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)
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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
.
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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)
=
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

π

=π


Fig 8.5
Fig 8.4
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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
=
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
π
=π
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 theregion AOBA the first quadrantbound
ed
bythecurve x axisand 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
π
=π
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
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.
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 by curve, 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
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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 x
2
= 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
0
24
x dx
−
(Why?)
=
2
3
0
24
3
x
x

−


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 x
2
+ y
2
= 32.
Solution The given equations are
y = x ... (1)
and x
2
+ y
2
= 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.
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 8.11
Y
O
A
yx=
Y'
B
M
(44),
X
X
'
(4 2 0),
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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
=
42
4
ydx
=
42
2
4
32
x dx
−
=
42
2 –1
4
1 1
32 32 sin
2 2
42
x
xx
−+××
=
= 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
22
22
1
xy
ab
+=
and the ordinates x = 0
and x = ae, where, b
2
= a
2
(1 – e
2
) 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.
Note that the area of the region BOB′RFSB
=
0
2
ae
ydx
=
22
0
2
ae
b
a x dx
a
−
=
2
2 2 –1
0
2
sin
22
ae
bx a x
ax
a a
−+
=
2 2 2 2 –1
2
sin
2
b
ae a a e a e
a
−+
=
2 –1
1 sin
ab e e e
−+
EXERCISE 8.1
1. Find the area of the region bounded by the curve y
2
= x and the lines x = 1,
x = 4 and the x-axis in the first quadrant.
2. Find the area of the region bounded by y
2
= 9x, x = 2, x = 4 and the x-axis in the
first quadrant.
Fig 8.12
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366 MATHEMATICS
3. Find the area of the region bounded by x
2
= 4y, y = 2, y = 4 and the y-axis in the
first quadrant.
4. Find the area of the region bounded by the ellipse
22
1
16 9
xy
+=
.
5. Find the area of the region bounded by the ellipse
22
1
49
xy
+=
.
6. Find the area of the region in the first quadrant enclosed by x-axis, line x =
3
y
and the circle x
2
+ y
2
= 4.
7. Find the area of the smaller part of the circle x
2
+ y
2
= a
2
cut off by the line
2
a
x=
.
8. The area between x = y
2
and x = 4 is divided into two equal parts by the line
x = a, find the value of a.
9. Find the area of the region bounded by the parabola y = x
2
and y =
x
.
10. Find the area bounded by the curve x
2
= 4y and the line x = 4y – 2.
11. Find the area of the region bounded by the curve y
2
= 4x and the line x = 3.
Choose the correct answer in the following Exercises 12 and 13.
12. 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) π (B)
2
π
(C)
3
π
(D)
4
π
13. 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
8.3 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 represented by
y = f (x), y = g (x), where f(x) ≥ g(x) in [a, b] 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
2019-20
APPLICATION OF INTEGRALS 367
y fx= ()
X
Y
y gx=()
xa=
xc=
y gx=()
y fx= ()
xb
=
A
B
R
C
D
Q
O
P
X
′
Y′
f(x) – g(x) and width dx so that the elementary area
Fig 8.13
Fig 8.14
dA = [f (x) – g(x)] dx, and the total area A can be taken as
A =
[() ()]
b
a
f x g x dx−
∫
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
=
() ()
b b
a a
f x dx g x dx
−
∫∫
=
[]
() () ,
b
a
f x g x dx
−
∫
where f (x) ≥ g (x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the
Fig 8.14, then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
=
[][]
() () () ()
c b
a c
f x g x dx g x f x dx
−+−
∫∫
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368 MATHEMATICS
Y
O
P(4,4)
C(4,0)
Y
X
X
Q(8,0)
Fig 8.16
Example 6 Find the area of the region bounded by the two parabolas y = x
2
and y
2
= x.
Solution The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8.15.
Here, we can set y
2
= x or y =
x
= f(x) and y = x
2
= g(x), where, f (x) ≥ g (x) in [0, 1].
Therefore, the required area of the shaded region
=
[]
1
0
() ()
f x g x dx−
=
1
2
0
x x dx

−

=
21 1
33 3
−=
Example 7 Find the area lying above x-axis and included between the circle
x
2
+ y
2
= 8x and inside of the parabola y
2
= 4x.
Solution The given equation of the circle x
2
+ y
2
= 8x can be expressed as
(x – 4)
2
+ y
2
= 16. Thus, the centre of the
circle is (4, 0) and radius is 4. Its intersection
with the parabola y
2
= 4x gives
x
2
+ 4x = 8x
or x
2
– 4x = 0
or x (x – 4) = 0
or x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0, 0) and P(4,4) above the
x-axis.
From the Fig 8.16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
=
4 8
0 4
ydx ydx
+

