Area And Volume Of The Region | Topic-wise Tests & Solved Examples for Mathematics PDF Download

 Page 1


 
 
     
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
 
                              
             S 
Introduction 
In geometry, we have learnt formulae to calculate area of various geometrical figures including triangle, rectangle, 
trapezium and circles. Such formulae are fundamental in the applications of mathematics to many fundamental in the 
applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate 
areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we 
shall need some concepts of Integral Calculus. 
 ua rature  y  ine  ntegral (Single  ntegral ) 
 he pro ess of fin ing the area of any  oun e  portion of a  urve is  alle  qua rature  
Area under Simple Curves 
In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral 
using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by 
the curve y = f(x), x-axis and the ordinates x = a and x = b. From fig. 8.1, we can think of area under the curve as 
composed of large number of very thin vertical strips cut by two lines at a distance x & x +dx from origin. Consider an 
arbitrary strip of height y and width dx, then dA (area of the elementary strip) = ydx, where y = f(x) by the formula of 
area of rectangle when strip is very-very thin. 
 This area is called the elementary area which is located at an arbitrary position within the region which is 
specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates  
x = a, x = b and the curve y = f(x) as the result of adding up the elementary area of thin strips across the region. 
Symbolically, we express 
  ?  
 
 
 ? y x
 
 
 ? f(x) x
 
 
 
The area A of the region bounded by the curve x = g(y), y-axis and the line y = c, y = d is given by 
  ? x y
 
 
 ? g(y) y
 
 
 
Here, we consider horizontal strips as shown in the fig. 
Remark: If the position of the curve under consideration is below the x-axis, then since f(x) < 0 from x = a to x = b, as 
shown in fig. 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is 
only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute  
value i e  |? f(x) x
 
 
| 
Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the  
fig      ere  
 
   an   
 
    herefore the area    oun e   y the  urve y f(x) x axis an  the or inates x a 
an  x     is given  y   | 
 
|  
 
  
Area between two curves 
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by cutting the region into a large 
number of small strips of elementary area and then adding up these elementary areas. Suppose we are given two 
curves represente   y y   f(x)  y   g(x)  where f(x) = g(x) in ,a  - as shown in fig 8.13. Here the points of intersection 
of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two 
curves.  
 For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. 
As indicated in the Fig. 8.13, elementary strip has height f(x) – g(x) and width dx so that elementary area 
   ,f(x) g(x)- x an  the total area    an  e taken as 
  ? ,f(x) g(x)- x
 
 
 
 lternatively  
  ,area  oun e   y y f(x) x axis an  the lines x a x  -  
,area  oun e   y y g(x) x axis an  the lines x a x  - 
 ? f(x) x
 
 
 ? g(x) x
 
 
 ? ,f(x) g(x)- x
 
 
 where f(x)= g(x) in ,a  - 
 f f(x)= g(x)in ,a  - an  f(x) g(x) in ,   - where a     as shown in the fig      then the area of the regions 
 oun e   y  urves  an  e written as  
Page 2


 
 
     
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
 
                              
             S 
Introduction 
In geometry, we have learnt formulae to calculate area of various geometrical figures including triangle, rectangle, 
trapezium and circles. Such formulae are fundamental in the applications of mathematics to many fundamental in the 
applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate 
areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we 
shall need some concepts of Integral Calculus. 
 ua rature  y  ine  ntegral (Single  ntegral ) 
 he pro ess of fin ing the area of any  oun e  portion of a  urve is  alle  qua rature  
Area under Simple Curves 
In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral 
using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by 
the curve y = f(x), x-axis and the ordinates x = a and x = b. From fig. 8.1, we can think of area under the curve as 
composed of large number of very thin vertical strips cut by two lines at a distance x & x +dx from origin. Consider an 
arbitrary strip of height y and width dx, then dA (area of the elementary strip) = ydx, where y = f(x) by the formula of 
area of rectangle when strip is very-very thin. 
 This area is called the elementary area which is located at an arbitrary position within the region which is 
specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates  
x = a, x = b and the curve y = f(x) as the result of adding up the elementary area of thin strips across the region. 
Symbolically, we express 
  ?  
 
 
 ? y x
 
 
 ? f(x) x
 
 
 
The area A of the region bounded by the curve x = g(y), y-axis and the line y = c, y = d is given by 
  ? x y
 
 
 ? g(y) y
 
 
 
Here, we consider horizontal strips as shown in the fig. 
Remark: If the position of the curve under consideration is below the x-axis, then since f(x) < 0 from x = a to x = b, as 
shown in fig. 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is 
only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute  
value i e  |? f(x) x
 
 
| 
Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the  
fig      ere  
 
   an   
 
    herefore the area    oun e   y the  urve y f(x) x axis an  the or inates x a 
an  x     is given  y   | 
 
|  
 
  
Area between two curves 
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by cutting the region into a large 
number of small strips of elementary area and then adding up these elementary areas. Suppose we are given two 
curves represente   y y   f(x)  y   g(x)  where f(x) = g(x) in ,a  - as shown in fig 8.13. Here the points of intersection 
of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two 
curves.  
 For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. 
As indicated in the Fig. 8.13, elementary strip has height f(x) – g(x) and width dx so that elementary area 
   ,f(x) g(x)- x an  the total area    an  e taken as 
  ? ,f(x) g(x)- x
 
 
 
 lternatively  
  ,area  oun e   y y f(x) x axis an  the lines x a x  -  
,area  oun e   y y g(x) x axis an  the lines x a x  - 
 ? f(x) x
 
