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1 | 4118-4121 | 2
(
)
px
q
x
a
+
−
=
2
A
B
(
)
x
a
x
a
+
−
−
3 2
(
) (
) (
)
px
qx
r
x
a
x
b
x
c
+
+
−
−
−
=
A
B
C
x
a
x
b
x
c
+
+
−
−
−
4 2
2
(
) (
)
px
qx
r
x
a
x
b
+
+
−
−
=
2
A
B
C
(
)
x
a
x
b
x
a
+
+
−
−
−
5 2
2
(
) (
)
px
qx
r
x
a
x
bx
c
+
+
−
+
+
=
2
A
B + C
x
x
a
x
bx
c
+
−
+
+
where x2 + bx + c can not be factorised further |
1 | 4119-4122 | 2
(
) (
) (
)
px
qx
r
x
a
x
b
x
c
+
+
−
−
−
=
A
B
C
x
a
x
b
x
c
+
+
−
−
−
4 2
2
(
) (
)
px
qx
r
x
a
x
b
+
+
−
−
=
2
A
B
C
(
)
x
a
x
b
x
a
+
+
−
−
−
5 2
2
(
) (
)
px
qx
r
x
a
x
bx
c
+
+
−
+
+
=
2
A
B + C
x
x
a
x
bx
c
+
−
+
+
where x2 + bx + c can not be factorised further � Integration by substitution
A change in the variable of integration often reduces an integral to one of the
fundamental integrals |
1 | 4120-4123 | 2
2
(
) (
)
px
qx
r
x
a
x
b
+
+
−
−
=
2
A
B
C
(
)
x
a
x
b
x
a
+
+
−
−
−
5 2
2
(
) (
)
px
qx
r
x
a
x
bx
c
+
+
−
+
+
=
2
A
B + C
x
x
a
x
bx
c
+
−
+
+
where x2 + bx + c can not be factorised further � Integration by substitution
A change in the variable of integration often reduces an integral to one of the
fundamental integrals The method in which we change the variable to some
other variable is called the method of substitution |
1 | 4121-4124 | 2
2
(
) (
)
px
qx
r
x
a
x
bx
c
+
+
−
+
+
=
2
A
B + C
x
x
a
x
bx
c
+
−
+
+
where x2 + bx + c can not be factorised further � Integration by substitution
A change in the variable of integration often reduces an integral to one of the
fundamental integrals The method in which we change the variable to some
other variable is called the method of substitution When the integrand involves
some trigonometric functions, we use some well known identities to find the
integrals |
1 | 4122-4125 | � Integration by substitution
A change in the variable of integration often reduces an integral to one of the
fundamental integrals The method in which we change the variable to some
other variable is called the method of substitution When the integrand involves
some trigonometric functions, we use some well known identities to find the
integrals Using substitution technique, we obtain the following standard
integrals |
1 | 4123-4126 | The method in which we change the variable to some
other variable is called the method of substitution When the integrand involves
some trigonometric functions, we use some well known identities to find the
integrals Using substitution technique, we obtain the following standard
integrals (i)
tan
log sec
C
x dx
x
=
+
∫
(ii)
cot
log sin
C
x dx
x
=
+
∫
(iii)
sec
log sec
tan
C
x dx
x
x
=
+
+
∫
(iv)
cosec
log cosec
cot
C
x dx
x
x
=
−
+
∫
� Integrals of some special functions
(i)
2
2
1 log
C
2
dx
x
a
a
x
a
x
a
−
=
+
+
−
∫
(ii)
2
2
1 log
C
2
dx
a
x
a
a
x
a
x
+
=
+
−
−
∫
(iii)
1
2
2
1 tan
C
dx
x
a
a
x
a
−
=
+
+
∫
INTEGRALS 357
(iv)
2
2
2
2
log
C
dx
x
x
a
x
a
=
+
−
+
−
∫
(v)
1
2
2
sin
C
dx
ax
a
x
−
=
+
−
∫
(vi)
2
2
2
2
log |
|
C
dx
x
x
a
x
a
=
+
+
+
+
∫
� Integration by parts
For given functions f1 and f2, we have
1
2
1
2
1
2
( )
( )
( )
( )
( )
( )
d
f
x
f
x dx
f
x
f
x dx
f x
f
x dx dx
dx
⋅
=
−
⋅
∫
∫
∫
∫
, i |
1 | 4124-4127 | When the integrand involves
some trigonometric functions, we use some well known identities to find the
integrals Using substitution technique, we obtain the following standard
integrals (i)
tan
log sec
C
x dx
x
=
+
∫
(ii)
cot
log sin
C
x dx
x
=
+
∫
(iii)
sec
log sec
tan
C
x dx
x
x
=
+
+
∫
(iv)
cosec
log cosec
cot
C
x dx
x
x
=
−
+
∫
� Integrals of some special functions
(i)
2
2
1 log
C
2
dx
x
a
a
x
a
x
a
−
=
+
+
−
∫
(ii)
2
2
1 log
C
2
dx
a
x
a
a
x
a
x
+
=
+
−
−
∫
(iii)
1
2
2
1 tan
C
dx
x
a
a
x
a
−
=
+
+
∫
INTEGRALS 357
(iv)
2
2
2
2
log
C
dx
x
x
a
x
a
=
+
−
+
−
∫
(v)
1
2
2
sin
C
dx
ax
a
x
−
=
+
−
∫
(vi)
2
2
2
2
log |
|
C
dx
x
x
a
x
a
=
+
+
+
+
∫
� Integration by parts
For given functions f1 and f2, we have
1
2
1
2
1
2
( )
( )
( )
( )
( )
( )
d
f
x
f
x dx
f
x
f
x dx
f x
f
x dx dx
dx
⋅
=
−
⋅
∫
∫
∫
∫
, i e |
1 | 4125-4128 | Using substitution technique, we obtain the following standard
integrals (i)
tan
log sec
C
x dx
x
=
+
∫
(ii)
cot
log sin
C
x dx
x
=
+
∫
(iii)
sec
log sec
tan
C
x dx
x
x
=
+
+
∫
(iv)
cosec
log cosec
cot
C
x dx
x
x
=
−
+
∫
� Integrals of some special functions
(i)
2
2
1 log
C
2
dx
x
a
a
x
a
x
a
−
=
+
+
−
∫
(ii)
2
2
1 log
C
2
dx
a
x
a
a
x
a
x
+
=
+
−
−
∫
(iii)
1
2
2
1 tan
C
dx
x
a
a
x
a
−
=
+
+
∫
INTEGRALS 357
(iv)
2
2
2
2
log
C
dx
x
x
a
x
a
=
+
−
+
−
∫
(v)
1
2
2
sin
C
dx
ax
a
x
−
=
+
−
∫
(vi)
2
2
2
2
log |
|
C
dx
x
x
a
x
a
=
+
+
+
+
∫
� Integration by parts
For given functions f1 and f2, we have
1
2
1
2
1
2
( )
( )
( )
( )
( )
( )
d
f
x
f
x dx
f
x
f
x dx
f x
f
x dx dx
dx
⋅
=
−
⋅
∫
∫
∫
∫
, i e , the
integral of the product of two functions = first function × integral of the
second function – integral of {differential coefficient of the first function ×
integral of the second function} |
1 | 4126-4129 | (i)
tan
log sec
C
x dx
x
=
+
∫
(ii)
cot
log sin
C
x dx
x
=
+
∫
(iii)
sec
log sec
tan
C
x dx
x
x
=
+
+
∫
(iv)
cosec
log cosec
cot
C
x dx
x
x
=
−
+
∫
� Integrals of some special functions
(i)
2
2
1 log
C
2
dx
x
a
a
x
a
x
a
−