=
4 8
2 2
0 4
2 4 ( 4)
x dx x dx
+ −−

(Why?)
Fig 8.15
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APPLICATION OF INTEGRALS 369
= (Why?)
=
= =
4
(8 3 )
3
+π
Example 8 In Fig 8.17, AOBA is the part of the ellipse 9x
2
+ y
2
= 36 in the first
quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the
chord AB.
Solution Given equation of the ellipse 9x
2
+ y
2
= 36 can be expressed as
22
1
4 36
xy
+=
or
22
22
1
26
xy
+=
and hence, its shape is as given in Fig 8.17.
Accordingly, the equation of the chord AB is
y – 0 =
60
( 2)
02
x
−
−
−
or y = – 3(x – 2)
or y = – 3x + 6
Area of the shaded region as shown in the Fig 8.17.
=
(Why?)
=
=
32 6
2
=×× −
= 3π – 6
Fig 8.17
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370 MATHEMATICS
Example 9 Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1).
Solution Let A (1, 0), B (2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8.18).
Area of ∆ABC
= Area of ∆ABD + Area of trapezium
BDEC – Area of ∆AEC
Now equation of the sides AB, BC and
CA are given by
y = 2 (x – 1), y = 4 – x, y =
1
2
(x – 1), respectively.
Hence, area of ∆ ABC =
2 3 3
1 2 1
1
2 ( 1) (4 )
2
x
x dx x dx dx
−
−+−−
∫ ∫∫
=
2 3 3
2 2 2
1 2 1
1
24
2 2 22
x xx
xx
x
 
−+− − −
 
 
=
2
22
21 3 2
2 2 143 42
22 2 2

 

− − − + ×− − ×−




–
2
13 1
31
22 2


−− −


=
3
2
Example 10 Find the area of the region enclosed between the two circles: x
2
+ y
2
= 4
and (x – 2)
2
+ y
2
= 4.
Solution Equations of the given circles are
x
2
+ y
2
= 4 ... (1)
and (x – 2)
2
+ y
2
= 4 ... (2)
Equation (1) is a circle with centre O at the
origin and radius 2. Equation (2) is a circle with
centre C (2, 0) and radius 2. Solving equations
(1) and (2), we have
(x –2)
2
+ y
2
= x
2
+ y
2
or x
2
– 4x + 4 + y
2
= x
2
+ y
2
or x = 1 which gives y =
3
±
Thus, the points of intersection of the given
circles are A(1,
3
) and A′(1,
–3
) as shown in
the Fig 8.19.
Fig 8.18
Fig 8.19
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APPLICATION OF INTEGRALS 371
Required area of the enclosed region O ACA′O between circles
= 2 [area of the region ODCAO] (Why?)
= 2 [area of the region ODAO + area of the region DCAD]
=
1 2
0 1
2
ydx ydx
+
∫∫
=
1 2
2 2
0 1
2 4 ( 2) 4
x dx x dx
−− + −
∫ ∫
(Why?)
=
1
2 –1
0
1 1 2
2 ( 2) 4 ( 2) 4sin
2
22
x
xx
−

− − − +×


+
2
2 –1
1
1 1
2 4 4sin
2 22
x
xx
− +×
=
1
2
2 –1 2 –1
1
0
2
( 2) 4 ( 2) 4sin 4 4sin
2 2
x x
xx xx
−

