3. Quadrature Complete Theory Module-5.pdf

41
QUADRATURE
Total No. of questions in Quadrature are-
In Chapter Examples................................................17
Solved Examples.....................................................19
Total No. of questions............................................36

42
1. INTRODUCTION
The process of finding area of some plane region
is called Quadrature. In this chapter we shall
find the area bounded by some simple plane
curves with the help of definite integral. For solving
the problems on quadrature easily, if possible
first draw the rough sketch of the required area.
2. CURVE TRACING
In chapter function, we have seen graphs of some
simple elementary curves. Here we introduce
some essential steps for curve tracing which will
enable us to determine the required area.
(i) Symmetry
The curve f(x, y) = 0 is symmetrical
 about x-axis if all terms of y contain even
powers.
 about y-axis if all terms of x contain even
powers.
 about the origin if f (– x, – y) = f (x, y).
For example, y2
= 4ax is symmetrical about
x-axis, and x2
= 4ay is symmetrical about
y- axis and the curve y = x3
is symmetrical about
the origin.
(ii) Origin
If the equation of the curve contains no constant
term then it passes through the origin.
For example x2
+ y2
+ 2ax = 0 passes through
origin.
(iii) Points of intersection with the axes
If we get real values of x on putting y = 0 in the
equation of the curve, then real values of x and
y = 0 give those points where the curve cuts the
x-axis. Similarly by putting x = 0, we can get the
points of intersection of the curve and y-axis.
For example, the curve x2
/a2
+ y2
/b2
= 1 intersects
the axes at points (± a, 0) and (0, ± b) .
(iv) Region
Write the given equation as y = f(x) , and find
minimum and maximum values of x which
determine the region of the curve.
For example for the curve xy2
= a2
(a – x)
y = a
a x
x

Now y is real, if 0 < x  a , so its region lies
between the lines x = 0 and x = a.
3. AREA BOUNDED BY A CURVE
(i) The area bounded by a cartesian curve
y = f(x), x-axis and ordinates x = a and
x = b is given by,
Area = 
b
a
dx
y = 
b
a
dx
)
x
(
f
Examples
based on Area Bounded by A Curve
Ex.1 Find the area bounded by the curve y = x3
,
x-axis and ordinates x = 1 and x = 2.
Sol. Required Area = ydx
1
2
z
= x dx
3
1
2
z =
x4
1
2
4
L
N
M
M
O
Q
P
P=
15
4
Ans.
Ex.2 Find the area bounded by the curve y= sec2
x,
x-axis and the line x =

4
Sol. Required Area = ydx
x
z
0
4
/
= sec
/
2
0
4
xdx

z = tan
/
x 0
4

= 1 Ans.
Ex.3 Find the area bounded by the curve y = mx,
x-axis and ordinates x = 1 and x = 2
Sol. Required area = ydx
1
2
z .
= mxdx
1
2
z =
mx2
1
2
2
L
N
M
M
O
Q
P
P
=
m
2
( 4 – 1) =
3
2
F
H
GI
K
Jm Ans.

43
Ex.4 Find the area bounded by the curve
y = x (1– x)2
and x-axis.
Sol. Clearly the given curve meets the x-axis at
(0,0) and (1,0) and for x = 0 to 1, y is positive
so required area-
= x x dx
1
2
0
1

zb g
= x x x dx
 
z 2 2 3
0
1
e j
=
x x x
2 3 4
0
1
2
2
3 4
 
L
N
M
M
O
Q
P
P=
1
2
–
2
3
+
1
4
=
1
12
Ans.
(ii) The area bounded by a cartesian curve
x = f(y), y-axis and abscissa y = c and
y = d
Area = x dy
c
d
z = f y dy
c
d
( )
z
Ex.5 Find the area bounded by the curve x2
=
1
4
y,
,
y-axis and between the lines y = 1 and
y = 4.
Sol. Required Area = x dy
1
4
z
=
1
2
1
4
y dy
z =
1
3
y3 2
1
4
/
=
1
3
( 8 – 1) = 7/3 Ans.
Ex.6 Find the area bounded by the curve y = logx;
x-axis and the line y = 2.
Sol. Required Area = xdy
0
2
z
= e dy
y
0
2
z
= ey
e j0
2
= e2
– 1 Ans.
Ex.7 Find the area bounded between the curve
y2
= 2y – x and y-axis.
Sol. The area between the given curve x = 2y– y2
and y-axis will be as shown in diagram.
 Required Area = 2 2
0
2
y y dy