 
 ? g(x) x
 
 
 ? ,f(x) g(x)- x
 
 
 where f(x)= g(x) in ,a  - 
 f f(x)= g(x)in ,a  - an  f(x) g(x) in ,   - where a     as shown in the fig      then the area of the regions 
 oun e   y  urves  an  e written as  
Free coa AM 
 
 otal area  rea of the region        rea of the region  P    
 ? ,f(x) g(x)- x
 
 
 ? ,g(x) f(x)- x
 
 
 
 reas of  uves given  y  artesian  quations  
 f  f(x) is a single value   an   ontinuous fun tion in the interval (a  ) then the area  oun e   y the  urve  
y f (x) the axis of  x an  the or inates at x a an  x   is 
                  ? f(x) x
 
 
                          or                       ? y x
 
 
 
 et     e the ar  of the  urve given  y the equation y f(x) an     an      e the two or inate at x a an  x    
respe tively   
 onsi er a point P(x y) on the  urve   et  (x  x y  y)  e a neigh ouring point to it   
 raw the or inates P  an     to the axis of x  
 hen P  y    y  y an      x   
 raw P 
 
an   P
 
perpen i ulars to    an   P pro u e  respe tively   
 et    enotes the area    P an        e the area        
 hen the area    P area      area    P 
 (    )      
 lso area    
 
P y  x an  area    P
 
 (y  y)  x Sin e the area    P lies  etween the areas P   
 
 
an     P
 
 therefore    lies  etween y x an  (y  y) x or     x lies  etween y an  (y  y)   
  s   P i e  x   an   y   we have  
  
 x
 y f(x)           or              x x 
 ntegrating  oth si es  etween the limits a to   we have  
? y x
 
 
 , -
 
 
 (area   when x  ) (area   when x a) 
or         ? y x
 
 
 area      
 hus area      ? y x
 
 
 ? f(x)
 
 
 x 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the a s issae y a an   
y   is given  y  
? x y
 
 
 ? f(y) y
 
 
 
 ou le  ntegration 
 n first  ase we  an also pro ee  as follows  et the area       e  ivi e  into su  areas  y  rawing lines  
parallel to x an  y  
axis respe tively an  the  istan e  etween two a joining lines parallel to x axis is  y an  parallel to y 
 axis is  x  
 et P(x y) an   (x  x y  y)  e two neigh ouring points on the given  urve    whose equation is  
y f (x)  et P  an      e the perpen i ulars to x axis  hen the area of the sha e  portion is  x  y   
 herefore the area of the strip P    is  
? x y
   ( )
   
 
 herefore the require  area      is ? ? x  y
   ( )
   
   
   
 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the  a s issae at y a  
an  y   is given  y   ? ? y  x
   ( )
   
   
   
 
 ules for  ra ing  artesian  urves  
  Symmetry   
(a)  f the equation of the  urve involves even an  only even powers of x then there is symmetry a out x axis  
( ) Similarly if the equation of the  urve involves  even an  only even powers of y then there is symmetry a out  
x axis  
( )  f the equation of the  urve involves even an  only even powers of x as well as of y then the  urve is symmetri al a out 
  oth the axes  
( )  f x  e  hange  into x an  y  e  hange  into y an  the equation remains un hange  then there is symmetry in 
opposites qua rants  
(e)  f x  e   hange  into y an  y  e  hange  into x an  the equation remains un hage  then the  urve is symmetri al a out 
the line y x  
Page 3


 
 
     
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
 
                              
             S 
Introduction 
In geometry, we have learnt formulae to calculate area of various geometrical figures including triangle, rectangle, 
trapezium and circles. Such formulae are fundamental in the applications of mathematics to many fundamental in the 
applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate 
areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we 
shall need some concepts of Integral Calculus. 
 ua rature  y  ine  ntegral (Single  ntegral ) 
 he pro ess of fin ing the area of any  oun e  portion of a  urve is  alle  qua rature  
Area under Simple Curves 
In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral 
using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by 
the curve y = f(x), x-axis and the ordinates x = a and x = b. From fig. 8.1, we can think of area under the curve as 
composed of large number of very thin vertical strips cut by two lines at a distance x & x +dx from origin. Consider an 
arbitrary strip of height y and width dx, then dA (area of the elementary strip) = ydx, where y = f(x) by the formula of 
area of rectangle when strip is very-very thin. 
 This area is called the elementary area which is located at an arbitrary position within the region which is 
specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates  
x = a, x = b and the curve y = f(x) as the result of adding up the elementary area of thin strips across the region. 
Symbolically, we express 
  ?  
 
 
 ? y x
 
 
 ? f(x) x
 
 
 
The area A of the region bounded by the curve x = g(y), y-axis and the line y = c, y = d is given by 
  ? x y
 
 
 ? g(y) y
 
 
 
Here, we consider horizontal strips as shown in the fig. 
Remark: If the position of the curve under consideration is below the x-axis, then since f(x) < 0 from x = a to x = b, as 
shown in fig. 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is 
only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute  
value i e  |? f(x) x
 
 
| 
Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the  
fig      ere  
 
   an   
 
    herefore the area    oun e   y the  urve y f(x) x axis an  the or inates x a 
an  x     is given  y   | 
 
|  
 
  
Area between two curves 
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by cutting the region into a large 
number of small strips of elementary area and then adding up these elementary areas. Suppose we are given two 
curves represente   y y   f(x)  y   g(x)  where f(x) = g(x) in ,a  - as shown in fig 8.13. Here the points of intersection 
of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two 
curves.  
 For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. 
As indicated in the Fig. 8.13, elementary strip has height f(x) – g(x) and width dx so that elementary area 
   ,f(x) g(x)- x an  the total area    an  e taken as 
  ? ,f(x) g(x)- x
 