=
+
+
−
∫
(ii)
2
2
1 log
C
2
dx
a
x
a
a
x
a
x
+
=
+
−
−
∫
(iii)
1
2
2
1 tan
C
dx
x
a
a
x
a
−
=
+
+
∫
INTEGRALS 357
(iv)
2
2
2
2
log
C
dx
x
x
a
x
a
=
+
−
+
−
∫
(v)
1
2
2
sin
C
dx
ax
a
x
−
=
+
−
∫
(vi)
2
2
2
2
log |
|
C
dx
x
x
a
x
a
=
+
+
+
+
∫
� Integration by parts
For given functions f1 and f2, we have
1
2
1
2
1
2
( )
( )
( )
( )
( )
( )
d
f
x
f
x dx
f
x
f
x dx
f x
f
x dx dx
dx
⋅
=
−
⋅
∫
∫
∫
∫
, i e , the
integral of the product of two functions = first function × integral of the
second function – integral of {differential coefficient of the first function ×
integral of the second function} Care must be taken in choosing the first
function and the second function |
1 | 4127-4130 | e , the
integral of the product of two functions = first function × integral of the
second function – integral of {differential coefficient of the first function ×
integral of the second function} Care must be taken in choosing the first
function and the second function Obviously, we must take that function as
the second function whose integral is well known to us |
1 | 4128-4131 | , the
integral of the product of two functions = first function × integral of the
second function – integral of {differential coefficient of the first function ×
integral of the second function} Care must be taken in choosing the first
function and the second function Obviously, we must take that function as
the second function whose integral is well known to us �
[ ( )
( )]
( )
C
x
x
e
f x
f
x
dx
e f x dx
′
+
=
+
∫
∫
� Some special types of integrals
(i)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
−
=
−
−
+
−
+
∫
(ii)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
+
=
+
+
+
+
+
∫
(iii)
2
2
2
2
2
sin1
C
2
2
x
a
x
a
x
dx
a
x
a
−
−
=
−
+
+
∫
(iv) Integrals of the types
2
2
or
dx
dx
ax
bx
c
ax
bx
c
+
+
+
+
∫
∫
can be
transformed into standard form by expressing
ax2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
a
a
a
a
a
+
+
=
+
+
−
(v) Integrals of the types
2
2
or
px
q dx
px
q dx
ax
bx
c
ax
bx
c
+
+
+
+
+
+
∫
∫
can be
358
MATHEMATICS
transformed into standard form by expressing
2
A
(
)
B
A (2
)
B
d
px
q
ax
bx
c
ax
b
dx
+
=
+
+
+
=
+
+
, where A and B are
determined by comparing coefficients on both sides |
1 | 4129-4132 | Care must be taken in choosing the first
function and the second function Obviously, we must take that function as
the second function whose integral is well known to us �
[ ( )
( )]
( )
C
x
x
e
f x
f
x
dx
e f x dx
′
+
=
+
∫
∫
� Some special types of integrals
(i)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
−
=
−
−
+
−
+
∫
(ii)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
+
=
+
+
+
+
+
∫
(iii)
2
2
2
2
2
sin1
C
2
2
x
a
x
a
x
dx
a
x
a
−
−
=
−
+
+
∫
(iv) Integrals of the types
2
2
or
dx
dx
ax
bx
c
ax
bx
c
+
+
+
+
∫
∫
can be
transformed into standard form by expressing
ax2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
a
a
a
a
a
+
+
=
+
+
−
(v) Integrals of the types
2
2
or
px
q dx
px
q dx
ax
bx
c
ax
bx
c
+
+
+
+
+
+
∫
∫
can be
358
MATHEMATICS
transformed into standard form by expressing
2
A
(
)
B
A (2
)
B
d
px
q
ax
bx
c
ax
b
dx
+
=
+
+
+
=
+
+
, where A and B are
determined by comparing coefficients on both sides � We have defined
( )
b
∫a f x dx
as the area of the region bounded by the curve
y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b |
1 | 4130-4133 | Obviously, we must take that function as
the second function whose integral is well known to us �
[ ( )
( )]
( )
C
x
x
e
f x
f
x
dx
e f x dx
′
+
=
+
∫
∫
� Some special types of integrals
(i)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
−
=
−
−
+
−
+
∫
(ii)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
+
=
+
+
+
+
+
∫
(iii)
2
2
2
2
2
sin1
C
2
2
x
a
x
a
x
dx
a
x
a
−
−
=
−
+
+
∫
(iv) Integrals of the types
2
2
or
dx
dx
ax
bx
c
ax
bx
c
+
+
+
+
∫
∫
can be
transformed into standard form by expressing
ax2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
a
a
a
a
a
+
+
=
+
+
−
(v) Integrals of the types
2
2
or
px
q dx
px
q dx
ax
bx
c
ax
bx
c
+
+
+
+
+
+
∫
∫
can be
358
MATHEMATICS
transformed into standard form by expressing
2
A
(
)
B
A (2
)
B
d
px
q
ax
bx
c
ax
b
dx
+
=
+
+
+
=
+
+
, where A and B are
determined by comparing coefficients on both sides � We have defined
( )
b
∫a f x dx
as the area of the region bounded by the curve
y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b Let x be a
given point in [a, b] |
1 | 4131-4134 | �
[ ( )
( )]
( )
C
x
x
e
f x
f
x
dx
e f x dx
′
+
=
+
∫
∫
� Some special types of integrals
(i)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
−
=
−
−
+
−
+
∫
(ii)
2
2
2
2
2
2
2
log
C
2
2
x
a
x
a
dx
x
a
x
x
a
+
=
+
+
+
+
+
∫
(iii)
2
2
2
2
2
sin1
C
2
2
x
a
x
a
x
dx
a
x
a
−
−
=
−
+
+
∫
(iv) Integrals of the types
2
2
or
dx
dx
ax
bx
c
ax
bx
c
+
+
+
+
∫
∫
can be
transformed into standard form by expressing
ax2 + bx + c =
2
2
2
2
2
4
b
c
b
c
b
a x
x
a
x
a
a
a
a
a
+
+
=
+
+
−
(v) Integrals of the types
2
2
or
px
q dx
px
q dx
ax
bx
c
ax
bx
c
+
+
+
+
+
+
∫
∫
can be
358
MATHEMATICS
transformed into standard form by expressing
2
A
(
)
B
A (2
)
B
d
px
q
ax
bx
c
ax
b
dx
+
=
+
+
+
=
+
+
, where A and B are