z
e j
= y
y
2
3
0
2
3

L
N
M
M
O
Q
P
P=
4
3
(iii) If the equation of a curve is in parametric
form, say x = f(t) , y = g(t), then the area
= ydx
a
b
z = g t f t dt
t
t
( ) '( )
1
2
z .
Where t1
and t2
are the values of t respectively
corresponding to the values of a & b of x.
Ex.8 Find the area bounded by the curves
x = a cost, y = b sint in the first quadrant.
Sol. Clearly the given equation are the parametric
equation of ellipse
x
a
y
b
2
2
2
2
 = 1. Curve meet
the x-axis in the first quadrant at (a,0)
 Required area = y dx
a
0
z
= b t
sin
/
b g
 2
0
z ( – a sin t ) dt
(  at x = 0, t =  / 2 and x = a, t = 0)
= ab sin
/
2
0
2
t dt

z =
ab
4
F
H
G I
K
J Ans.
4. SYMMETRICAL AREA
If the curve is symmetrical about a coordinate
axis (or a line or origin), then we find the area of
one symmetrical portion and multiply it by the
number of symmetrical portions to get the
required area.
Y
O (0,0) (1,0) X
Y
O (0,0) (1,0) X

44
Examples
based on Symmetric Area
Ex.9 Find the area bounded by the parabola
y2
= 4x and its latus rectum.
Sol. Since the curve is symmetrical about x- axis
therefore the required Area
= 2 ydx
0
1
z
= 2 4
0
1
x dx
z
= 4.
2
3
x3 2
0
1
/
= 8/3 Ans.
Ex.10 Find the whole area of circle x2
+ y2
=a2
Sol. The required area is symmetric about both
the axis as shown in figure.
 Required Area
= 4 a x dx
a
2 2
0

z
= 4
x
a x
a x
a
a
2 2
2 2
2
1
0
 
L
N
M
M
O
Q
P
P

sin
= 4.

2 2
2

L
N
M
M
O
Q
P
P
a
= a2
Ans.
Ex.11 Find the area bounded by the curve
xy2
= a2
(a – x) and x-axis.
Sol. xy2
= a2
(a – x)  y2
= a2
(a – x)/x
or y = a
a x
x

...(1)
Here the power of y is even so curve is
symmetric w.r.t x-axis. Putting x = a in
equation (1) we get
y = 0 and at x = 0, we get y = 
 required area
A = A1
+ A2
= 2A1
A = 2 y dx
x
x a


z0
= 2 a
a
0
z a x
x

dx
(from Eq.(1))
Putting x = a sin2

dx = 2a sin  cos  d 
When x = 0   = 0  x = a   =

2
 area =2 a
0
2
/
z cos
sin


. 2a cos sin  d 
= 4a2 cos
/
2
0
2

z  d  = 4a2
.
1
2
.