 
 
 lternatively  
  ,area  oun e   y y f(x) x axis an  the lines x a x  -  
,area  oun e   y y g(x) x axis an  the lines x a x  - 
 ? f(x) x
 
 
 ? g(x) x
 
 
 ? ,f(x) g(x)- x
 
 
 where f(x)= g(x) in ,a  - 
 f f(x)= g(x)in ,a  - an  f(x) g(x) in ,   - where a     as shown in the fig      then the area of the regions 
 oun e   y  urves  an  e written as  
Free coa AM 
 
 otal area  rea of the region        rea of the region  P    
 ? ,f(x) g(x)- x
 
 
 ? ,g(x) f(x)- x
 
 
 
 reas of  uves given  y  artesian  quations  
 f  f(x) is a single value   an   ontinuous fun tion in the interval (a  ) then the area  oun e   y the  urve  
y f (x) the axis of  x an  the or inates at x a an  x   is 
                  ? f(x) x
 
 
                          or                       ? y x
 
 
 
 et     e the ar  of the  urve given  y the equation y f(x) an     an      e the two or inate at x a an  x    
respe tively   
 onsi er a point P(x y) on the  urve   et  (x  x y  y)  e a neigh ouring point to it   
 raw the or inates P  an     to the axis of x  
 hen P  y    y  y an      x   
 raw P 
 
an   P
 
perpen i ulars to    an   P pro u e  respe tively   
 et    enotes the area    P an        e the area        
 hen the area    P area      area    P 
 (    )      
 lso area    
 
P y  x an  area    P
 
 (y  y)  x Sin e the area    P lies  etween the areas P   
 
 
an     P
 
 therefore    lies  etween y x an  (y  y) x or     x lies  etween y an  (y  y)   
  s   P i e  x   an   y   we have  
  
 x
 y f(x)           or              x x 
 ntegrating  oth si es  etween the limits a to   we have  
? y x
 
 
 , -
 
 
 (area   when x  ) (area   when x a) 
or         ? y x
 
 
 area      
 hus area      ? y x
 
 
 ? f(x)
 
 
 x 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the a s issae y a an   
y   is given  y  
? x y
 
 
 ? f(y) y
 
 
 
 ou le  ntegration 
 n first  ase we  an also pro ee  as follows  et the area       e  ivi e  into su  areas  y  rawing lines  
parallel to x an  y  
axis respe tively an  the  istan e  etween two a joining lines parallel to x axis is  y an  parallel to y 
 axis is  x  
 et P(x y) an   (x  x y  y)  e two neigh ouring points on the given  urve    whose equation is  
y f (x)  et P  an      e the perpen i ulars to x axis  hen the area of the sha e  portion is  x  y   
 herefore the area of the strip P    is  
? x y
   ( )
   
 
 herefore the require  area      is ? ? x  y
   ( )
   
   
   
 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the  a s issae at y a  
an  y   is given  y   ? ? y  x
   ( )
   
   
   
 
 ules for  ra ing  artesian  urves  
  Symmetry   
(a)  f the equation of the  urve involves even an  only even powers of x then there is symmetry a out x axis  
( ) Similarly if the equation of the  urve involves  even an  only even powers of y then there is symmetry a out  
x axis  
( )  f the equation of the  urve involves even an  only even powers of x as well as of y then the  urve is symmetri al a out 
  oth the axes  
( )  f x  e  hange  into x an  y  e  hange  into y an  the equation remains un hange  then there is symmetry in 
opposites qua rants  
(e)  f x  e   hange  into y an  y  e  hange  into x an  the equation remains un hage  then the  urve is symmetri al a out 
the line y x  
Fr AM 
For more notes visit santoshifamily.com or call 8130648819 
  Points   
 o fin  the points where  the  urve   uts the x axis we put y   in the eqution of the  urve an  solve the resulting  
eqution for x Similarly the points of interse tion with the y axis are o taine   y putting x   an  solving the  
resulting equation for y  
     angents at origin   
 f origin is a point on the given  urve then equating to zero the lowest  egree terms o  urring in the equation of the 
 urve we get the tangent at the origin   
   angents at any point other than origin    
 o fin  the tangents at (h k) to  urve we shift the origin to (h k) an  thus transform the equation of the  urve  y  
putting x x h an  y y k an  then equating to zero the lowest   egree terms  we get the  
tangents at the new origin i e at (h k) 
   symptotes whi h are parallel to the axes  
 o fin  the asymptotes  whi h are parallel to the axes we first  etermine the  egree of the equations of the  urve   
 ssume the  egree of the equation is n  ow if the given equation  ontains the terms x
 
an  y
 
 then there is no 
asymptote parallel to the axes  
 f the given equation  oes  ontain x
 
  then equate to zero the total  oeffi ient of the next lower  egr ee power 
of x  t may happen that the equation  oes not  ontain the terms x
   
 then equate to zero the next lower  
 egree power of x i e x
   
 an  we shall get the asymptotes parallel to x axis   
Similarly we  an fin  asymptotes parallel to y axis   
   e gion   
 f for some value of x greater than some quantity say a the  orrespon ing values of y  ome out to  e imaginary then no 
part of the  urve will lie  eyon  x a 
Similarly if for some value of y greater than some quantity say   the  orrespon ing values of x  ome out to  
 e imaginary then no part of the  urve will lie  eyon  y     
 ules for  ra ing Polar  urves 
  Symmetry   
 f    e  hange  into   an  the equation of the given  urve remains un hage  then there is symmetry  
a out the initial line   
 f r  e  hange  into r an  the equation of the given  urve remains un hange  then there is symmetry a out the pole  
  Points   
 he  urve will pass through the pole if for some value of   the value of r  omes to  e zero   
 e form a ta le of  orrespon ing values of r an    whi h give us a num er of points on the  urve Plotting 
these points we get the require   urve   
 n the polar equations in whi h only perio i  fun tions (sin   os  tan ) et  o  ur the values of   from  
  to    nee   e  onsi ere  as the remaining values of    o not give any new  ran h of the  urve   
   egion   
 f r is imaginary when       then the  urve  oes not exist in the region  oun e   y the lines     an   
     
   symptotes   
 in  the asymptotes of the  urve  y the metho  as alrea y mentione  a ove   
   angents   
 in tan i e r(
  
 r
)  t will in i ate the  ire tion of trangent at any point of the  urve   
 
 x    in  the area en lose   y the ellipse (
x
 
a
 
) (
y
 
 
 