determined by comparing coefficients on both sides � We have defined
( )
b
∫a f x dx
as the area of the region bounded by the curve
y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b Let x be a
given point in [a, b] Then
( )
x
∫a f x dx
represents the Area function A (x) |
1 | 4132-4135 | � We have defined
( )
b
∫a f x dx
as the area of the region bounded by the curve
y = f (x), a ≤ x ≤ b, the x-axis and the ordinates x = a and x = b Let x be a
given point in [a, b] Then
( )
x
∫a f x dx
represents the Area function A (x) This concept of area function leads to the Fundamental Theorems of Integral
Calculus |
1 | 4133-4136 | Let x be a
given point in [a, b] Then
( )
x
∫a f x dx
represents the Area function A (x) This concept of area function leads to the Fundamental Theorems of Integral
Calculus � First fundamental theorem of integral calculus
Let the area function be defined by A(x) =
( )
x
∫a f x dx
for all x ≥ a, where
the function f is assumed to be continuous on [a, b] |
1 | 4134-4137 | Then
( )
x
∫a f x dx
represents the Area function A (x) This concept of area function leads to the Fundamental Theorems of Integral
Calculus � First fundamental theorem of integral calculus
Let the area function be defined by A(x) =
( )
x
∫a f x dx
for all x ≥ a, where
the function f is assumed to be continuous on [a, b] Then A′(x) = f (x) for all
x ∈ [a, b] |
1 | 4135-4138 | This concept of area function leads to the Fundamental Theorems of Integral
Calculus � First fundamental theorem of integral calculus
Let the area function be defined by A(x) =
( )
x
∫a f x dx
for all x ≥ a, where
the function f is assumed to be continuous on [a, b] Then A′(x) = f (x) for all
x ∈ [a, b] � Second fundamental theorem of integral calculus
Let f be a continuous function of x defined on the closed interval [a, b] and
let F be another function such that
F( )
( )
d
x
f x
dx
=
for all x in the domain of
f, then
[
]
( )
F( )
C
F ( )
F ( )
b
b
a
a f x dx
x
b
a
=
+
=
−
∫ |
1 | 4136-4139 | � First fundamental theorem of integral calculus
Let the area function be defined by A(x) =
( )
x
∫a f x dx
for all x ≥ a, where
the function f is assumed to be continuous on [a, b] Then A′(x) = f (x) for all
x ∈ [a, b] � Second fundamental theorem of integral calculus
Let f be a continuous function of x defined on the closed interval [a, b] and
let F be another function such that
F( )
( )
d
x
f x
dx
=
for all x in the domain of
f, then
[
]
( )
F( )
C
F ( )
F ( )
b
b
a
a f x dx
x
b
a
=
+
=
−
∫ This is called the definite integral of f over the range [a, b], where a and b
are called the limits of integration, a being the lower limit and b the
upper limit |
1 | 4137-4140 | Then A′(x) = f (x) for all
x ∈ [a, b] � Second fundamental theorem of integral calculus
Let f be a continuous function of x defined on the closed interval [a, b] and
let F be another function such that
F( )
( )
d
x
f x
dx
=
for all x in the domain of
f, then
[
]
( )
F( )
C
F ( )
F ( )
b
b
a
a f x dx
x
b
a
=
+
=
−
∫ This is called the definite integral of f over the range [a, b], where a and b
are called the limits of integration, a being the lower limit and b the
upper limit —�
�
�
�
�—
APPLICATION OF INTEGRALS 359
Fig 8 |
1 | 4138-4141 | � Second fundamental theorem of integral calculus
Let f be a continuous function of x defined on the closed interval [a, b] and
let F be another function such that
F( )
( )
d
x
f x
dx
=
for all x in the domain of
f, then
[
]
( )
F( )
C
F ( )
F ( )
b
b
a
a f x dx
x
b
a
=
+
=
−
∫ This is called the definite integral of f over the range [a, b], where a and b
are called the limits of integration, a being the lower limit and b the
upper limit —�
�
�
�
�—
APPLICATION OF INTEGRALS 359
Fig 8 1
� One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form |
1 | 4139-4142 | This is called the definite integral of f over the range [a, b], where a and b
are called the limits of integration, a being the lower limit and b the
upper limit —�
�
�
�
�—
APPLICATION OF INTEGRALS 359
Fig 8 1
� One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form – BIRKHOFF �
8 |
1 | 4140-4143 | —�
�
�
�
�—
APPLICATION OF INTEGRALS 359
Fig 8 1
� One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form – BIRKHOFF �
8 1 Introduction
In geometry, we have learnt formulae to calculate areas
of various geometrical figures including triangles,
rectangles, trapezias and circles |
1 | 4141-4144 | 1
� One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form – BIRKHOFF �
8 1 Introduction
In geometry, we have learnt formulae to calculate areas
of various geometrical figures including triangles,
rectangles, trapezias and circles Such formulae are
fundamental in the applications of mathematics to many
real life problems |
1 | 4142-4145 | – BIRKHOFF �
8 1 Introduction
In geometry, we have learnt formulae to calculate areas
of various geometrical figures including triangles,
rectangles, trapezias and circles Such formulae are
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 |
1 | 4143-4146 | 1 Introduction
In geometry, we have learnt formulae to calculate areas
of various geometrical figures including triangles,
