1. SOLVED SUBJECTIVE EXAMPLES
Example 1 :
Find the area bounded by y = x |sinx| and x-axis between x = 0, x = 2.
Solution:
xsin x, if sin x 0, i.e., 0 x
y
xsin x,if sin x 0, i.e., x 2
  

 
     

Required area =
2
0
xsin x dx ( xsin x)dx
 

 
  0
0
x ( cos x) ( cos x)dx


   

2
2
(x( cos x)) ( cos x)dx




   

y
y= x
O x
2

2
0
sin x (2 ) sin x 4



       sq. units
Example 2 :
If the line y = mx divides the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve
y= 1 + 4x – x2
into two equalparts, then find the value ofm.
Solution:
The given curve is y – 5 = – (x – 2)2
Thus givencurve is a parabola with vertex at (2, 5) and axis x = 2
Given that area CBFC =Area CDEBC
So area CDEBFC = 2 Area CBFC
Area CDEBFC
3/2
2
0
(1 4x x )dx
  

3/2
3
2
0
x 3 9 9 39
x 2x 2
3 2 4 8 8
 
      
 
 
sq. units
Area CBFC
3/ 2
0
9m
mxdx
4
 

y = mx
y
x
F
x = 3/2
y = 1 + 4x – x
2
B
E
D
C
So we must have
39 18m
8 8
 or
13
m
6


2. Example 3 :
Find out the area enclosed bythe polynomialfunctionofleast degree satisfying
1/ x
3
x 0
f(x)
lim 1
x

 

 
 
= e and the circle x2
+ y2
= 2 above the axis
Solution:
Since
1/ x
3
x 0
f(x)
lim 1
x

 

 
 
exists and f(x) is of least degree, so f(x) must be of the formf(x) = ax4
,
solving the limit we get a = 1, so f(x) = x4
. Hence we have two curves
y = x4
... (i)
2
y 2 x
  ... (ii)
Solving we get, 2 4 8 2
2 x x x x 2 0
     
 (x2
– 1) (x6
+ x4
+ x2
+ 2) = 0
y = x4
x
–1 1
2
y 2 x
= -
 x2
– 1 = 0  x =  1
Required area
 
1
2 4
0
2 2 x x dx
  

1
5
2 1
0
x 2 x x
2 2 x sin
2 2 5
2

 
   
 
 
 
1 1 3
2
2 4 5 5 2
 
   
    
 
 
   
sq. units.
Example 4 :
Find out the area enclosed byy = x2
+ cosx and its normal at x=
2

in the first quadrant.
Solution:
f(x) = x2
+ cos x  f (x)
 = 2x – sinx
 f ( ) 1
2

  
Equation ofnormalat x =
2

is
2
1
y x
4 1 2
 
 
 
  
   
     
 
x
(0, 1)A
P
Q R
O

3. at x axis, y = 0 
2
( 1)
x
4 2
   
 
Required area = area OAPQO + area of triangle PQR
/2 2 2
2
0
1 ( 1)
(x cos x)dx
2 4 2 2 4
  
     
     
 
 
 

/2
3 2
0
x ( 1)
sin x
3 8

    
  
 
 
 
=
 
 
3 2 3 4
1
1 1 1
24 8 4 24 32

    
   
       
 
 
 
=
5 4 3
1
32 32 24
  
   .
Example 5 :
Find out the ratio of areas in which the function f(x) =
3
x x
100 35
 

 
 
 
divides the circle
x2
+ y2
– 4x + 2y + 1 = 0 ([.] denotes the greatest integer function).
Solution:
Circle is x2
+ y2
– 4x + 2y + 1 = 0
X
Y
(x –2) + (y + 1) = 4
2 2
f(x) = 0 for 0 < x < 4
(2 3,0)
-
(2 3,
+ 0)
or, (x – 2)2
+ (y + 1)2
= 4 = 22
... (i)
Now for 0 x 4
 