)   
Solution   he equation of the  urve is 
x
 
a
 
 
y
 
 
 
   
or             y
 
  
 
(  
x
 
a
 
) 
 
 
a
 
(a
 
 x
 
) 
or              y (
 
a
)v(a
 
 x
 
) 
Sin e the  urve is symmetri al a out  oth the axes therefore the require  area is four times the area in first 
qua rant   
hen e the require  area  ? y x
 
 
               taking verti al strip 
  ?
 
a
v(a
 
 x
 
) x
 
 
 
 
  
a
 
 
 
0 
v
 
 
  
 
  
 
sin
  
 
 
1
 
 
           
 
  
a
,     
 
sin
  
     - 
  
a
0 
 
 
 
 
1  a   
 
Page 4


 
 
     
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
 
                              
             S 
Introduction 
In geometry, we have learnt formulae to calculate area of various geometrical figures including triangle, rectangle, 
trapezium and circles. Such formulae are fundamental in the applications of mathematics to many fundamental in the 
applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate 
areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we 
shall need some concepts of Integral Calculus. 
 ua rature  y  ine  ntegral (Single  ntegral ) 
 he pro ess of fin ing the area of any  oun e  portion of a  urve is  alle  qua rature  
Area under Simple Curves 
In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral 
using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by 
the curve y = f(x), x-axis and the ordinates x = a and x = b. From fig. 8.1, we can think of area under the curve as 
composed of large number of very thin vertical strips cut by two lines at a distance x & x +dx from origin. Consider an 
arbitrary strip of height y and width dx, then dA (area of the elementary strip) = ydx, where y = f(x) by the formula of 
area of rectangle when strip is very-very thin. 
 This area is called the elementary area which is located at an arbitrary position within the region which is 
specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates  
x = a, x = b and the curve y = f(x) as the result of adding up the elementary area of thin strips across the region. 
Symbolically, we express 
  ?  
 
 
 ? y x
 
 
 ? f(x) x
 
 
 
The area A of the region bounded by the curve x = g(y), y-axis and the line y = c, y = d is given by 
  ? x y
 
 
 ? g(y) y
 
 
 
Here, we consider horizontal strips as shown in the fig. 
Remark: If the position of the curve under consideration is below the x-axis, then since f(x) < 0 from x = a to x = b, as 
shown in fig. 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is 
only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute  
value i e  |? f(x) x
 
 
| 
Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the  
fig      ere  
 
   an   
 
    herefore the area    oun e   y the  urve y f(x) x axis an  the or inates x a 
an  x     is given  y   | 
 
|  
 
  
Area between two curves 
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by cutting the region into a large 
number of small strips of elementary area and then adding up these elementary areas. Suppose we are given two 
curves represente   y y   f(x)  y   g(x)  where f(x) = g(x) in ,a  - as shown in fig 8.13. Here the points of intersection 
of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two 
curves.  
 For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. 
As indicated in the Fig. 8.13, elementary strip has height f(x) – g(x) and width dx so that elementary area 
   ,f(x) g(x)- x an  the total area    an  e taken as 
  ? ,f(x) g(x)- x
 
 
 
 lternatively  
  ,area  oun e   y y f(x) x axis an  the lines x a x  -  
,area  oun e   y y g(x) x axis an  the lines x a x  - 
 ? f(x) x
 
 
 ? g(x) x
 
 
 ? ,f(x) g(x)- x
 
 
 where f(x)= g(x) in ,a  - 
 f f(x)= g(x)in ,a  - an  f(x) g(x) in ,   - where a     as shown in the fig      then the area of the regions 
 oun e   y  urves  an  e written as  
Free coa AM 
 
 otal area  rea of the region        rea of the region  P    
 ? ,f(x) g(x)- x
 
 
 ? ,g(x) f(x)- x
 
 
 
 reas of  uves given  y  artesian  quations  
 f  f(x) is a single value   an   ontinuous fun tion in the interval (a  ) then the area  oun e   y the  urve  
y f (x) the axis of  x an  the or inates at x a an  x   is 
                  ? f(x) x
 
 
                          or                       ? y x
 
 
 
 et     e the ar  of the  urve given  y the equation y f(x) an     an      e the two or inate at x a an  x    
respe tively   
 onsi er a point P(x y) on the  urve   et  (x  x y  y)  e a neigh ouring point to it   
 raw the or inates P  an     to the axis of x  
 hen P  y    y  y an      x   
 raw P 
 
an   P
 
perpen i ulars to    an   P pro u e  respe tively   
 et    enotes the area    P an        e the area        
 hen the area    P area      area    P 
 (    )      
 lso area    
 
P y  x an  area    P
 
 (y  y)  x Sin e the area    P lies  etween the areas P   
 
 
an     P
 
 therefore    lies  etween y x an  (y  y) x or     x lies  etween y an  (y  y)   
  s   P i e  x   an   y   we have  
  
 x
 y f(x)           or              x x 
 ntegrating  oth si es  etween the limits a to   we have  
? y x
 