rectangles, trapezias and circles Such formulae are
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 |
1 | 4144-4147 | Such formulae are
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 |
1 | 4145-4148 | 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 In the previous chapter, we have studied to find the
area bounded by the curve y = f (x), the ordinates x = a,
x = b and x-axis, while calculating definite integral as the
limit of a sum |
1 | 4146-4149 | However, they are inadequate for calculating the areas
enclosed by curves For that we shall need some concepts
of Integral Calculus In the previous chapter, we have studied to find the
area bounded by the curve y = f (x), the ordinates x = a,
x = b and x-axis, while calculating definite integral as the
limit of a sum Here, in this chapter, we shall study a specific
application of integrals to find the area under simple curves,
area between lines and arcs of circles, parabolas and
ellipses (standard forms only) |
1 | 4147-4150 | For that we shall need some concepts
of Integral Calculus In the previous chapter, we have studied to find the
area bounded by the curve y = f (x), the ordinates x = a,
x = b and x-axis, while calculating definite integral as the
limit of a sum Here, in this chapter, we shall study a specific
application of integrals to find the area under simple curves,
area between lines and arcs of circles, parabolas and
ellipses (standard forms only) We shall also deal with finding
the area bounded by the above said curves |
1 | 4148-4151 | In the previous chapter, we have studied to find the
area bounded by the curve y = f (x), the ordinates x = a,
x = b and x-axis, while calculating definite integral as the
limit of a sum Here, in this chapter, we shall study a specific
application of integrals to find the area under simple curves,
area between lines and arcs of circles, parabolas and
ellipses (standard forms only) We shall also deal with finding
the area bounded by the above said curves 8 |
1 | 4149-4152 | Here, in this chapter, we shall study a specific
application of integrals to find the area under simple curves,
area between lines and arcs of circles, parabolas and
ellipses (standard forms only) We shall also deal with finding
the area bounded by the above said curves 8 2 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 |
1 | 4150-4153 | We shall also deal with finding
the area bounded by the above said curves 8 2 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 |
1 | 4151-4154 | 8 2 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 | 4152-4155 | 2 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 |
1 | 4153-4156 | 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 Consider
an arbitrary strip of height y and width dx,
then dA (area of the elementary strip)= ydx,
where, y = f(x) |
1 | 4154-4157 | From Fig 8 1, we can think of area
under the curve as composed of large
number of very thin vertical strips Consider
an arbitrary strip of height y and width dx,
then dA (area of the elementary strip)= ydx,
where, y = f(x) Chapter 8
APPLICATION OF INTEGRALS
A |
1 | 4155-4158 | 1, we can think of area
under the curve as composed of large
number of very thin vertical strips Consider
an arbitrary strip of height y and width dx,
then dA (area of the elementary strip)= ydx,
where, y = f(x) Chapter 8
APPLICATION OF INTEGRALS
A L |
1 | 4156-4159 | Consider
an arbitrary strip of height y and width dx,
then dA (area of the elementary strip)= ydx,
where, y = f(x) Chapter 8
APPLICATION OF INTEGRALS
A L Cauchy
(1789-1857)
360
MATHEMATICS
Fig 8 |
1 | 4157-4160 | Chapter 8
APPLICATION OF INTEGRALS
A L Cauchy
(1789-1857)
360
MATHEMATICS
Fig 8 2
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 |
1 | 4158-4161 | L Cauchy
(1789-1857)
360
MATHEMATICS
Fig 8 2
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 areas of thin strips across the region
PQRSP |
1 | 4159-4162 | Cauchy
(1789-1857)
360
MATHEMATICS
Fig 8 2
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 areas of thin strips across the region
PQRSP Symbolically, we express
A =
A
( )
b
b
b
a
a
a
d
ydx
f x dx
=
=
∫
∫
∫
The area A of the region bounded by
the curve x = g (y), y-axis and the lines y = c,
y = d is given by
A =
( )
d
d
c
c
xdy
g y dy
=
∫
∫
Here, we consider horizontal strips as shown in
the Fig 8 |
1 | 4160-4163 | 2
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 areas of thin strips across the region
PQRSP Symbolically, we express
A =
A
( )
b
b
b
a
a
a
d
ydx
f x dx
=
=
∫
∫
∫
The area A of the region bounded by
the curve x = g (y), y-axis and the lines y = c,
y = d is given by
A =
( )
d
d
c
c
xdy
g y dy
=
∫
∫
Here, we consider horizontal strips as shown in
the Fig 8 2
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 |
1 | 4161-4164 | 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 areas of thin strips across the region
PQRSP Symbolically, we express
A =
A
( )
b
b
b
a
a
a
d
ydx
f x dx
=
=
∫
∫
∫
The area A of the region bounded by
the curve x = g (y), y-axis and the lines y = c,
y = d is given by
A =
( )
d
d
c
c
xdy
g y dy