− −− + + −+


=
–1 1 –1 1
1 1
3 4sin 4sin ( 1) 4sin 1 3 4sin
2 2
− −
−

−+ − −+ − −


=
34 4 4 34
62 2 6
ππ π π

− −× +× + × − −×



=
2 2
3 2 23
3
3
π π

− − + π + π− −


=
8
23
3
π
−
EXERCISE 8.2
1. Find the area of the circle 4x
2
+ 4y
2
= 9 which is interior to the parabola x
2
= 4y.
2. Find the area bounded by curves (x
– 1)
2
+ y
2
= 1 and x
2
+ y
2
= 1.
3. Find the area of the region bounded by the curves y = x
2
+ 2, y = x, x = 0 and
x = 3.
4. Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2).
5. Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4.
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372 MATHEMATICS
Choose the correct answer in the following exercises 6 and 7.
6. Smaller area enclosed by the circle x
2
+ y
2
= 4 and the line x + y = 2 is
(A) 2 (π – 2) (B) π – 2 (C) 2π – 1 (D) 2 (π + 2)
7. Area lying between the curves y
2
= 4x and y = 2x is
(A)
2
3
(B)
1
3
(C)
1
4
(D)
3
4
Miscellaneous Examples
Example 11 Find the area of the parabola y
2
= 4ax bounded by its latus rectum.
Solution From Fig 8.20, the vertex of the parabola
y
2
= 4ax is at origin (0, 0). The equation of the
latus rectum LSL′ is x = a. Also, parabola is
symmetrical about the x-axis.
The required area of the region OLL′O
= 2 (area of the region OLSO)
=
0
2
a
ydx
=
0
24
a
ax dx
=
0
22
a
a xdx
×
=
3
2
0
2
4
3
a
ax

×



=
3
2
8
3
aa




=
2
8
3
a
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1.
Solution As shown in the Fig 8.21, the line
y = 3x + 2 meets x-axis at x =
2
3
−
and its graph
lies below x-axis for
and above
x-axis for
.
Fig 8.21
X
′
O
Y′
X
Y
S
L
L'
(,0)a
Fig 8.20
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APPLICATION OF INTEGRALS 373
Fig 8.23
The required area = Area of the region ACBA + Area of the region ADEA
=
2
1
3
2
1
3
(3 2) (3 2)
x dx x dx
−
−
−
++ +
∫∫
=
2
1
2 2
3
2
1
3
3 3
2 2
2 2
x x
x
x
−
−
−

+ ++


=
1 25 13
66 3
+=
Example 13 Find the area bounded by
the curve y = cos x between x = 0 and
x = 2π.
Solution From the Fig 8.22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD.
Thus, we have the required area
=
3
2
2
2
3
0
22
cos cos cos
xdx xdx xdx
+ +
∫∫ ∫
=
[]
[]
[]
3
2
2 2
3
0
2
2
sin sin sin
xx x
π π
π
π
π
++
= 1 + 2 + 1 = 4
Example 13 Prove that the curves y
2
= 4x and x
2
= 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts.
Solution Note that the point of intersection of the
parabolas y
2
= 4x and x
2
= 4y are (0, 0) and (4, 4) as
Fig 8.22
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374 MATHEMATICS
Y'
R
O
XX
'
x
=2
T S
P(0,1)
x =1
Y
Q(1,2)
shown in the Fig 8.23.
Now, the area of the region OAQBO bounded by curves y
2
= 4x and x
2
= 4y.
=
=
=
32 16 16
333
−=
... (1)
Again, the area of the region OPQAO bounded by the curves x
2
= 4y, x = 0, x = 4
and x-axis
=
... (2)
Similarly, the area of the region OBQRO bounded by the curve y
2
= 4x, y-axis,
y = 0 and y = 4
=
... (3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i.e., area bounded by parabolas
y
2
= 4x and x
2
= 4y divides the area of the square in three equal parts.
Example 14 Find the area of the region
{(x, y) : 0 ≤ y ≤ x
2
+ 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}
Solution Let us first sketch the region whose area is to
be found out. This region is the intersection of the
following regions.
A
1
= {(x, y) : 0 ≤ y ≤ x
2
+ 1},
A
2
= {(x, y) : 0 ≤ y ≤ x + 1}
and A
3
= {(x, y) : 0 ≤ x ≤ 2}
The points of intersection of y = x
2
+ 1 and y = x + 1 are points P(0, 1) and Q(1, 2).
From the Fig 8.24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
=
1 2
2
0 1
( 1) ( 1)
x dx x dx
++ +

(Why?)
Fig 8.24
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APPLICATION OF INTEGRALS 375
=
1 2
3 2
1
0
3 2
x x
x x