 
 , -
 
 
 (area   when x  ) (area   when x a) 
or         ? y x
 
 
 area      
 hus area      ? y x
 
 
 ? f(x)
 
 
 x 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the a s issae y a an   
y   is given  y  
? x y
 
 
 ? f(y) y
 
 
 
 ou le  ntegration 
 n first  ase we  an also pro ee  as follows  et the area       e  ivi e  into su  areas  y  rawing lines  
parallel to x an  y  
axis respe tively an  the  istan e  etween two a joining lines parallel to x axis is  y an  parallel to y 
 axis is  x  
 et P(x y) an   (x  x y  y)  e two neigh ouring points on the given  urve    whose equation is  
y f (x)  et P  an      e the perpen i ulars to x axis  hen the area of the sha e  portion is  x  y   
 herefore the area of the strip P    is  
? x y
   ( )
   
 
 herefore the require  area      is ? ? x  y
   ( )
   
   
   
 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the  a s issae at y a  
an  y   is given  y   ? ? y  x
   ( )
   
   
   
 
 ules for  ra ing  artesian  urves  
  Symmetry   
(a)  f the equation of the  urve involves even an  only even powers of x then there is symmetry a out x axis  
( ) Similarly if the equation of the  urve involves  even an  only even powers of y then there is symmetry a out  
x axis  
( )  f the equation of the  urve involves even an  only even powers of x as well as of y then the  urve is symmetri al a out 
  oth the axes  
( )  f x  e  hange  into x an  y  e  hange  into y an  the equation remains un hange  then there is symmetry in 
opposites qua rants  
(e)  f x  e   hange  into y an  y  e  hange  into x an  the equation remains un hage  then the  urve is symmetri al a out 
the line y x  
Fr AM 
For more notes visit santoshifamily.com or call 8130648819 
  Points   
 o fin  the points where  the  urve   uts the x axis we put y   in the eqution of the  urve an  solve the resulting  
eqution for x Similarly the points of interse tion with the y axis are o taine   y putting x   an  solving the  
resulting equation for y  
     angents at origin   
 f origin is a point on the given  urve then equating to zero the lowest  egree terms o  urring in the equation of the 
 urve we get the tangent at the origin   
   angents at any point other than origin    
 o fin  the tangents at (h k) to  urve we shift the origin to (h k) an  thus transform the equation of the  urve  y  
putting x x h an  y y k an  then equating to zero the lowest   egree terms  we get the  
tangents at the new origin i e at (h k) 
   symptotes whi h are parallel to the axes  
 o fin  the asymptotes  whi h are parallel to the axes we first  etermine the  egree of the equations of the  urve   
 ssume the  egree of the equation is n  ow if the given equation  ontains the terms x
 
an  y
 
 then there is no 
asymptote parallel to the axes  
 f the given equation  oes  ontain x
 
  then equate to zero the total  oeffi ient of the next lower  egr ee power 
of x  t may happen that the equation  oes not  ontain the terms x
   
 then equate to zero the next lower  
 egree power of x i e x
   
 an  we shall get the asymptotes parallel to x axis   
Similarly we  an fin  asymptotes parallel to y axis   
   e gion   
 f for some value of x greater than some quantity say a the  orrespon ing values of y  ome out to  e imaginary then no 
part of the  urve will lie  eyon  x a 
Similarly if for some value of y greater than some quantity say   the  orrespon ing values of x  ome out to  
 e imaginary then no part of the  urve will lie  eyon  y     
 ules for  ra ing Polar  urves 
  Symmetry   
 f    e  hange  into   an  the equation of the given  urve remains un hage  then there is symmetry  
a out the initial line   
 f r  e  hange  into r an  the equation of the given  urve remains un hange  then there is symmetry a out the pole  
  Points   
 he  urve will pass through the pole if for some value of   the value of r  omes to  e zero   
 e form a ta le of  orrespon ing values of r an    whi h give us a num er of points on the  urve Plotting 
these points we get the require   urve   
 n the polar equations in whi h only perio i  fun tions (sin   os  tan ) et  o  ur the values of   from  
  to    nee   e  onsi ere  as the remaining values of    o not give any new  ran h of the  urve   
   egion   
 f r is imaginary when       then the  urve  oes not exist in the region  oun e   y the lines     an   
     
   symptotes   
 in  the asymptotes of the  urve  y the metho  as alrea y mentione  a ove   
   angents   
 in tan i e r(
  
 r
)  t will in i ate the  ire tion of trangent at any point of the  urve   
 
 x    in  the area en lose   y the ellipse (
x
 
a
 
) (
y
 
 
 
)   
Solution   he equation of the  urve is 
x
 
a
 
 
y
 
 
 
   
or             y
 
  
 
(  
x
 
a
 
) 
 
 
a
 
(a
 
 x
 
) 
or              y (
 
a
)v(a
 
 x
 
) 
Sin e the  urve is symmetri al a out  oth the axes therefore the require  area is four times the area in first 
qua rant   
hen e the require  area  ? y x
 
 
               taking verti al strip 
  ?
 
a
v(a
 
 x
 
) x
 
 
 
 
  
a
 
 
 
0 
v
 
 
  
 
  
 
sin
  
 
 
1
 
 
           
 
  
a
,     
 
sin
  
     - 
  
a
0 
 
 
 
 
1  a   
 
Free coach AM 
 
 x    in  the area  oun e   y the para ola y
 
  ax an  its latus re tum   
Solution  he equation of the latus re tum is x a   
 he given  urve is symmetri al a out x axis 
  he require  area    
 
 
    rea  S   ? y x
 
 
  ? v( ax ) x
 
 
    va 
 
 
(x
 
 
)
 