3 3
x x x x
0 1 0
100 35 100 35
 
     
 
 
 
So, the circle have to find out the ratio inwhichx axis divides the circle (i). Now, at x-axis, y= 0.
So, (x–2)2
= 3

4. So, the circle cuts the x axis at the points (2 – 3 , 0) and (2 + 3 , 0)
Let  
2 3
2
2 3
4 3 3
A 4 (x 2) 1 dx
3


 
    

The required ratio is
A 4 3 3
4 A 8 3 3
 

   
Example 6 :
Find the area of the figure enclosed by the curve 5x2
+ 6xy + 2y2
+ 7x + 6y + 6 = 0
Solution:
Equationofcurve canbe re-written as
2y2
+ 6(1 + x) y + 5x2
+ 7x + 6 = 0
1
3(1 x) (3 x)(x 1)
y
2
    
 , 2
3(1 x) (3 x)(x 1)
y
2
    

Therefore the curves (y1
and y2
) are defined for values of xfor which (3 – x) (x – 1)  0
i.e., 1 x 3
 
(Actuallythe given equationdenotes an ellipse, because 0
  and h2
< ab).
Required area willbe given by
3 3
1 2
1 1
A (y y )dx A (3 x)(x 1) dx
     
 
y
x
O
y1
y2
1 2 3
–3
–6
Put x = 3 cos2
 + sin2
 i.e., dx = – 2sin2 d
/ 2
2
0
A 2 sin 2 d sq. units
2


  

Example 7 :
Find the area enclosed between the curves
y = ln(x + e), x = ln
1
y
 
 
 
and x-axis.
Solution:
Given curves are y= ln(x + e)
and x = ln
1
y
 
 
 
or, x = –ln y
or, y = e–x
or,
x
1
y
e
 
 
 

5. Clearlythe two curves cut at x = 0
Graph ofcurves (i) and (ii) willbe as showninthe figure.
x
C
A O
(1 – e, 0)
x = –e
Y
y = ln(x + e)
y = e–x
Y’
B
x’
Required area = shaded area
1
1 2
0
(x x )dy
 

1
y
0
[ ln y (e e)]dy
   

1
y
0
(yln y – y) e ey
 
   
 
1
y
0
y ln y y e ey
 
    
 
y 0
(0 1 e e) (0 0 1 0) lim ylog y 0

 
        
 
 
 = 1 + 1 = 2 sq. units.
Example 8 :
Find the area enclosed by the parabola (y – 2)2
= x– 1, the tangent to the parabola at (2, 3) and
the x-axis.
Solution:
Given parabola is (y – 2)2
= x – 1 ... (i)
Its axis is y = 2 and vertex is (1, 2). Let P (2, 3)
 .
From (i), 2(y – 2)
dy
1
dx

dy 1
dx 2(y 2)
 

At
dy 1
P(2, 3),
dx 2

 Equation of tangent at P(2, 3) is
y
x
O
y = 3
y = 2
Q(–4, 0)
P (2, 3)
A (1, 2)
R (5, 0)
y – 3 =
1
(x 2)
2

or, x 2y 4 0
   ... (ii)
Line (ii) cuts the x-axis at (–4, 0) and y-axis at (0, 2).
Required area, RQPAR =
3
1 2
0
(x x )dy


3 3
2 2
0 0
[(y 2) 1 (2y 4)]dy (y 6y 9)dy
       
 
3
3
2
0
y
3y 9y
3
 
  
 
 
 
= (9 – 27 + 27) – 0 = 9 sq. units.