=
∫
∫
Here, we consider horizontal strips as shown in
the Fig 8 2
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 |
1 | 4162-4165 | Symbolically, we express
A =
A
( )
b
b
b
a
a
a
d
ydx
f x dx
=
=
∫
∫
∫
The area A of the region bounded by
the curve x = g (y), y-axis and the lines y = c,
y = d is given by
A =
( )
d
d
c
c
xdy
g y dy
=
∫
∫
Here, we consider horizontal strips as shown in
the Fig 8 2
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 |
1 | 4163-4166 | 2
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 |
1 | 4164-4167 | 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 |
1 | 4165-4168 | 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 ,
( )
b
∫a f x dx |
1 | 4166-4169 | Thus, if the area is negative, we
take its absolute value, i e ,
( )
b
∫a f x dx Fig 8 |
1 | 4167-4170 | e ,
( )
b
∫a f x dx Fig 8 3
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 8 |
1 | 4168-4171 | ,
( )
b
∫a f x dx Fig 8 3
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 8 4 |
1 | 4169-4172 | Fig 8 3
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 8 4 Here, A1 < 0 and A2 > 0 |
1 | 4170-4173 | 3
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 8 4 Here, A1 < 0 and A2 > 0 Therefore, the area
A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given
by A = |A1| + A2 |
1 | 4171-4174 | 4 Here, A1 < 0 and A2 > 0 Therefore, the area
A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given
by A = |A1| + A2 APPLICATION OF INTEGRALS 361
Example 1 Find the area enclosed by the circle x2 + y2 = a2 |
1 | 4172-4175 | Here, A1 < 0 and A2 > 0 Therefore, the area
A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given
by A = |A1| + A2 APPLICATION OF INTEGRALS 361
Example 1 Find the area enclosed by the circle x2 + y2 = a2 Solution From Fig 8 |
1 | 4173-4176 | Therefore, the area
A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given
by A = |A1| + A2 APPLICATION OF INTEGRALS 361
Example 1 Find the area enclosed by the circle x2 + y2 = a2 Solution From Fig 8 5, the whole area enclosed
by the given circle
= 4 (area of the region AOBA bounded by
the curve, x-axis and the ordinates x = 0 and
x = a) [as the circle is symmetrical about both
x-axis and y-axis]
=
40
∫a ydx
(taking vertical strips)
=
2
2
40
a
a
−x dx
∫
Since x2 + y2 = a2 gives y =
2
2
a
x
±
−
As the region AOBA lies in the first quadrant, y is taken as positive |
1 | 4174-4177 | APPLICATION OF INTEGRALS 361
Example 1 Find the area enclosed by the circle x2 + y2 = a2 Solution From Fig 8 5, the whole area enclosed
by the given circle
= 4 (area of the region AOBA bounded by
the curve, x-axis and the ordinates x = 0 and
x = a) [as the circle is symmetrical about both
x-axis and y-axis]
=
40
∫a ydx
(taking vertical strips)
=
2
2
40
a
a
−x dx
∫
Since x2 + y2 = a2 gives y =
2
2
a
x
±
−
As the region AOBA lies in the first quadrant, y is taken as positive Integrating, we get
the whole area enclosed by the given circle
=
2
2
2
–1
0
4
sin
2
2
a
x
a
x
a
x
a
−
+
=
2
1
4
0
sin 1
0
2
2
a
a
−
× +
−
=
2
2
4
2
2
a
a
π
=π
Fig 8 |
1 | 4175-4178 | Solution From Fig 8 5, the whole area enclosed
by the given circle
= 4 (area of the region AOBA bounded by
the curve, x-axis and the ordinates x = 0 and
x = a) [as the circle is symmetrical about both
x-axis and y-axis]
=
40
∫a ydx
(taking vertical strips)
=
2
2
40
a
a
−x dx
∫
Since x2 + y2 = a2 gives y =
2
2
a
x
±
−
As the region AOBA lies in the first quadrant, y is taken as positive Integrating, we get
the whole area enclosed by the given circle
=
2
2
2
–1
0
4
sin
2
2
a
x
a
x
a
x
a
−
+
=
2
1
4
0
sin 1
0
2
2
a
a
−
× +
−
=
2
2
4
2
2
a
a
π
=π
Fig 8 5
Fig 8 |
1 | 4176-4179 | 5, the whole area enclosed
by the given circle
= 4 (area of the region AOBA bounded by
the curve, x-axis and the ordinates x = 0 and
x = a) [as the circle is symmetrical about both
x-axis and y-axis]
=
40
∫a ydx
(taking vertical strips)
=
2
2
40
a
a
−x dx
∫
Since x2 + y2 = a2 gives y =
2
2
a
x
±
−
As the region AOBA lies in the first quadrant, y is taken as positive Integrating, we get
the whole area enclosed by the given circle
=
2
2
2
–1
0
4
sin
2
2
a
x
a
x
a
x
a
−
+
=
2
1
4
0
sin 1
0
2
2
a
a
−
× +
−
=
2
2
4
2
2
a
a
π
=π
Fig 8 5
Fig 8 4
362
MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8 |
1 | 4177-4180 | Integrating, we get
the whole area enclosed by the given circle
=
2
2
2
–1
0
4
sin
2
2
a
x
a
x
a
x
a
−
+
=
2
1
4
0
sin 1
0
2
2
a
a
−
× +
−
=
2
2
4
2
2
a
a
π
=π
Fig 8 5
Fig 8 4
362
MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8 6, the whole area of the
region enclosed by circle
=
40
∫a xdy
=
2
2
40
a
a
y
dy
−
∫
(Why |
1 | 4178-4181 | 5
Fig 8 4
362
MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8 6, the whole area of the
region enclosed by circle
=
40
∫a xdy
=
2
2
40
a
a
y
dy
−
∫
(Why )
=
2
2
2
1
0
4
sin
2
2
a
a
y
y
a
y
a
−
−
+
=
2
1
4
0
sin 1
0
2
a2
a
−
× +
−
=
2
2
4
a2 2
a
π = π
Example 2 Find the area enclosed by the ellipse
2
2
2
2
1
x
y
a
b
+
=
Solution From Fig 8 |
1 | 4179-4182 | 4
362
MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8 6, the whole area of the
region enclosed by circle
=
40
∫a xdy
=
2