 
 
 
 
a
 
  
 x   Sket h roughly an  prove that the area of its loop ay
 
 x
 
(a x) is
 
  
a
 
  
Solution  he require  area of the loop  ? y x
 
 
  ?
xv(a x)
va
 x
 
 
 
 
 ?
asin
 
 va os 
va
 
 
 
 a sin  os    
(Put x asin
 
   x  a sin  os   ) 
 a
 
? sin
 
  os
 
   
 
 
 
 
 
 
  
a
 
  
use? sin
 
x os
 
x x
 
 
 
 
 .
m  
 
/ .
n  
 
/
  .
m n  
 
/
 
 
 
 (
m  
 
 
n  
 
) 
 x    in  the area of the loop of the  urve y
 
 x(x  )
 
 
Solution  he given  urve is symmetri al a out x axis   
Putting y   we get x   an  x     
 herefore the loop is forme   etween x   an  x     
 en e the require  area  
  ? y x
 
 
  
  ? (x  )
vx x
 
 
 
  ? (x
 
 
 x
 
 
)
 
 
 x 
 
  
(numeri ally)           (verify) 
 x   Sket h roughly an  fin  the whole area of the  urve a
 
y
 
 x
 
( a x) 
Solution  he  urve is symmetri al a out x axis 
putting y   we get x   an  x  a 
 he require  area 
  ? y x
  
 
  ?
x
   
v( a x)
a
 x
  
 
 
 
 
a
? ( a sin
 
 )
   
   
 
v( a )v(  sin
 
 ) a sin  os     
(Put x  a sin
 
   x  a sin  os   ) 
   a
 
? sin
 
 
 
 
 
 os
 
     a
 
                              (verify) 
Question framing, replace x by y 
 x    in  the whole area of the  urve a
 
x
 
 y
 
( a y) 
Solution  he  urve is symmetri al a out y axis 
 
 
Page 5


 
 
     
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
  
 
 
 
 
 
 
  
 
  
 
 
 
 
 
 
 
                              
             S 
Introduction 
In geometry, we have learnt formulae to calculate area of various geometrical figures including triangle, rectangle, 
trapezium and circles. Such formulae are fundamental in the applications of mathematics to many fundamental in the 
applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate 
areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we 
shall need some concepts of Integral Calculus. 
 ua rature  y  ine  ntegral (Single  ntegral ) 
 he pro ess of fin ing the area of any  oun e  portion of a  urve is  alle  qua rature  
Area under Simple Curves 
In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral 
using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by 
the curve y = f(x), x-axis and the ordinates x = a and x = b. From fig. 8.1, we can think of area under the curve as 
composed of large number of very thin vertical strips cut by two lines at a distance x & x +dx from origin. Consider an 
arbitrary strip of height y and width dx, then dA (area of the elementary strip) = ydx, where y = f(x) by the formula of 
area of rectangle when strip is very-very thin. 
 This area is called the elementary area which is located at an arbitrary position within the region which is 
specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates  
x = a, x = b and the curve y = f(x) as the result of adding up the elementary area of thin strips across the region. 
Symbolically, we express 
  ?  
 
 
 ? y x
 
 
 ? f(x) x
 
 
 
The area A of the region bounded by the curve x = g(y), y-axis and the line y = c, y = d is given by 
  ? x y
 
 
 ? g(y) y
 
 
 
Here, we consider horizontal strips as shown in the fig. 
Remark: If the position of the curve under consideration is below the x-axis, then since f(x) < 0 from x = a to x = b, as 
shown in fig. 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is 
only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute  
value i e  |? f(x) x
 
 
| 
Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the  
fig      ere  
 
   an   
 
    herefore the area    oun e   y the  urve y f(x) x axis an  the or inates x a 
an  x     is given  y   | 
 
|  
 
  
Area between two curves 
Intuitively, true in the sense of Leibnitz, integration is the act of calculating the area by cutting the region into a large 
number of small strips of elementary area and then adding up these elementary areas. Suppose we are given two 
curves represente   y y   f(x)  y   g(x)  where f(x) = g(x) in ,a  - as shown in fig 8.13. Here the points of intersection 
of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two 
curves.  
 For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. 
As indicated in the Fig. 8.13, elementary strip has height f(x) – g(x) and width dx so that elementary area 
   ,f(x) g(x)- x an  the total area    an  e taken as 
  ? ,f(x) g(x)- x
 
 
 
 lternatively  
  ,area  oun e   y y f(x) x axis an  the lines x a x  -  
,area  oun e   y y g(x) x axis an  the lines x a x  - 
 ? f(x) x
 
 
 ? g(x) x
 
 
 ? ,f(x) g(x)- x
 
 
 where f(x)= g(x) in ,a  - 
 f f(x)= g(x)in ,a  - an  f(x) g(x) in ,   - where a     as shown in the fig      then the area of the regions 
 oun e   y  urves  an  e written as  
Free coa AM 
 
 otal area  rea of the region        rea of the region  P    
 ? ,f(x) g(x)- x
 
 
 ? ,g(x) f(x)- x
 
 
 
 reas of  uves given  y  artesian  quations  
 f  f(x) is a single value   an   ontinuous fun tion in the interval (a  ) then the area  oun e   y the  urve  
y f (x) the axis of  x an  the or inates at x a an  x   is 
                  ? f(x) x
 
