2  C (2)

At the beginning, the car covers no distance, that is; when t=0, s=0, so 2 becomes
0  11 0   C  C  0.
2

Substituting C = 0 into  2  , we get s  11t 2 (3)

(c) At t = 6 sec, (3) becomes, s  11 6   11 36  396 feet. Thus, the car goes 396feet far in 6 seconds.
2

EXAMPLE 06: On the moon the magnitude of the acceleration due to gravity is less than on the earth; it is
approximately -5.3 feet per second per second. Consider a ball thrown upward from the surface of the moon
with a velocity of 120feet per second.
(a) Obtain a function that gives the velocity of the ball at any time t.
(b) Determine a function that shows the distance of the ball from the moon’s surface at any time t.
Solution: (a) We know that the acceleration is the derivative of velocity. That is;
a  dv / dt  5.3
Integrating both sides with respect to t, we get
v
  5.3 dt  5.31dt  v  5.3t  C (1)
Since initial velocity of the ball is 120feet/sec, that is; when t = 0, so (1) becomes,

28
 CALCULUS AND ANALYTICAL GEOMETRY

120  5.3  0   C  C  120.

Substituting C =120 into (1), we get v  5.3 t  120,
which is the required function that gives the velocity of the ball at any time t.
(b) Also, we know that velocity is the derivative of distance, that is;
v  ds / dt  5.3t  120
Integrating both sides with respect to t, we get
t2

s   5.3t  120  dt  5.3 120t  C  s  2.65t 2  120t  C
2
(2)
Initially the ball covers no distance, that is; when t = 0, s = 0. Substituting these values into (2), we get
0  2.65  0   120  0   C  C  0.
Now equation (2) becomes s  2.65t 2  120 t ,
which is the required function that shows the distance of the ball from the moon’s surface at any time t.
EXAMPLE 07: The height h (in feet) of a tree is a function of time t(in years). Suppose you begin
(t = 0) by planting a 5-foot tree in your yard and the tree grows to maturity according to the formula
dh 1
 0.3  , t  0.
dt t
(a) Determine a formula for the height of the tree at any time t.
(b) Find the height of the tree after 1year, 4years, 9years, and 16years.
dh 1
Solution: (a) Here  0.3 
dt t
Integrating both sides with respect to t, we get
 1 

t1 2
h   0.3  dt  0.3t   C  h  0.3t  2 t  C (1)
 t 12
At t = 0 and h = 5. Substituting these values into (1), we get 5  0.3  0   2 0  C  C  5.
Again, (1) becomes h  .3t  2 t  5 (2)
which is the required formula for the height of the tree at any time t.
(b) When t = 1, h  0.31  2 1  5  h  7.3 feet,
When t = 4, h  0.3 4   2 4  5  h  10.2 feet,
When t = 9, h  0.3 9   2 9  5  h  13.7 feet, and
When t = 16, h  0.316   2 16  5  h  17.8 feet.
EXAMPLE 08: Find the volume of the solid of revolution obtained by revolving the curve y  x 2 about the x
– axis between x  1 and x  3 
b
V     f  x  dx . Substituting the values, we get
2
Solution: Here
a
3 3
1 3 1
V   x  dx   x 4 dx     x5    35  15     242   48.4 
 
2 2 1
1 1
5 1 5 5
Thus, the volume of the solid of revolution obtained by revolving the curve y  x 2 about the
x – axis between x  1 and x  3 is 48.4 cubic units.
EXAMPLE 09: Find the volume of the solid of revolution obtained by revolving the curve y  1 / x about the
x – axis between x  1 and x  2 
b


Solution: Since V    f  x  dx . Substituting the values, we get
2

29
 CALCULUS AND ANALYTICAL GEOMETRY

2 2 2
x 1
2 2
1 1   1 1
 
1
V     dx   x 2 dx         1        
 x 1 1 x1 2   2 2
1 1

1
 cubic units.
Hence the volume is
2
EXAMPLE 10: Find the volume of the solid of revolution obtained by revolving the curve y  x 2  1 about
the x – axis on the interval 0,3 
b


Solution: Here V    f  x  dx . Substituting the values, we get
2

a
3 3 3

V    x 2  1 dx  
   x 4  2 x 2  1 dx    15 x5  23 x3  x 
2

0 0 0

1 5 2 3 
V     3   3  3  0     48.6  18  3  69.6 
5 3 
Hence the volume is 69.6 cubic units.