2
=  a2
Ans.
5. POSITIVE AND NEGATIVE AREA
Area is always taken as positive. If some part of
the area lies in the positive side i.e. above x-
axis and some part lies in the negative side i.e.
below x- axis then the area of two parts should
be calculated separately and then add their
numerical values to get the desired area.
Examples
based on Positive and Negative Area
Ex.12 Find the area between the curve y = cos x
and x-axis when  /4 < x <  .
Sol. Here some part of the required area lies above
x-axis and some part lies below x- axis as
shown in the diagram. So
Required area = cos
/
/
x dx


4
2
z + cos
/
x dx


2
z
= sin /
/
x 

4
2
+ sin /
x 

2
= (1– 1/ 2 ) + | 0 – 1| =
2 2 1
2

Ans.
Y
O
S
(1,0)
Y
O
S
(1,0)

45
O x= 4a
y 2
= 4ax
x2
=4ay
Y
X
O x= 4a
y 2
= 4ax
x2
=4ay
Y
X
Ex.13 Find the area between the curve
y = x (x – 1) (x – 2) and x-axis.
Sol. Required area
= x x x dx
( )( )
 
z 1 2
0
1
+ x x x dx
( )( )
 
z 1 2
1
2
=
1
4
x x x
4 3 2
0
1
4 4
 
e j +
1
4
x x x
4 3 2
1
2
4 4
 
e j
=
1
4
[1+ |(16 – 32 + 16) – (1 – 4 + 4)| ] =
1
2
Ans.
6. AREA BETWEEN TWO CURVES
I. When two curves intersect at two points and
their common area lies between these points.
If y = f1
(x) and y = f2
(x) are two curves where
f1
(x) > f2
(x) which intersect at two points
A (x = a) and B (x = b) and their common
area lies between A & B, then their
Common area = y y dx
a
b
1 2

z
b g
y=f
(x)
2
O
A
Y
y=f (x)
1 B
X
dx x=b
x=a
= f x f x dx
a
b
1 2
( ) ( )

z
Examples
based on Area between two curves
Ex.14 Find the area between two curves y = x3
and
y = 4x in the first quadrant.
Sol. Solving the given equations, we get x = 0
and x =  2, but we have to obtain area in
the first quadrant only, therefore required area
= 4 3
0
2
x x dx

z
e j
= 2
4
2
4
0
2
x
x

F
H
G
I
K
J= 4 units. Ans.
Ex.15 Find the area between two curves y2
= 4ax
and x2
= 4ay.
Sol. By solving the equation of two given curves
y2
= 4ax and x2
= 4 ay
we get x = 4a
So required area
= 4
4
2
0
4
ax
x
a
dx
a

F
H
G
I
K
J
z
= 2
3 2 12
3 2 3
0
4
a
x x
a
a
.
/
/

L
N
M
M
O
Q
P
P
=
4
3
4
3 2
a
a
b g/
–
64
12
3
a
a
=
16
3
a2
Ans.
Ex.16 Find the area bounded between curve y = x3
and line y = x.
Sol. By solving y = x3
and y = x, we get
x = 0, – 1, 1.It is
clear from the
given figure that
the required area
is symmetrical
about the origin.
 Required area
= 2 x x dx

z 3
0
1
e j = 2
x x
2 4
2 4

L
N
M
M
O
Q
P
P
= 1/2 Ans.
II. When two curves intersect at a point and
the area between them is bounded by
x-axis. If y = f1
(x) and y = f2
(x) are two
curves which intersect at P (  ,  ) and
meet x-axis at A(a, 0), B(b, 0) respectively,
then area between them and x-axis is
given by
Area = f x dx
a
1( )

z + f x dx
b
2( )

z
Y
O
(– 1,–1)
(1,1)
X
A
B
Y
O
(– 1,–1)
(1,1)
X
A
B

46
Ex.17 Find the area between the curves y = 2x,
x + y = 1 and x-axis.
Sol. Two given curves intersect at P where
x = 1/3 and they meet x-axis at O and
A(x = 1) So
required area = 2
0
1 3
x dx
/
z + 1
1 3
1

z x dx
b g
/
= x2
0
1 3
/
+ x
x

L
N
M
M
O
Q
P
P
2
1 3
1
2
/
x
+
y
=
1
y
=
2x
P(x = 1/3)
Y
X
O
=
1
9
+
1
2
1
3
1
18
 
F
H
G I
K
J
L
N
M O
Q
P=
1
3
Ans.