 
                          or                       ? y x
 
 
 
 et     e the ar  of the  urve given  y the equation y f(x) an     an      e the two or inate at x a an  x    
respe tively   
 onsi er a point P(x y) on the  urve   et  (x  x y  y)  e a neigh ouring point to it   
 raw the or inates P  an     to the axis of x  
 hen P  y    y  y an      x   
 raw P 
 
an   P
 
perpen i ulars to    an   P pro u e  respe tively   
 et    enotes the area    P an        e the area        
 hen the area    P area      area    P 
 (    )      
 lso area    
 
P y  x an  area    P
 
 (y  y)  x Sin e the area    P lies  etween the areas P   
 
 
an     P
 
 therefore    lies  etween y x an  (y  y) x or     x lies  etween y an  (y  y)   
  s   P i e  x   an   y   we have  
  
 x
 y f(x)           or              x x 
 ntegrating  oth si es  etween the limits a to   we have  
? y x
 
 
 , -
 
 
 (area   when x  ) (area   when x a) 
or         ? y x
 
 
 area      
 hus area      ? y x
 
 
 ? f(x)
 
 
 x 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the a s issae y a an   
y   is given  y  
? x y
 
 
 ? f(y) y
 
 
 
 ou le  ntegration 
 n first  ase we  an also pro ee  as follows  et the area       e  ivi e  into su  areas  y  rawing lines  
parallel to x an  y  
axis respe tively an  the  istan e  etween two a joining lines parallel to x axis is  y an  parallel to y 
 axis is  x  
 et P(x y) an   (x  x y  y)  e two neigh ouring points on the given  urve    whose equation is  
y f (x)  et P  an      e the perpen i ulars to x axis  hen the area of the sha e  portion is  x  y   
 herefore the area of the strip P    is  
? x y
   ( )
   
 
 herefore the require  area      is ? ? x  y
   ( )
   
   
   
 
Similarly we  an prove that the area  oun e   y the  urve x f(y) the y axis an  the  a s issae at y a  
an  y   is given  y   ? ? y  x
   ( )
   
   
   
 
 ules for  ra ing  artesian  urves  
  Symmetry   
(a)  f the equation of the  urve involves even an  only even powers of x then there is symmetry a out x axis  
( ) Similarly if the equation of the  urve involves  even an  only even powers of y then there is symmetry a out  
x axis  
( )  f the equation of the  urve involves even an  only even powers of x as well as of y then the  urve is symmetri al a out 
  oth the axes  
( )  f x  e  hange  into x an  y  e  hange  into y an  the equation remains un hange  then there is symmetry in 
opposites qua rants  
(e)  f x  e   hange  into y an  y  e  hange  into x an  the equation remains un hage  then the  urve is symmetri al a out 
the line y x  
Fr AM 
For more notes visit santoshifamily.com or call 8130648819 
  Points   
 o fin  the points where  the  urve   uts the x axis we put y   in the eqution of the  urve an  solve the resulting  
eqution for x Similarly the points of interse tion with the y axis are o taine   y putting x   an  solving the  
resulting equation for y  
     angents at origin   
 f origin is a point on the given  urve then equating to zero the lowest  egree terms o  urring in the equation of the 
 urve we get the tangent at the origin   
   angents at any point other than origin    
 o fin  the tangents at (h k) to  urve we shift the origin to (h k) an  thus transform the equation of the  urve  y  
putting x x h an  y y k an  then equating to zero the lowest   egree terms  we get the  
tangents at the new origin i e at (h k) 
   symptotes whi h are parallel to the axes  
 o fin  the asymptotes  whi h are parallel to the axes we first  etermine the  egree of the equations of the  urve   
 ssume the  egree of the equation is n  ow if the given equation  ontains the terms x
 
an  y
 
 then there is no 
asymptote parallel to the axes  
 f the given equation  oes  ontain x
 
  then equate to zero the total  oeffi ient of the next lower  egr ee power 
of x  t may happen that the equation  oes not  ontain the terms x
   
 then equate to zero the next lower  
 egree power of x i e x
   
 an  we shall get the asymptotes parallel to x axis   
Similarly we  an fin  asymptotes parallel to y axis   
   e gion   
 f for some value of x greater than some quantity say a the  orrespon ing values of y  ome out to  e imaginary then no 
part of the  urve will lie  eyon  x a 
Similarly if for some value of y greater than some quantity say   the  orrespon ing values of x  ome out to  
 e imaginary then no part of the  urve will lie  eyon  y     
 ules for  ra ing Polar  urves 
  Symmetry   
 f    e  hange  into   an  the equation of the given  urve remains un hage  then there is symmetry  
a out the initial line   
 f r  e  hange  into r an  the equation of the given  urve remains un hange  then there is symmetry a out the pole  
  Points   
 he  urve will pass through the pole if for some value of   the value of r  omes to  e zero   
 e form a ta le of  orrespon ing values of r an    whi h give us a num er of points on the  urve Plotting 
these points we get the require   urve   
 n the polar equations in whi h only perio i  fun tions (sin   os  tan ) et  o  ur the values of   from  
  to    nee   e  onsi ere  as the remaining values of    o not give any new  ran h of the  urve   
   egion   
 f r is imaginary when       then the  urve  oes not exist in the region  oun e   y the lines     an   
     
   symptotes   
 in  the asymptotes of the  urve  y the metho  as alrea y mentione  a ove   
   angents   
 in tan i e r(
  
 r
)  t will in i ate the  ire tion of trangent at any point of the  urve   
 
 x    in  the area en lose   y the ellipse (
x
 
a
 
) (
y
 
 
 
)   
Solution   he equation of the  urve is 
x
 
a
 
 
y
 
 
 
   
or             y
 
  
 
