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 )
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