47
Ex.1 The area bounded by the curve y = 3/x2,,
x-axis and the lines x = 1 and x = 2 is-
(A) 3/2 (B) 1/2
(C) 2 (D) 1
Sol. Area = y dx
1
2
z =
3
2
1
2
x
dx
z
= –
3
1
2
x
L
N
M
O
Q
P= – 3
1
2
1

F
H
G I
K
J
= 3/2 Ans.[A]
Ex.2 The area between the curve y = sin2 x,
x-axis and the ordinates x = 0 and x =

2
is-
(A)  (B)  /2
(C)  /4 (D)  /8
Sol. Required area = sin
/
2
0
2
x dx

z
=
1 2
2
0
2

z cos
/
x
dx

=
1
2
x
x

L
N
M O
Q
P
sin
/
2
2 0
2

=

4
Ans.[C]
Ex.3 The area between the curve y = 4 + 3x – x2
and x-axis is-
(A) 125/6 (B) 125/3
(C) 125/2 (D) None of these
Sol. Putting y = 0, we get,
x2 – 3x – 4 = 0
 (x – 4) (x + 1) = 0
 x = – 1 or x = 4
 Required area = 4 3 2
1
4
 

z x x dx
e j
= 4
3
2 3
2 3
1
4
x
x x
 
F
H
G
I
K
J

=
125
6
Ans.[A]
Ex.4 The area bounded by the curve y2 = 4x,
y-axis and y = 3 is-
(A) 2 units (B) 9/4 units
(C) 7/3 units (D) 3 units
Sol. Area = x dy
0
3
z =
y
dy
2
0
3
4
z
=
1
4
y3
0
3
3
L
N
M
M
O
Q
P
P=
1
12
( 27 – 0)
= 9/4 units Ans.[B]
Ex.5 The area bounded by the curve x = a cos3 t,
y = a sin3 t, is -
(A)
a2
8
(B)
a2
4
(C)
3
8
2
a
(D)
2
3
2
a
Sol. Given curve
x
a
F
H
GI
K
J
1 3
/
= cos t,
y
a
F
H
GI
K
J
1 3
/
= sin t
Squarring and adding x2/3 + y2/3 = a2/3
Clearly it is symmetric with respect to both
the axis, so whole area is
= 4 ydx
a
0
z
= 4 a t
sin .
/
3
2
0

z 3a cos2 t ( – sint) dt
By given equation at x = 0; t =

2
, at x= a;
t = 0
= 12a2 sin cos
/
4 2
0
2
t t dt

z
= 12a2
3 11
6 4 2
. .
. .
.

2
=
3
8
2
a
Ans.[C]
Ex.6 The area between the curve y = sech x and
x-axis is-
(A)  (B) 
(C) 2  (D)  /2
Sol. Given curve is symmetrical about y-axis as
shown in the diagram.
Reqd. area = 2 sech xdx
0

z
SOLVED EXAMPLE

48
= 2
2
0 e e
dx
x x
 

z = 4
e
e
dx
x
x
2
0 1


z
= 4 tan

1
0
ex
e j = 4
 
2 4

L
N
M O
Q
P
= 
Ans.[B]
Ex.7 The area bounded by the circle x2 + y2 = 1
and the curve |x| + |y| = 1 is -
(A)  – 2 (B)  – 2 2
(C) 2(  – 2 2 ) (D) None of these
Sol. By changing x as – x and y as – y, both the
given equation remains unchanged so required
area will be symmetric w.r.t both the axis,
which is shown in the fig., so required area
is
= 4 1 1
2
0
1
  
L
N
M O
Q
P
z x x dx
b g
= 4
x
x x x
x
2
1
1
2 2
2 1
2
0
1
   
L
N
M
M
O
Q
P
P

sin
= 4 0
1
2 2
1
1
2
  
L
N
M O
Q
P
.