(  
x
 
a
 
) 
 
 
a
 
(a
 
 x
 
) 
or              y (
 
a
)v(a
 
 x
 
) 
Sin e the  urve is symmetri al a out  oth the axes therefore the require  area is four times the area in first 
qua rant   
hen e the require  area  ? y x
 
 
               taking verti al strip 
  ?
 
a
v(a
 
 x
 
) x
 
 
 
 
  
a
 
 
 
0 
v
 
 
  
 
  
 
sin
  
 
 
1
 
 
           
 
  
a
,     
 
sin
  
     - 
  
a
0 
 
 
 
 
1  a   
 
Free coach AM 
 
 x    in  the area  oun e   y the para ola y
 
  ax an  its latus re tum   
Solution  he equation of the latus re tum is x a   
 he given  urve is symmetri al a out x axis 
  he require  area    
 
 
    rea  S   ? y x
 
 
  ? v( ax ) x
 
 
    va 
 
 
(x
 
 
)
 
 
 
 
 
a
 
  
 x   Sket h roughly an  prove that the area of its loop ay
 
 x
 
(a x) is
 
  
a
 
  
Solution  he require  area of the loop  ? y x
 
 
  ?
xv(a x)
va
 x
 
 
 
 
 ?
asin
 
 va os 
va
 
 
 
 a sin  os    
(Put x asin
 
   x  a sin  os   ) 
 a
 
? sin
 
  os
 
   
 
 
 
 
 
 
  
a
 
  
use? sin
 
x os
 
x x
 
 
 
 
 .
m  
 
/ .
n  
 
/
  .
m n  
 
/
 
 
 
 (
m  
 
 
n  
 
) 
 x    in  the area of the loop of the  urve y
 
 x(x  )
 
 
Solution  he given  urve is symmetri al a out x axis   
Putting y   we get x   an  x     
 herefore the loop is forme   etween x   an  x     
 en e the require  area  
  ? y x
 
 
  
  ? (x  )
vx x
 
 
 
  ? (x
 
 
 x
 
 
)
 
 
 x 
 
  
(numeri ally)           (verify) 
 x   Sket h roughly an  fin  the whole area of the  urve a
 
y
 
 x
 
( a x) 
Solution  he  urve is symmetri al a out x axis 
putting y   we get x   an  x  a 
 he require  area 
  ? y x
  
 
  ?
x
   
v( a x)
a
 x
  
 
 
 
 
a
? ( a sin
 
 )
   
   
 
v( a )v(  sin
 
 ) a sin  os     
(Put x  a sin
 
   x  a sin  os   ) 
   a
 
? sin
 
 
 
 
 
 os
 
     a
 
                              (verify) 
Question framing, replace x by y 
 x    in  the whole area of the  urve a
 
x
 
 y
 
( a y) 
Solution  he  urve is symmetri al a out y axis 
 
 
Free co 
 
Putting x   we get y   an  y  a  
 en e the  urve is forme   etween (   ) an  (   a )   
 o  urve will lie  eyon  y  a an   elow y     
  he require  area  
  ? x y
  
 
 
  ?
y
   
v( a y)
a
 y
  
 
   
 
 
a
? ( a sin
 
 )
   
v( a  a sin
 
 ) a sin  os   
   
 
           (Put y  a sin
 
    y  a sin  os   ) 
   a
 
? sin
 
  os
 
     a
 
 
 
 
                 (verify)  
 x   Sket h roughly the  urve a
 
y
 
 a
 
x
 
 x
 
an  fin  the whole area within it   
Sol
 
   he require  area  
    area of half a loop  
   ? y x
 
 
 
   ?
 v(a
 
 x
 
)
a
 
 
 x 
 
a
 ? xv(a
 
 x
 
) x
 
 
 
 
 
 a
 
( erify ) 
 
 x    in  the area  oun e   y the   urve xy
 
  a
 
( a x) an  its asymptote  
Solution   he given  urve is symmetri al a out the x axis an   uts the x axis at the point ( a  )  
 quating to zero the  oeff ient of  highest powers of y in the equation of the  urve we get x   i e the  
y axis is the asymptote   
 en e the require  area  
  ? y x
  
 
 
  ?  
 a v( a x) 
vx
   x
  
 
 
  a ?
v a  a sin
 
   a sin  os 
 v( a )sin
 
  
  /
 
                             (put x  a sin
 
    x  a sin   ons    ) 
   a
 
  ? os
 
     a
 
  /
 
                                 (verify)                     
 
 
 x   Prove that the area of a loop of the  urve a
 
y
 
 x
 
(a
 
 x
 
) is
 a
 
 
   
Sol
 
   he  urve is symmetri al a out  oth the axes Putting y   we get x   x  a  hus the limits for a  
loop varies from   to a   
 en e the requaire  area  
  ? y x
 
 
 
  ?
x
 
v(a
 
 x
 
)
a
 
  x
 
 
 
 
 
a
 
? a
 
sin
 
 
  /
 
 v(a
 
 a
 
sin
 
 ) a os                 (Put x asin    x a os   ) 
  a
 
? sin
 
 
  /
 
 os
 
    
 a
 
 
                                       (verify) 
 x     in  the area  etween the  urve y
 
 (a x) x
 
 an  its asymptotes   lso fin  the ratio in whi h the 
or inate x a /  ivi es this area   
Sol
 
   he  urve is symmetri al a out the x axis  quating to zero the  oeffi ent of the highest power of y  
we get a x    i e x a  is an asymptote of the  urve parallel to the y axis   
 et    e the whole area  etween x   an  x a /  hen  
   ? y x
 
 
  ?
x
  /
v(a x)
 
 
 x 
   ?
a
  /
sin
 
  a sin  os 
v(a) os 
  /
 
                   (putting x asin
 
     x  a sin  os   )