2
40
a
a
y
dy
−
∫
(Why )
=
2
2
2
1
0
4
sin
2
2
a
a
y
y
a
y
a
−
−
+
=
2
1
4
0
sin 1
0
2
a2
a
−
× +
−
=
2
2
4
a2 2
a
π = π
Example 2 Find the area enclosed by the ellipse
2
2
2
2
1
x
y
a
b
+
=
Solution From Fig 8 7, the area of the region ABA′B′A bounded by the ellipse
=
in
4
,
0,
area of theregion AOBA
the first quadrant bounded
bythecurve x
axis and theordinates x
x
a
−
=
=
(as the ellipse is symmetrical about both x-axis and y-axis)
=
40
(takingverticalstrips)
a ydx
∫
Now
2
2
2
2
x
y
a
+b
= 1 gives
2
2
b
y
a
x
=±a
−
, but as the region AOBA lies in the first
quadrant, y is taken as positive |
1 | 4180-4183 | 6, the whole area of the
region enclosed by circle
=
40
∫a xdy
=
2
2
40
a
a
y
dy
−
∫
(Why )
=
2
2
2
1
0
4
sin
2
2
a
a
y
y
a
y
a
−
−
+
=
2
1
4
0
sin 1
0
2
a2
a
−
× +
−
=
2
2
4
a2 2
a
π = π
Example 2 Find the area enclosed by the ellipse
2
2
2
2
1
x
y
a
b
+
=
Solution From Fig 8 7, the area of the region ABA′B′A bounded by the ellipse
=
in
4
,
0,
area of theregion AOBA
the first quadrant bounded
bythecurve x
axis and theordinates x
x
a
−
=
=
(as the ellipse is symmetrical about both x-axis and y-axis)
=
40
(takingverticalstrips)
a ydx
∫
Now
2
2
2
2
x
y
a
+b
= 1 gives
2
2
b
y
a
x
=±a
−
, but as the region AOBA lies in the first
quadrant, y is taken as positive So, the required area is
=
2
2
40
a b
a
x dx
a
−
∫
=
2
2
2
–1
0
4
sin
2
2
a
b x
a
x
a
x
a
a
−
+
(Why |
1 | 4181-4184 | )
=
2
2
2
1
0
4
sin
2
2
a
a
y
y
a
y
a
−
−
+
=
2
1
4
0
sin 1
0
2
a2
a
−
× +
−
=
2
2
4
a2 2
a
π = π
Example 2 Find the area enclosed by the ellipse
2
2
2
2
1
x
y
a
b
+
=
Solution From Fig 8 7, the area of the region ABA′B′A bounded by the ellipse
=
in
4
,
0,
area of theregion AOBA
the first quadrant bounded
bythecurve x
axis and theordinates x
x
a
−
=
=
(as the ellipse is symmetrical about both x-axis and y-axis)
=
40
(takingverticalstrips)
a ydx
∫
Now
2
2
2
2
x
y
a
+b
= 1 gives
2
2
b
y
a
x
=±a
−
, but as the region AOBA lies in the first
quadrant, y is taken as positive So, the required area is
=
2
2
40
a b
a
x dx
a
−
∫
=
2
2
2
–1
0
4
sin
2
2
a
b x
a
x
a
x
a
a
−
+
(Why )
=
2
1
4
0
sin
1
0
2
2
b
a
a
a
−
× +
−
=
2
4
2 2
b a
ab
a
π =π
Fig 8 |
1 | 4182-4185 | 7, the area of the region ABA′B′A bounded by the ellipse
=
in
4
,
0,
area of theregion AOBA
the first quadrant bounded
bythecurve x
axis and theordinates x
x
a
−
=
=
(as the ellipse is symmetrical about both x-axis and y-axis)
=
40
(takingverticalstrips)
a ydx
∫
Now
2
2
2
2
x
y
a
+b
= 1 gives
2
2
b
y
a
x
=±a
−
, but as the region AOBA lies in the first
quadrant, y is taken as positive So, the required area is
=
2
2
40
a b
a
x dx
a
−
∫
=
2
2
2
–1
0
4
sin
2
2
a
b x
a
x
a
x
a
a
−
+
(Why )
=
2
1
4
0
sin
1
0
2
2
b
a
a
a
−
× +
−
=
2
4
2 2
b a
ab
a
π =π
Fig 8 6
Fig 8 |
1 | 4183-4186 | So, the required area is
=
2
2
40
a b
a
x dx
a
−
∫
=
2
2
2
–1
0
4
sin
2
2
a
b x
a
x
a
x
a
a
−
+
(Why )
=
2
1
4
0
sin
1
0
2
2
b
a
a
a
−
× +
−
=
2
4
2 2
b a
ab
a
π =π
Fig 8 6
Fig 8 7
APPLICATION OF INTEGRALS 363
Alternatively, considering horizontal strips as
shown in the Fig 8 |
1 | 4184-4187 | )
=
2
1
4
0
sin
1
0
2
2
b
a
a
a
−
× +
−
=
2
4
2 2
b a
ab
a
π =π
Fig 8 6
Fig 8 7
APPLICATION OF INTEGRALS 363
Alternatively, considering horizontal strips as
shown in the Fig 8 8, the area of the ellipse is
=
0
4∫
b xdy =
2
2
0
4
−
∫
ab
b
y dy
b
(Why |
1 | 4185-4188 | 6
Fig 8 7
APPLICATION OF INTEGRALS 363
Alternatively, considering horizontal strips as
shown in the Fig 8 8, the area of the ellipse is
=
0
4∫
b xdy =
2
2
0
4
−
∫
ab
b
y dy
b
(Why )
=
2
2
2
–1
0
4
sin
2
2
b
a
y
b
y
b
y
b
b
−
+
=
2
–1
4
0
sin 1
0
2
2
a
b
b
b
× +
−
=
2
4
2 2
a b
ab
b
π =π
8 |
1 | 4186-4189 | 7
APPLICATION OF INTEGRALS 363
Alternatively, considering horizontal strips as
shown in the Fig 8 8, the area of the ellipse is
=
0
4∫
b xdy =
2
2
0
4
−
∫
ab
b
y dy
b
(Why )
=
2
2
2
–1
0
4
sin
2
2
b
a
y
b
y
b
y
b
b
−
+
=
2
–1
4
0
sin 1
0
2
2
a
b
b
b
× +
−
=
2
4
2 2
a b
ab
b
π =π
8 2 |
1 | 4187-4190 | 8, the area of the ellipse is
=
0
4∫
b xdy =
2
2
0
4
−
∫
ab
b
y dy
b
(Why )
=
2
2
2
–1
0
4
sin
2
2
b
a
y
b
y
b
y
b
b
−
+
=
2
–1
4
0
sin 1
0
2
2
a
b
b
b
× +
−
=
2
4
2 2
a b
ab
b
π =π
8 2 1 The area of the region bounded by a curve and a line
In this subsection, we will find the area of the region bounded by a line and a circle,
a line and a parabola, a line and an ellipse |
1 | 4188-4191 | )
=
2
2
2
–1
0
4
sin
2
2
b
a
y
b
y
b
y
b
b
−
+
=
2
–1
4
0
sin 1
0
2
2
a
b
b
b
× +
−
=
2
4
2 2
a b
ab
b
π =π
8 2 1 The area of the region bounded by a curve and a line
In this subsection, we will find the area of the region bounded by a line and a circle,
a line and a parabola, a line and an ellipse Equations of above mentioned curves will be
in their standard forms only as the cases in other forms go beyond the scope of this
textbook |
1 | 4189-4192 | 2 1 The area of the region bounded by a curve and a line
In this subsection, we will find the area of the region bounded by a line and a circle,
a line and a parabola, a line and an ellipse Equations of above mentioned curves will be
in their standard forms only as the cases in other forms go beyond the scope of this
