# NCERT Textbook - Application of Integrals JEE Notes | EduRev

## JEE : NCERT Textbook - Application of Integrals JEE Notes | EduRev

``` Page 1

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. 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.
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.
8.2 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. 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)
2019-20
Page 2

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. 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.
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.
8.2 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. 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)
2019-20
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 8.2
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.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 8.4. Here, A
1
< 0 and A
2
> 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 = |A
1
| + A
2
.
2019-20
Page 3

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. 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.
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.
8.2 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. 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)
2019-20
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 8.2
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.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 8.4. Here, A
1
< 0 and A
2
> 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 = |A
1
| + A
2
.
2019-20
APPLICATION OF INTEGRALS         361
Example 1 Find the area enclosed by the circle x
2
+ y
2
= a
2
.
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]
=
0
4
a
ydx
?
(taking vertical strips)
=
22
0
4
a
a x dx -
?
Since x
2
+ y
2
= a
2
gives    y =
22
ax ±-
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 ax
ax
a
? ?
-+
? ?
? ?
=  =
2
2
4
22
a
a
??
p ??
=p
? ?? ?
? ?? ?
Fig 8.5
Fig  8.4
2019-20
Page 4

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. 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.
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.
8.2 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. 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)
2019-20
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 8.2
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.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 8.4. Here, A
1
< 0 and A
2
> 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 = |A
1
| + A
2
.
2019-20
APPLICATION OF INTEGRALS         361
Example 1 Find the area enclosed by the circle x
2
+ y
2
= a
2
.
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]
=
0
4
a
ydx
?
(taking vertical strips)
=
22
0
4
a
a x dx -
?
Since x
2
+ y
2
= a
2
gives    y =
22
ax ±-
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 ax
ax
a
? ?
-+
? ?
? ?
=  =
2
2
4
22
a
a
??
p ??
=p
? ?? ?
? ?? ?
Fig 8.5
Fig  8.4
2019-20
362 MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the
region enclosed by circle
=
0
4
a
xdy
?
=
22
0
4
a
a y dy -
?
(Why?)
=
2
22 1
0
4 sin
2 2
a
a y y
ay
a
-
? ?
-+
? ?
? ?
=
=
2
2
4
22
a
a
p
=p
Example 2 Find the area enclosed by the ellipse
22
22
1
xy
ab
+=
Solution From Fig 8.7, the area of the region ABA'B'A bounded by the ellipse
=
in
4
, 0,
area of the region AOBA the first quadrant bounded
by thecurve x axis and theordinates x x a
? ?
? ?
- ==
? ?
(as the ellipse is symmetrical about both x-axis and y-axis)
=
0
4 (taking verticalstrips)
a
ydx
?
Now
22
22
xy
ab
+
= 1 gives
22
b
y ax
a
=± -
, but as the region AOBA lies in the first
quadrant, y is taken as positive. So, the required area is
=
22
0
4
a b
a x dx
a
-
?
=
2
2 2 –1
0
4
sin
22
a
bx a x
ax
a a
? ?
-+
? ?
? ?
(Why?)
=
2
1
4
0 sin 1 0
22
ba a
a
-
? ? ? ?
×+ -
? ? ? ?
? ? ? ? ? ?
=
2
4
22
ba
ab
a
p
=p
Fig 8.6
Fig 8.7
2019-20
Page 5

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. 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.
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.
8.2 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. 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)
2019-20
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 8.2
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.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 8.4. Here, A
1
< 0 and A
2
> 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 = |A
1
| + A
2
.
2019-20
APPLICATION OF INTEGRALS         361
Example 1 Find the area enclosed by the circle x
2
+ y
2
= a
2
.
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]
=
0
4
a
ydx
?
(taking vertical strips)
=
22
0
4
a
a x dx -
?
Since x
2
+ y
2
= a
2
gives    y =
22
ax ±-
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 ax
ax
a
? ?
-+
? ?
? ?
=  =
2
2
4
22
a
a
??
p ??
=p
? ?? ?
? ?? ?
Fig 8.5
Fig  8.4
2019-20
362 MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the
region enclosed by circle
=
0
4
a
xdy
?
=
22
0
4
a
a y dy -
?
(Why?)
=
2
22 1
0
4 sin
2 2
a
a y y
ay
a
-
? ?
-+
? ?
? ?
=
=
2
2
4
22
a
a
p
=p
Example 2 Find the area enclosed by the ellipse
22
22
1
xy
ab
+=
Solution From Fig 8.7, the area of the region ABA'B'A bounded by the ellipse
=
in
4
, 0,
area of the region AOBA the first quadrant bounded
by thecurve x axis and theordinates x x a
? ?
? ?
- ==
? ?
(as the ellipse is symmetrical about both x-axis and y-axis)
=
0
4 (taking verticalstrips)
a
ydx
?
Now
22
22
xy
ab
+
= 1 gives
22
b
y ax
a
=± -
, but as the region AOBA lies in the first
quadrant, y is taken as positive. So, the required area is
=
22
0
4
a b
a x dx
a
-
?
=
2
2 2 –1
0
4
sin
22
a
bx a x
ax
a a
? ?
-+
? ?
? ?
(Why?)
=
2
1
4
0 sin 1 0
22
ba a
a
-
? ? ? ?
×+ -
? ? ? ?
? ? ? ? ? ?
=
2
4
22
ba
ab
a
p
=p
Fig 8.6
Fig 8.7
2019-20
APPLICATION OF INTEGRALS         363
Alternatively, considering horizontal strips as
shown in the Fig 8.8, the area of the ellipse is
= 4
0
xdy
b
?
= 4
22
0
a
b
b y dy
b
-
?
(Why?)
=
=
2
–1
4
0 sin 1 0
22
ab b
b
? ? ? ?
×+ -
? ? ? ?
? ? ? ? ? ?
=
2
4
22
ab
ab
b
p
=p
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.
Example 3 Find the area of the region bounded
by the curve y = x
2
and the line y = 4.
Solution Since the given curve represented by
the equation y = x
2
is a parabola symmetrical
about y-axis only, therefore, from Fig 8.9, the
required area of the region AOBA is given by
2
0
4
xdy
?
=
area of theregion BONBbounded bycurve, axis
2
and thelines 0and = 4
y
yy
- ? ?
? ?
=
? ?
=
4
0
2 ydy
?
= 2
2
3
3
2
4
×
?
?
?
?
?
?
?
?
y
0

4 32
8
33
= ×=
(Why?)
Here, we have taken horizontal strips as indicated in the Fig 8.9.
Fig 8.8
Fig 8.9
2019-20
```
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## Mathematics (Maths) Class 12

209 videos|222 docs|124 tests

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