PRACTICE PROBLEMS
1. Evaluate the following:
ax 2  bx  c
3
 1
x  ax 2  bx  c  dx
i   x2
dx  ii    iii    x   dx
 x
3
 1   1  
 iv    v    a2  x2    x  x  x
1
 vi 
3
x  dx dx 2
  dx
 x x2 
 2 x  1
3
x3  3x 2  4
 vii    x 1  x  2  ix  
dx
dx  viii  dx
x 4x  4

 1  cos x dx  xi   sin x 1  cos 2 xdx  xii    ea ln x  e x ln a  dx

1
 x
 2  3
sin 2  2  3sin x
 xiii   1  x  1  xdx  xiv  0
1  cos 
d  xv  0
cos 2 x
dx

2. Evaluate the following:

e x  e x
   x ln x  x 1  x 
sin x dx dx
i  dx  ii  dx  iii   iv 
4  5cos x e x  e x

 3sin x  4sin x  vii  

dx dx dx
v  vi 
 2 x  3 1
3 2
x x a2 2

x5 e x
cos e  dx x

  
dx
 viii   ix  dx  x 
1  3sin x  8cos x
2 2
1 x 2
x
3. Evaluate the following:
ln 1  x 2  1 x
i
 x2 
dx  ii  x cot 1 xdx  iii  tan 1
1 x
dx

1  sin x sin  ln x 
 
 iv  x3e x dx  v  e x dx  vi 

2
dx
1  cos x x3
 vii   e sec x 1  tan x  dx
x
 viii   cos ln x  dx
7. Evaluate the following:

30
 CALCULUS AND ANALYTICAL GEOMETRY

2x  3
 2x  x x
dx x
i   ii  dx  iii  dx
 x 1  1  2 x  3  12 x  35
2 2 2

 2
3x  1
 1  cos x  2  cos x  dx  1  3e   x  1  x  3 dx
sin x dx
 iv  v  vi 
x
 2e 2 x 2
0
1
1  x2
  x  1  x  1  x  x
xdx xdx
 vii   viii  dx  ix 
 4 2 2  x2  1
2 4
0

Evaluate:
 2  2

 
cot x sin x
i  dx  ii  dx
0
cot x  tan x 0
sin x  cos x

31
 CALCULUS AND ANALYTICAL GEOMETRY

5. Show that:
 2 
2
 
x sin x
i  ln  tan x  dx  0  ii  dx 
1  cos 2 x 4
0 0

ln 1  x 
1 1

 iii  ln   1 dx  0
 
1
 iv  dx  ln 2 
x  1 x 2
8
0 0
6. Discuss the convergence of
  

 1  x    1  x 
x dx
i   ii  xe  x dx  iii 
2

3
dx
1 0 1
x
2  2

 
xdx cos x
 iv  v dx 
1
x 1 0
1  sin x
7. Evaluate:
   2  2  2

i   x3 e  x dx  ii   xe  x dx iii    iv   v  tan  d 

3 3
sin 3 x cos5 2 xdx sin 7 xdx
0 0 0 0 0

  
dx cos xdx
 x  xi  5  4 x  x 2 dx  xii 
2 x  3x  4
2
4sin x  4sin x  5
2

 x  1 x  2 
  x   2ax  x 
dx dx
 xiii  dx  xiv   xv 
x  x2 2 2
 1 x 2 32

32