textbook Example 3 Find the area of the region bounded
by the curve y = x2 and the line y = 4 |
1 | 4190-4193 | 1 The area of the region bounded by a curve and a line
In this subsection, we will find the area of the region bounded by a line and a circle,
a line and a parabola, a line and an ellipse Equations of above mentioned curves will be
in their standard forms only as the cases in other forms go beyond the scope of this
textbook Example 3 Find the area of the region bounded
by the curve y = x2 and the line y = 4 Solution Since the given curve represented by
the equation y = x2 is a parabola symmetrical
about y-axis only, therefore, from Fig 8 |
1 | 4191-4194 | Equations of above mentioned curves will be
in their standard forms only as the cases in other forms go beyond the scope of this
textbook Example 3 Find the area of the region bounded
by the curve y = x2 and the line y = 4 Solution Since the given curve represented by
the equation y = x2 is a parabola symmetrical
about y-axis only, therefore, from Fig 8 9, the
required area of the region AOBA is given by
4
20
∫xdy
=
areaof theregionBONB boundedbycurve,
axis
2 andthelines
0and
= 4
y
y
y
−
=
=
4
20
ydy
∫
=
4
3
2
0
2
2
3
y
×
4
32
8
3
3
=
×
=
(Why |
1 | 4192-4195 | Example 3 Find the area of the region bounded
by the curve y = x2 and the line y = 4 Solution Since the given curve represented by
the equation y = x2 is a parabola symmetrical
about y-axis only, therefore, from Fig 8 9, the
required area of the region AOBA is given by
4
20
∫xdy
=
areaof theregionBONB boundedbycurve,
axis
2 andthelines
0and
= 4
y
y
y
−
=
=
4
20
ydy
∫
=
4
3
2
0
2
2
3
y
×
4
32
8
3
3
=
×
=
(Why )
Here, we have taken horizontal strips as indicated in the Fig 8 |
1 | 4193-4196 | Solution Since the given curve represented by
the equation y = x2 is a parabola symmetrical
about y-axis only, therefore, from Fig 8 9, the
required area of the region AOBA is given by
4
20
∫xdy
=
areaof theregionBONB boundedbycurve,
axis
2 andthelines
0and
= 4
y
y
y
−
=
=
4
20
ydy
∫
=
4
3
2
0
2
2
3
y
×
4
32
8
3
3
=
×
=
(Why )
Here, we have taken horizontal strips as indicated in the Fig 8 9 |
1 | 4194-4197 | 9, the
required area of the region AOBA is given by
4
20
∫xdy
=
areaof theregionBONB boundedbycurve,
axis
2 andthelines
0and
= 4
y
y
y
−
=
=
4
20
ydy
∫
=
4
3
2
0
2
2
3
y
×
4
32
8
3
3
=
×
=
(Why )
Here, we have taken horizontal strips as indicated in the Fig 8 9 Fig 8 |
1 | 4195-4198 | )
Here, we have taken horizontal strips as indicated in the Fig 8 9 Fig 8 8
Fig 8 |
1 | 4196-4199 | 9 Fig 8 8
Fig 8 9
364
MATHEMATICS
Alternatively, we may consider the vertical
strips like PQ as shown in the Fig 8 |
1 | 4197-4200 | Fig 8 8
Fig 8 9
364
MATHEMATICS
Alternatively, we may consider the vertical
strips like PQ as shown in the Fig 8 10 to
obtain the area of the region AOBA |
1 | 4198-4201 | 8
Fig 8 9
364
MATHEMATICS
Alternatively, we may consider the vertical
strips like PQ as shown in the Fig 8 10 to
obtain the area of the region AOBA To this
end, we solve the equations x2 = y and y = 4
which gives x = –2 and x = 2 |
1 | 4199-4202 | 9
364
MATHEMATICS
Alternatively, we may consider the vertical
strips like PQ as shown in the Fig 8 10 to
obtain the area of the region AOBA To this
end, we solve the equations x2 = y and y = 4
which gives x = –2 and x = 2 Thus, the region AOBA may be stated as
the region bounded by the curve y = x 2, y = 4
and the ordinates x = –2 and x = 2 |
1 | 4200-4203 | 10 to
obtain the area of the region AOBA To this
end, we solve the equations x2 = y and y = 4
which gives x = –2 and x = 2 Thus, the region AOBA may be stated as
the region bounded by the curve y = x 2, y = 4
and the ordinates x = –2 and x = 2 Therefore, the area of the region AOBA
=
2
2 ydx
−∫
[ y = (y-coordinate of Q) – (y-coordinate of P) = 4 – x 2]
=
(
)
2
2
20
4
x
dx
−
∫
(Why |
1 | 4201-4204 | To this
end, we solve the equations x2 = y and y = 4
which gives x = –2 and x = 2 Thus, the region AOBA may be stated as
the region bounded by the curve y = x 2, y = 4
and the ordinates x = –2 and x = 2 Therefore, the area of the region AOBA
=
2
2 ydx
−∫
[ y = (y-coordinate of Q) – (y-coordinate of P) = 4 – x 2]
=
(
)
2
2
20
4
x
dx
−
∫
(Why )
=
2
3
0
2 4
x3
x
−
8
2 4 2
3
=
× −
32
3
=
Remark From the above examples, it is inferred that we can consider either vertical
strips or horizontal strips for calculating the area of the region |
1 | 4202-4205 | Thus, the region AOBA may be stated as
the region bounded by the curve y = x 2, y = 4
and the ordinates x = –2 and x = 2 Therefore, the area of the region AOBA
=
2
2 ydx
−∫
[ y = (y-coordinate of Q) – (y-coordinate of P) = 4 – x 2]
=
(
)
2
2
20
4
x
dx
−
∫
(Why )
=
2
3
0
2 4
x3
x
−
8
2 4 2
3
=
× −
32
3
=
Remark From the above examples, it is inferred that we can consider either vertical
strips or horizontal strips for calculating the area of the region Henceforth, we shall
consider either of these two, most preferably vertical strips |
1 | 4203-4206 | Therefore, the area of the region AOBA
=
2
2 ydx
−∫
[ y = (y-coordinate of Q) – (y-coordinate of P) = 4 – x 2]
=
(
)
2
2
20
4
x
dx
−
∫
(Why )
=
2
3
0
2 4