=  – 2
Ans.[A]
Ex.8 The area bounded by the curve y = sin x,
x = 0 and x = 2  is -
(A) 4 units (B) 0 units
(C) 4  units (D) 2 units
Sol. f(x) = y = sin x
when x [0,  ], sin x  0
and when x [  ,2  ], sin x  0
 required area = y dx
0

z + ( )

zy dx


2
= sinx dx
0

z + ( sin )

z x dx


2
= [–cos x] 0

+ [cos x] 

2
= (– cos  + cos 0)+(cos2  – cos  )
= (1 + 1) + (1 + 1)
= 4 units Ans.[A]
Ex.9 The area between the curves y = x and
y = x is-
(A) 1/3 (B) 1/6
(C) 2/3 (D) 1
Sol. The points of intersection of curves are
x = 0 and x = 1.
 Required area = x x dx

z
e j
0
1
=
2
3 2
3 2 2
0
1
x x
/

L
N
M
M
O
Q
P
P
=
2
3
–
1
2
=
1
6
Ans.[B]
Ex.10 The area between the parabola x2 = 4y and
line x = 4y – 2 is-
(A) 9/4 (B) 9/8
(C) 9/2 (D) 9
Sol. Solving the equation of the given curves for x,
we get
x2 = x + 2
 (x – 2) (x + 1) = 0
 x = – 1 , 2
So, reqd. area
=
x x
dx


L
N
M
M
O
Q
P
P

z 2
4 4
2
1
2
=
1
4
x
x
x
2 3
1
2
2
2
3
 
L
N
M
M
O
Q
P
P

=
1
4
[(2 + 4 – 8/3) – (1/2 – 2 + 1/3)] = 9/8
Ans.[B]
Y
X
O
P
Q
x=-1
x=2
Y
X
O
P
Q
x=-1
x=2
x = –1
x = 2

49
Ex.11 The area between the curve y = cos2 x,
x-axis and ordinates x = 0 and x =  in the
interval (0,  ) is-
(A)  (B)  /4
(C)  /2 (D) 2 
Sol. Required area = cos2
0
x dx

z
= cos
/
2
0
2
x dx

z + cos
/
2
2
x dx


z
=
1
2
×

2
+
1
2
1 2
2

z cos
/
x dx
b g


=

4
+
1
2
x
x

L
N
M O
Q
P
sin
/
2
2 2


=

4
+
1
2



F
H
G I
K
J
L
N
M O
Q
P
2
=

4
+

4
=

2
Ans.[C]
Ex.12 The area between the curves y = tan x,
y = cot x and x-axis in the interval [0,  / 2 ]
is-
(A) log 2 (B) log 3
(C) log 2 (D) None of these
Sol. From the fig. it is clear that
= tan
/
x dx
0
4

z – cot
/
/
x dx


4
2
z
=[log sec x] 0
4
/
+[ log sin x] 

/
/
4
2
= log 2 – log
1
2
= log 2 Ans.[A]
Ex.13 The area between the curves y = cos x and
the line y = x + 1 in the second quadrant is-
(A) 1 (B) 2
(C) 3/2 (D) 1/2
Sol. Let the line y = x + 1, meets x-axis at the
point A(0,1). Also suppose that the curve
y = cos x meets x-axis and y-axis
respectively at the points C and A. From the
adjoint figure it is obvious that
Required area = area of ABC
= area of OAC – area of OAB
= cos
/
xdx

z
 2
0
–
1
2
× OB × OA
A
Y
O
(0,1)
X
( /2,0)
C B O X
= [sin x] /2
0
–
1
2
× 1 × 1
= 1 – (1/2) = (1/2). Ans.[D]
Ex.14 The area bounded by the curves y = sin x,
y = cos x and y-axis in first quadrant is-
(A) 2 – 1 (B) 2
(C) 2 + 1 (D) None of these
Sol. In first quadrant sin x and cos x meet at
x =  /4. The required area is as shown in
the diagram. So
Required area = cos sin
/
x x dx

z
b g
0
4

= [sin x + cos x] 0
4
/
= 1 2 1 2
/ /

e j– ( 0 + 1)
= 2 – 1 Ans.[A]
Y
O
x=/4
X
x=/2
Y
O
x=/4
X
x=/2
x = 4 x=2

50
Ex.15 The area bounded by curve y = | x – 1| and
y = 1 is-
(A) 1 (B) 2
(C) 1/2 (D) None of these
Sol. y = |x – 1| =
x when x
x when x
 