6. Example 9 :
Find the area of the region bounded by the curves y = f(x), y = |g(x)| and the lines x = 0, x = 2,
where fand g are continuous functions satisfying f(x+ y) = f(x) + f(y) – 8xyfor all x, y R

and g(x + y) = g(x) + g(y) + 3xy (x + y) for all x, y R
 . also f (0) 8 and g (0) 4
 
 
Solution:
Given, f(x + y) = f(x) + f(y) – 8xyfor all x, y
Putting x = 0 and y = 0, we get
f(0) = f(0) + f(0)  f(0) = 0
h 0
f (x h) f (x)
f (x) lim
h

 
 
h 0
f (x h) f (x 0)
lim
h

  

h 0
f (x) f(h) 8xh f(x) f(0) 8x.0
lim
h

    

h 0
f (h) f (0)
lim 8x f (0) 8x
h



   
Thus f (x) 8 8x [ f (0) 8]
 
  

Integrating bothsides, we get
f(x) = 8x – 4x2
+ c ... (2)
Putting x = 0, we get f(0) = 0 + c  c = 0 [ f (0) 0]


Hence f(x) = 8x – 4x2
... (3)
Given g(x + y) = g(x) + g(y) + 3xy(x + y) for allx and y
Putting x = y= 0, we get g(0) = 0 ... (4)
Now
h 0 h 0
g(x h) g(x) g(x h) g(x 0)
g (x) lim lim
h h
 
    
  
2
h 0
g(h) g(0)
lim 3x 3xh
h


 
  
 
 
2 2
g (0) 3x 4 3x [ g (0) 4]
 
     

Thus g (x )
 = – 4 + 3x2
... (5)
 g(x) = – 4x + x3
+ k
Putting x = 0, we get g(0) = k
 k = 0 [ g(0) 0]


 g(x) = x3
– 4x ... (6)
For points where y= f(x) and y= g(x) intersect,

7. 8x – 4x2
= x3
– 4x
 x3
+ 4x2
– 12 x = 0
 x = 0, 2, – 6
Sign scheme for f(x) i.e., for (8x – 4x2
) is
–
–ve +ve
0 2
–ve

Y
O
X
y = f(x)
(2, 0)
y = |g(x)|
Sign scheme for g(x) i.e. for x(x2
– 4) is
–
–ve +ve +ve
–2 0 2
–ve

f(x) – |g(x)| = 8x – 4x2
– (4x – x3
) ( g(x)  0 in [0, 2])
= x3
– 4x2
+ 4x = x(x – 2)2  0
Area bounded byy = f(x) and y = |g(x)|
between x = 0 and x = 2
2 2
2 3
1 2
0 0
(y y )dx [(8x 4x ) (4x x )]dx
     
 
2
3 2
0
4
(x 4x 4x)dx sq. units
3
   
 .
Example 10 :
Find the area enclosed by the circle x2
+ y2
= 4, the parabola y = x2
+ x + 1, the curve
2 x x
y sin cos
4 4
 
 
 
 
and the x-axis, (where [x] denotes the integralpart ofx).
Solution:
Equation ofgiven circle is x2
+ y2
= 4 ... (1)
 – 2  x  2 and – 2  y  2
Let z = sin2
x x
cos
4 4
 = 1 – cos2
x x
cos
4 4

= 1 + t – t2
, where t = cos
x
4
Now –
1 x 1
2 4 2
 
 0 < t  1 
5
1 z
4
 
for x [ 2, 2],
  curve y =
2 x x
sin cos
4 4
 

 
 
becomes y = 1 ... (2)

8. Given parabola is
y = x2
+ x + 1 ... (3)
or,
2
1 3
x y
2 4
 
  
 
 
... (4)
Its axis is
1
x
2
  and vertex is
1 3
,
2 4
 

 
 
.
Now, we have to find out the area enclosed by the circle x2
+ y2
= 4,
parabola
2
3 1
y x ,
4 2
   