+++







=
()
1 1
1 0 22 1
3 2

 
+−+ +− +






=
23
6
Miscellaneous Exercise on Chapter 8
1. Find the area under the given curves and given lines:
(i) y = x
2
, x = 1, x = 2 and x-axis
(ii) y = x
4
, x = 1, x = 5 and x-axis
2. Find the area between the curves y = x and y = x
2
.
3. Find the area of the region lying in the first quadrant and bounded by y = 4x
2
,
x = 0, y = 1 and y = 4.
4. Sketch the graph of y =
3
x
+
and evaluate .
5. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
6. Find the area enclosed between the parabola y
2
= 4ax and the line y = mx.
7. Find the area enclosed by the parabola 4y = 3x
2
and the line 2y = 3x + 12.
8. Find the area of the smaller region bounded by the ellipse
22
1
94
xy
+=
and the
line
1
32
xy
+=
.
9. Find the area of the smaller region bounded by the ellipse
22
22
1
xy
ab
+=
and the
line
1
xy
ab
+=
.
10. Find the area of the region enclosed by the parabola x
2
= y, the line y = x + 2 and
the x-axis.
11. Using the method of integration find the area bounded by the curve
1
xy
+=
.
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1].
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376 MATHEMATICS
12. Find the area bounded by curves {(x, y) : y ≥ x
2
and y = | x|}.
13. Using the method of integration find the area of the triangle ABC, coordinates of
whose vertices are A(2, 0), B (4, 5) and C (6, 3).
14. Using the method of integration find the area of the region bounded by lines:
2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
15. Find the area of the region {(x, y) : y
2
≤ 4x, 4x
2
+ 4y
2
≤ 9}
Choose the correct answer in the following Exercises from 16 to 20.
16. Area bounded by the curve y = x
3
, the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9 (B)
15
4
−
(C)
15
4
(D)
17
4
17. The area bounded by the curve y = x | x| , x-axis and the ordinates x = – 1 and
x = 1 is given by
(A) 0 (B)
1
3
(C)
2
3
(D)
4
3
[Hint : y = x
2
if x > 0 and y = – x
2
if x < 0].
18. The area of the circle x
2
+ y
2
= 16 exterior to the parabola y
2
= 6x is
(A)
4
(4 3)
3
π−
(B)
4
(4 3)
3
π+
(C)
4
(8 3)
3
π−
(D)
4
(8 3)
3
π+
19. The area bounded by the y-axis, y = cos x and y = sin x when
0
2
x
π
≤≤
is
(A)
2 ( 2 1)
−
(B)
21
−
(C)
21
+
(D)
2
Summary
The area of the region bounded by the curve y = f (x), x-axis and the lines
x = a and x = b (b > a) is given by the formula:
Area ()
b b
a a
ydx f x dx
==
∫∫
.
The area of the region bounded by the curve x = φ (y), y-axis and the lines
y = c, y = d is given by the formula:
Area ()
d d
c c
xdy y dy
= =φ
∫∫
.
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APPLICATION OF INTEGRALS 377
The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
[]
Area () ()
b
a
f x g x dx
=−
∫
, where, f (x) ≥ g (x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], a < c < b, then
[][]
Area () () () ()
c b
a c
f x g x dx g x f x dx= −+−
∫∫
.
Historical Note
The origin of the Integral Calculus goes back to the early period of development
of Mathematics and it is related to the method of exhaustion developed by the
mathematicians of ancient Greece. This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc. In this sense, the method of exhaustion can be regarded as an early method
of integration. The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B.C.) and Archimedes
(300 B.C.)
Systematic approach to the theory of Calculus began in the 17th century.
In 1665, Newton began his work on the Calculus described by him as the theory
of fluxions and used his theory in finding the tangent and radius of curvature at
any point on a curve. Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents.
During 1684-86, Leibnitz published an article in the Acta Eruditorum
which he called Calculas summatorius, since it was connected with the summation
of a number of infinitely small areas, whose sum, he indicated by the symbol ‘∫’.
In 1696, he followed a suggestion made by J. Bernoulli and changed this article to
Calculus integrali. This corresponded to Newton’s inverse method of tangents.
Both Newton and Leibnitz adopted quite independent lines of approach which
was radically different. However, respective theories accomplished results that
were practically identical. Leibnitz used the notion of definite integral and what is
quite certain is that he first clearly appreciated tie up between the antiderivative
and the definite integral.
Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P.de Fermat, I. Newton and G. Leibnitz at the end of 17th century.
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378 MATHEMATICS
—

—
However, this justification by the concept of limit was only developed in the
works of A.L. Cauchy in the early 19th century. Lastly, it is worth mentioning the
following quotation by Lie Sophie’s:
“It may be said that the conceptions of differential quotient and integral which
in their origin certainly go back to Archimedes were introduced in Science by the
investigations of Kepler, Descartes, Cavalieri, Fermat and Wallis .... The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz”.
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