 
R
S
T
1 1
1 1
Point of intersection of y = x – 1, y = 1 is
(2, 1)
Point of intersection of y = 1 – x, y = 1 is
(0, 1)
Required area = Area of  PQR
=
1
2
(PQ) . (RT)
=
1
2
. 2.1 = 1
Ans.[A]
Ex.16 If area bounded by the curve y = 8x2 – x5
and ordinates x = 1, x = k is
16
3
then k =
(A) 2 (B) [8 – 17 ]1/3
(C) [ 17 – 8]1/3 (D) –1
Sol. ( )
8 2 5
1
x x dx
k

z =
16
3

8
3 6
3 6
1
x x
k

L
N
M
M
O
Q
P
P=
16
3

8
3
(k3 – 1) –
k6
1
6

F
H
G
I
K
J=
16
3
 16 k3 – k6 – 15 = 32
 k6 – 16k3 + 47 = 0
 k3 = 8 ± 17
 k = (8 ± 17 )1/3
Ans.[B]
Ex.17 The area bounded by curve y = ex log x and
y =
log x
ex
is-
(A)
e2
5
4

(B)
e
e
2
5
4

(C)
e
4
–
5
4e
(D) None of these
Sol. Solving the equation of curves
ex log x =
log x
ex
 log x ex
ex

F
H
G I
K
J
1
= 0
 x = 1, 1/e
 Required area =
log
log
/
x
ex
ex x dx
e

F
H
G I
K
J
z
1
1
=
1
2 2 4
2 2 2
1
1
e
x
e x
x x
e
log
log .
/
b g b g
 
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
=
1
2e
[0– (–1)2] – e 0
1
4
1
2
1
4
2 2
   
F
H
G I
K
J
L
N
M O
Q
P
e e
= –
1
2e
–
1
2e
+
e
4
–
1
4e
=
e
4
–
5
4e
Ans.[C]
Ex.18 If 0  x   ; then the area bounded by the
curve y = x and y = x + sin x is-
(A) 2 (B) 4
(C) 2  (D) 4 
Sol. For the points of intersection of the given
curves
x = x + sin x
 sin x = 0
 x = 0, 
 required area
= x x x dx
 
z sin
b g
0

= sin x dx
0

z = – cos x 0

= 2
Ans.[A]
O
P (x= )
Y
X
)
O
P (x= )
Y
X
)

51
Ex.19 The area bounded by curves 3x2 + 5y = 32
and y = |x – 2| is-
(A) 25 (B) 17/2
(C) 33/2 (D) 33
Sol. Here the first curve can be written in the
following form
x2 = –
5
3
y 
F
H
G I
K
J
32
5
which is a parabola whose vertex lies on the
y-axis.
Again second curve is given by
y =
x x
x x
 
  
R
S
|
T
|
2 2
2 2
,
,
b g
which consists of two perpendicular lines AB
and AC as shown in the fig.
M A
(–2, 0) (2, 0)
L
(3, 0)
C B
(3, 1)
X
Y
These lines meet the parabola at B(3,1) and
C(–2,4) .
Hence the reqd. area A is given by
A = y dx ABL ACM
 

z  
2
3
=
1
5
32 3
2
3
2

z 
( )
x dx –
1
2
. 1.1 –
1
2
(4 . 4)
=
1
5
32 3
2
3
x x


–
17
2
=
1
5
[69+ 56] –
17
2
=
33
2
Ans.[C]

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