  
   
   
line y= 1 and x-axis
y
x
O
(0, 2)
B
A
E
Q
R U
S
T
P
D
C (2, 0)
(–2, 0) – 3
1
1
2
- -
T S R U D
(–2, 0) ( 3, 0)
 (–1, 0) (–1/2, 0) (2, 0)
Required area isshown as shaded regionin the figure.
Hence required area = area OABCO + area PQRS + area RQEAOR + 2 area CBDC.
=  
0 2
2 2
1 3
3 1 ( 3 1) 1 (x x 1)dx 2 4 x dx

        
 
 
0 2
3 2
2 1
3
1
x x x x
2 3 1 x 2 4 x 2sin
3 2 2 2


   
 
       
   
 
 
 
 
 
  1 1 3 2
2 3 1 0 1 2 (0 )
3 2 2 3
 
 
  
 
           
 
 
 
   
 
   
 
 
  5 2
2 3 1 3
6 3

    
2 1
3 sq. units
3 6

 
  
 
 
.

9. SOLVED OBJECTIVE EXAMPLES
Example 1 :
Area enclosed by the curve |x –2| + |y + 1| = 1 is equal to
(A) 4 sq. units (B) 6 sq. units
(C) 2 sq. units (D) 8 sq. units
Solution:
After shiftingthe origin at the point (2, –1) theequationofcurve becomes, |x| + |y| =1. This curve
willrepresent a square as shown inthe adjacent figure.
y
C
O
A
D
y - x = 1 x + y = 1
x + y = –1 x – y = 1
Area ofthis square is clearlyequalto 4 times the area oftriangleOAB. Thus required area = 2 sq.
units.
Example 2 :
Area bounded by the curves y = |x| – 2 and y = 1 – |x–1| is equal to
(A) 4 sq. units (B) 6 sq. units
(C) 2 sq. units (D) 8 sq. units
Solution:
Bounded figureABCD is a rectangle.
AB 1 1 2
  
BC 4 4 2 2
  
A(1, 1)
B
C
D
2
0 1
–1
–2
–2
y = |x| –2
y = 1 – |x –1|
Thus, bounded area = ( 2) (2 2) = 4 sq. units.
Example 3 :
Area bounded bythe curve y= max{sinx, cosx} and x-axis, betweenthe lines x
4

 and x = 2
is equalto
(A)
(4 2 1)
2

sq. units (B) (4 2 1)sq.units

(C)
(4 2 1)
sq. units
2

(D) None ofthese

10. Solution:
Bold lines represents the graph ofy= max{sinx, cosx}.
Required area,
/4
 


y
x
O
y = sin x
y = cos x
5 /4
/4
sin x dx sin x dx
 
 
 
 
3 /2 2
5 / 4 3 /2
cos xdx cos x dx
 
 
 
 
(4 2 1)
sq.units
2


Example 4 :
Area bounded by the parabola y = x2
– 2x + 3 and tangents drawn to it fromthe point P(1, 0) is
equalto
(A) 4 2 sq. units (B)
4 2
sq. units
3
(C)
8 2
sq. units
3
(D)
16
2 sq. units
3
Solution:
Let thedrawntangents be PAand PB.AB isclearlythe chord ofcontact ofpoint P.Thus equation
ofAB is
1
. (y 0)
2
 = x.1 – (2 + 1) + 3 i.e., y = 4
x coordinates ofpointsAand B willbe given by,
x2
– 2x + 3 = 4 i.e., x2
– 2x –1 = 0
 x = 1 2

ThusAB = 2 2 units.
Hence PAB
1
(2 2).4 4 2
2
   sq. units
A B
x
y = 4
P(1, 0)
y = x – 2x + 3
2
Now area bounded by lineAB and parabola is equalto
1 2
2
1 2
(4 2 (x 2x 3))dx


  

=
4 2
3
sq. units.
Thus required area =
4 2 8 2
4 2
3 3
  sq. units.