x3
x
−
8
2 4 2
3
=
× −
32
3
=
Remark From the above examples, it is inferred that we can consider either vertical
strips or horizontal strips for calculating the area of the region Henceforth, we shall
consider either of these two, most preferably vertical strips Example 4 Find the area of the region in the first quadrant enclosed by the x-axis,
the line y = x, and the circle x2 + y2 = 32 |
1 | 4204-4207 | )
=
2
3
0
2 4
x3
x
−
8
2 4 2
3
=
× −
32
3
=
Remark From the above examples, it is inferred that we can consider either vertical
strips or horizontal strips for calculating the area of the region Henceforth, we shall
consider either of these two, most preferably vertical strips Example 4 Find the area of the region in the first quadrant enclosed by the x-axis,
the line y = x, and the circle x2 + y2 = 32 Solution The given equations are
y = x |
1 | 4205-4208 | Henceforth, we shall
consider either of these two, most preferably vertical strips Example 4 Find the area of the region in the first quadrant enclosed by the x-axis,
the line y = x, and the circle x2 + y2 = 32 Solution The given equations are
y = x (1)
and
x2 + y2 = 32 |
1 | 4206-4209 | Example 4 Find the area of the region in the first quadrant enclosed by the x-axis,
the line y = x, and the circle x2 + y2 = 32 Solution The given equations are
y = x (1)
and
x2 + y2 = 32 (2)
Solving (1) and (2), we find that the line
and the circle meet at B(4, 4) in the first
quadrant (Fig 8 |
1 | 4207-4210 | Solution The given equations are
y = x (1)
and
x2 + y2 = 32 (2)
Solving (1) and (2), we find that the line
and the circle meet at B(4, 4) in the first
quadrant (Fig 8 11) |
1 | 4208-4211 | (1)
and
x2 + y2 = 32 (2)
Solving (1) and (2), we find that the line
and the circle meet at B(4, 4) in the first
quadrant (Fig 8 11) Draw perpendicular
BM to the x-axis |
1 | 4209-4212 | (2)
Solving (1) and (2), we find that the line
and the circle meet at B(4, 4) in the first
quadrant (Fig 8 11) Draw perpendicular
BM to the x-axis Therefore, the required area = area of
the region OBMO + area of the region
BMAB |
1 | 4210-4213 | 11) Draw perpendicular
BM to the x-axis Therefore, the required area = area of
the region OBMO + area of the region
BMAB Now, the area of the region OBMO
=
4
4
0
0
ydx
xdx
=
∫
∫ |
1 | 4211-4214 | Draw perpendicular
BM to the x-axis Therefore, the required area = area of
the region OBMO + area of the region
BMAB Now, the area of the region OBMO
=
4
4
0
0
ydx
xdx
=
∫
∫ (3)
=
4
2
0
1
2 x
= 8
Fig 8 |
1 | 4212-4215 | Therefore, the required area = area of
the region OBMO + area of the region
BMAB Now, the area of the region OBMO
=
4
4
0
0
ydx
xdx
=
∫
∫ (3)
=
4
2
0
1
2 x
= 8
Fig 8 10
Fig 8 |
1 | 4213-4216 | Now, the area of the region OBMO
=
4
4
0
0
ydx
xdx
=
∫
∫ (3)
=
4
2
0
1
2 x
= 8
Fig 8 10
Fig 8 11
Y
O
A
y
x
=
Y'
B
M
(4�4)
,
X
X'
(4
2 0)
,
APPLICATION OF INTEGRALS 365
O
F (
o)
ae,
B
Y
Y′
B'
S
R
X
X′
x
ae
=
Again, the area of the region BMAB
=
4 2
4
ydx
∫
=
4 2
2
4
32
−x dx
∫
=
4 2
2
–1
4
1
1
32
32
sin
2
2
4 2
x
x
x
−
+
×
×
=
–1
–1
1
1
4
1
1
4 2
0
32
sin 1
32
16
32
sin
2
2
2
2
2
×
+
×
×
−
−
+
×
×
= 8 π – (8 + 4π) = 4π – 8 |
1 | 4214-4217 | (3)
=
4
2
0
1
2 x
= 8
Fig 8 10
Fig 8 11
Y
O
A
y
x
=
Y'
B
M
(4�4)
,
X
X'
(4
2 0)
,
APPLICATION OF INTEGRALS 365
O
F (
o)
ae,
B
Y
Y′
B'
S
R
X
X′
x
ae
=
Again, the area of the region BMAB
=
4 2
4
ydx
∫
=
4 2
2
4
32
−x dx
∫
=
4 2
2
–1
4
1
1
32
32
sin
2
2
4 2
x
x
x
−
+
×
×
=
–1
–1
1
1
4
1
1
4 2
0
32
sin 1
32
16
32
sin
2
2
2
2
2
×
+
×
×
−
−
+
×
×
= 8 π – (8 + 4π) = 4π – 8 (4)
Adding (3) and (4), we get, the required area = 4π |
1 | 4215-4218 | 10
Fig 8 11
Y
O
A
y
x
=
Y'
B
M
(4�4)
,
X
X'
(4
2 0)
,
APPLICATION OF INTEGRALS 365
O
F (
o)
ae,
B
Y
Y′
B'
S
R
X
X′
x
ae
=
Again, the area of the region BMAB
=
4 2
4
ydx
∫
=
4 2
2
4
32
−x dx
∫
=
4 2
2
–1
4
1
1
32
32
sin
2
2
4 2
x
x
x
−
+
×
×
=
–1
–1
1
1
4
1
1
4 2
0
32
sin 1
32
16
32
sin
2
2
2
2
2
×
+
×
×
−
−
+
×
×
= 8 π – (8 + 4π) = 4π – 8 (4)
Adding (3) and (4), we get, the required area = 4π Example 5 Find the area bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
= and the ordinates x = 0
and x = ae, where, b2 = a2 (1 – e2) and e < 1 |
1 | 4216-4219 | 11
Y
O
A
y
x
=
Y'
B
M
(4�4)
,
X
X'
(4
2 0)
,
APPLICATION OF INTEGRALS 365
O
F (
o)
ae,
B
Y
Y′
B'
S
R
X
X′
x
ae
=
Again, the area of the region BMAB
=
4 2
4
ydx
∫
=
4 2
2
4
32
−x dx
∫
=
4 2
2
–1
4
1
1
32
32
sin
2
2
4 2
x
x
x
−
+
×
×
=
–1
–1
1
1
4
1
1
4 2
0
32
sin 1
32
16
32
sin
2
2
2
2
2
×
+
×
×
−
−
+
×
×
= 8 π – (8 + 4π) = 4π – 8 (4)
Adding (3) and (4), we get, the required area = 4π Example 5 Find the area bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
= and the ordinates x = 0
and x = ae, where, b2 = a2 (1 – e2) and e < 1 Solution The required area (Fig 8 |
1 | 4217-4220 | (4)
Adding (3) and (4), we get, the required area = 4π Example 5 Find the area bounded by the ellipse
2
2
2
2
1
x
y
a
b
+
= and the ordinates x = 0
and x = ae, where, b2 = a2 (1 – e2) and e < 1 Solution The required area (Fig 8 12) of the region BOB′RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae |
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