NCERT Textbook: Application of Integrals

NCERT Textbook: Application of Integrals | NCERT Textbooks (Class 6 to Class 12) - CTET & State TET PDF Download

``` Page 1

292 MATHEMATICS
Fig 8.1
v One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form. – BIRKHOFF v
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). W e 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)
Rationalised 2023-24
Page 2

292 MATHEMATICS
Fig 8.1
v One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form. – BIRKHOFF v
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). W e 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)
Rationalised 2023-24
APPLICATION OF INTEGRALS         293
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
.
Rationalised 2023-24
Page 3

292 MATHEMATICS
Fig 8.1
v One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form. – BIRKHOFF v
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). W e 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)
Rationalised 2023-24
APPLICATION OF INTEGRALS         293
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
.
Rationalised 2023-24
294 MATHEMATICS
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)
=
2 2
0
4
a
a x dx -
?
Since x
2
+ y
2
= a
2
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
2
4
2 2
a
a
? ?
p ? ?
=p
? ? ? ?
? ? ? ?
Fig 8.5
Fig  8.4
Rationalised 2023-24
Page 4

292 MATHEMATICS
Fig 8.1
v One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form. – BIRKHOFF v
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). W e 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)
Rationalised 2023-24
APPLICATION OF INTEGRALS         293
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
.
Rationalised 2023-24
294 MATHEMATICS
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)
=
2 2
0
4
a
a x dx -
?
Since x
2
+ y
2
= a
2
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
2
4
2 2
a
a
? ?
p ? ?
=p
? ? ? ?
? ? ? ?
Fig 8.5
Fig  8.4
Rationalised 2023-24
APPLICATION OF INTEGRALS         295
Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the
region enclosed by circle
=
0
4
a
xdy
?
=
2 2
0
4
a
a y dy -
?
(Why?)
=
2
2 2 1
0
4 sin
2 2
a
a y y
a y
a
-
? ?
- +
? ?
? ?
=
=
2
2
4
2 2
a
a
p
= p
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
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
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
0
4
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
p
= p
Fig 8.6
Fig 8.7
Rationalised 2023-24
Page 5

292 MATHEMATICS
Fig 8.1
v One should study Mathematics because it is only through Mathematics that
nature can be conceived in harmonious form. – BIRKHOFF v
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). W e 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)
Rationalised 2023-24
APPLICATION OF INTEGRALS         293
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
.
Rationalised 2023-24
294 MATHEMATICS
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)
=
2 2
0
4
a
a x dx -
?
Since x
2
+ y
2
= a
2
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
2
4
2 2
a
a
? ?
p ? ?
=p
? ? ? ?
? ? ? ?
Fig 8.5
Fig  8.4
Rationalised 2023-24
APPLICATION OF INTEGRALS         295
Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the
region enclosed by circle
=
0
4
a
xdy
?
=
2 2
0
4
a
a y dy -
?
(Why?)
=
2
2 2 1
0
4 sin
2 2
a
a y y
a y
a
-
? ?
- +
? ?
? ?
=
=
2
2
4
2 2
a
a
p
= p
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
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
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
0
4
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
p
= p
Fig 8.6
Fig 8.7
Rationalised 2023-24
296 MATHEMATICS
Alternatively, considering horizontal strips as
shown in the Fig 8.8, the area of the ellipse is
= 4
0
xdy
b
?
= 4
2 2
0
a
b
b y dy
b
-
?
(Why?)
=
=
2
–1
4
0 sin 1 0
2 2
a b b
b
? ? ? ?
× + -
? ? ? ?
? ? ? ? ? ?
=
2
4
2 2
a b
ab
b
p
=p
EXERCISE 8.1
1. Find the area of the region bounded by the ellipse
2 2
1
16 9
x y
+ = .
2. Find the area of the region bounded by the ellipse
2 2
1
4 9
x y
+ = .
Choose the correct answer in the following Exercises 3 and 4.
3. Area lying in the first quadrant and bounded by the circle x
2
+ y
2
= 4 and the lines
x = 0 and x = 2 is
(A) p (B)
2
p
(C)
3
p
(D)
4
p
4. Area of the region bounded by the curve y
2
= 4x, y-axis and the line y = 3 is
(A) 2 (B)
9
4
(C)
9
3
(D)
9
2
Miscellaneous Examples
Example 3  Find the area of the region bounded by the line y = 3x + 2, the x-axis and
the ordinates x = –1 and x = 1.
Fig 8.8
Rationalised 2023-24
```

NCERT Textbooks (Class 6 to Class 12)

681 docs|672 tests

FAQs on NCERT Textbook: Application of Integrals - NCERT Textbooks (Class 6 to Class 12) - CTET & State TET

 1. What is the concept of integration and how is it applied in real-life situations?
Ans. Integration is a mathematical concept used to calculate the area under a curve or the accumulation of a quantity over a given interval. It is widely used in various real-life situations such as calculating the total distance covered by an object, finding the area of irregular shapes, determining the volume of a solid, and analyzing population growth or decay.
 2. How can integrals be used to solve problems related to motion and displacement?
Ans. Integrals are used in physics to solve problems related to motion and displacement. By integrating the velocity function over a given interval, we can find the total displacement of an object during that interval. This is because the integral of velocity gives us the area under the velocity-time graph, which represents the displacement.
 3. Can integrals be used to determine the average value of a function over a given interval?
Ans. Yes, integrals can be used to determine the average value of a function over a given interval. By integrating the function over that interval and dividing it by the length of the interval, we can find the average value. This is useful in various applications such as calculating average temperature, average speed, or average rate of change.
 4. How are definite integrals used to find the area between two curves?
Ans. Definite integrals are used to find the area between two curves by subtracting the integral of the lower curve from the integral of the upper curve over a given interval. This gives us the net area between the curves. By setting up the integral properly and evaluating it, we can determine the exact value of the area.
 5. Can integrals be used to solve optimization problems?
Ans. Yes, integrals can be used to solve optimization problems. By formulating the problem as an optimization question and finding an appropriate function to optimize, we can use calculus and integrals to find the maximum or minimum value of that function. This technique is often used in economics, engineering, and other fields to maximize profits, minimize costs, or optimize various parameters.

NCERT Textbooks (Class 6 to Class 12)

681 docs|672 tests

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