11. Example 5 :
Area bounded bythe curves y= sinx, tangent drawn to it at x = 0 and the line x =
2

, is equalto
(A)
2
4
sq. units
2
 
(B)
2
4
sq. units
4
 
(C)
2
2
sq.units
4
 
(D)
2
2
sq. units
2
 
Solution:
The tangent drawnto y= sinxat x = 0 is the line y= x. Clearly the line y= x lies above the graph
of y= sinx x 0,
2

 
  
 
.
Thus required area
/ 2
0
(x sin x)dx

 

/ 2
2 2
0
x 4
cos x
2 4

   
  
 
 
 
sq. units.
Example 6 :
IfA(n) represents the area bounded bythe curve y= n. lnx, where n N
 and n> 1, the x-axis and
the lines x = 1 and x= e, then the value ofA(n) + nA(n–1) is equalto
(A)
2
n
e 1

(B)
2
n
e 1

(C) n2
(D) en2
Solution:
A(n) = n
e e
e
1
1 1
n x dx n n x.x dx n
 
 
  
 
 
 
l l  A(n – 1) = (n – 1)
 A(n) + nA(n – 1) = n + n(n – 1) = n2
Example 7 :
Value ofthe parametera such that the area bounded byy= a2
x2
+ ax + 1, coordinateaxes and the
line x = 1, attains it’s least value, is equalto
(A) –
1
4
(B) –
1
2
(C)
3
4
 (D) –1

12. Solution:
a2
x2
+ ax + 1 is clearlypositive for allreal values of x.Area under consideration
1 2
2 2
0
a a
(a x ax 1)dx 1
3 2
     

2
1
(2a 3a 6)
6
  
2
1 3 9 18
2 a a 6
6 2 16 16
 
 
    
 
 
 
 
=
2
1 3 39
2 a
6 4 8
 
 
 
 
 
 
 
 
which is clearlyminimumfor a = –
3
4
.
Example 8 :
Area oftheregion which consists ofallthe points satisfying the conditions|x– y| + |x+ y|  8 and
xy  2, is equalto
(A) 4(7 – ln8) sq. units (B) 4 (9 – ln8) sq. units
(C) 2(7 – ln8) sq. units (D) 2 (9 – ln 8) sq. units
Solution:
The expression|x– y| + |x+ y|  8, represents the interior regionofthe square formedbythe lines
x 4, y 4 and xy 2
   represents the regionlying inside the hyperbola xy= 2.
Required area,
 
4
4
1/2
1/2
2
2 4 dx 2 4x 2 nx
x
 
 
    
   
 
 l
= 4(7 – 3 ln2) sq. units
x = –4
x = 4
y = 4
y = x
D A
x
C B
y
y = –4
y = –x
Example 9 :
Apoint Pmoves inxyplaneinsuchawaythat [|x|]+ [|y|]=1, where [.] denotesthe greatest integer
function.Areaofthe region representing allpossible positions ofthe point Pis equalto
(A) 4 sq. units (B) 16 sq. units
(C) 2 2 sq. units (D) 8 sq. units

13. Solution:
If [|x|] = 1 and [|y|] = 0
then 1 | x | 2, 0 | y | 1
   
x ( 2, 1] [1, 2), y ( 1,1)
      
if [|x|] = 0, [|y|] = 1
Then x ( 1, 1), y ( 2, 1] [1,2]
     
y
x
–2 2
–2
–1
1
1
–1
O
2
Area ofrequired region
= 4(2 – 1) (1 – (–1)) = 8 sq. units.
Example 10 :
Area enclosed bythe curve y= f(x) defined parametricallyas
2
2 2
1 t 2t
x , y
1 t 1 t

 
 
is equalto
(A) sq.units
 (B) / 2 sq. units

(C)
3
sq. units
4

(D)
3
2

sq. units
Solution:
Clearlyt canbe any real number
Let
2
2
1 tan
t tan x
1 tan
 
   
 
 x cos2
 
and 2
2tan
y sin 2
1 tan

  
 
 x2
+ y2
= 1
Thus required area = 2
.1
  sq. units.