Ans. Equation of the curve (rightward parabola) is  

  ……….(i)

Required area (shaded region)

=  [From eq. (i)]

=  =  = 

=  =   =  sq. units

Ans. Equation of the curve (rightward parabola) is  

  ……….(i)

Required area (shaded region) bounded by curve  (vertical lines ) and axis in first quadrant.

= ∣∣∣∫24ydx∣∣∣=∣∣∣∫243x−−√dx∣∣∣|∫24ydx|=|∫243xdx|       [From eq. (i)]

=  = 

=  = 

=   =  sq. units

Ans. Equation of curve (parabola) is    ……….(i) 

 Required (shaded) area bounded by curve  (Horizontal lines ) and axis in first quadrant.

= 

=  =  =  sq. units

Ans. Equation of ellipse is  ……….(i) 

Here 

From eq. (i),

 

 y=±3416−x2−−−−−−√y=±3416−x2……….(ii)

For arc of ellipse in first quadrant.

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Intersections of ellipse (i) with axis 

Putting  in eq. (i),   

Therefore, Intersections of ellipse (i) with axis are (0, 4) and .

Intersections of ellipse (i) with axis 

Putting  in eq. (i),   

Therefore, Intersections of ellipse (i) with axis are (0, 3) and .

Now Area of region bounded by ellipse (i) = Total shaded area

= 4 x Area OAB of ellipse in first quadrant

=  [ At end B of arc AB of ellipse;  and at end A of arc AB ; ]

=  = 

=  

=  = 

=  sq. units

Ans. Equation of ellipse is  

Here 

From eq. (i),

 

 y=±324−x2−−−−−√y=±324−x2……….(ii)

for arc of ellipse in first quadrant.

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Ellipse (i) ia symmetrical about axis,

( On changing  in eq. (i), it remains unchanged)

Intersections of ellipse (i) with axis 

Putting  in eq. (i), 

Therefore, Intersections of ellipse (i) with axis are (0, 2) and .

Intersections of ellipse (i) with axis 

Putting  in eq. (i), 

 

Therefore, Intersections of ellipse (i) with axis are (0, 3) and .

Now Area of region bounded by ellipse (i) = Total shaded area

= 4 x Area OAB of ellipse in first quadrant

=  [ At end B of arc AB of ellipse;  and at end A of arc AB ; ]

=  = 

=  

= 

=   sq. units

Ans. Step I. To draw the graphs and shade the region whose area we are to find. 

Equation of the circle is  ……….(i)

We know that eq. (i) represents a circle whose centre is (0, 0) and radius is 2.

Equation of the given line is 

  ……….(ii)

We know that eq. (ii) being of the form  where   represents a straight line passing through the origin and making angle of  with axis.

Step II. To find values of  and  Putting  from eq. (ii) in eq. (i),

  

 

Putting  in , and 

Therefore, the two points of intersections of circle (i) and line (ii) are A and D.

Step III. Now shaded area OAM between segment OA of line (ii) and axis

=  

=  =  =  =  =  sq. units……….(iii)

Step IV. Now shaded area AMB between are AB of circle and axis.

= =  

=  From eq. (ii),

=  = 

=  =  =   sq. units……….(iv)

Step V. Required shaded area OAB = Area of OAM + Area of AMB

=  sq. units

Ans. Equation of the circle is  ……….(i) 

 

  ……….(ii)

Here Area of smaller part of the circle  cut off by the line  = Area of ABMC = 2 x Area of ABM

=  = [From eq. (ii)]

= 

= 

= 2[0+a22.π2−a2√×a2√−a22sin−112√]2[0+a22.π2−a2×a2−a22sin−112]

= 2[πa24−a2√×a2√−a22.π4]2[πa24−a2×a2−a22.π4]

= 2[πa24−a22−πa28]2[πa24−a22−πa28]

= 2a2[π4−12−π8]2a2[π4−12−π8]

= 2a2[2π−2−π8]2a2[2π−2−π8]

= a24(π−2)=πa24−a22a24(π−2)=πa24−a22

=>a22(π2−1)=>a22(π2−1)sq. units

Ans. Equation of the curve (parabola) is  ……(i) 

Now area bounded by parabola (i) and vertical line  is divided into two equal parts by the vertical line 

 Area OAMB = Area AMBDNC

   = 

 = 

       = 

 

 units

Ans. The required area is the area included between the parabola  and the modulus function  

To find: Area between the parabola  and the ray  for 

Here, Limits of integration  

  

  

Now, for  , table of values,

 if   if 

	0	1	2
	0	1	2

 

	0
	0	1	2

Now, Area between parabola  and axis between limits  and 

=  =  =  ………..(i)

And Area of ray  and axis,

=  =  =  ………..(ii)

  Required shaded area in first quadrant = Area between ray  for  and axis Area between parabola  and axis in first quadrant

= Area given by eq. (ii) – Area given by eq. (i) =  sq. units

Similarly, shaded area in second quadrant = 1616sq. units

Therefore, total area of shaded region in the above figure = 1616+ 1616= 1313sq. units

Ans. Step I. Graphs and region of Integration 

Equation of the given curve is

 ……..(i)

Equation of the given line is

 ……….(ii)

 

	0	1
	0		0

Step II. Putting  from eq. (i) in eq. (ii),

  

  x(x−2)+1(x−2)=0x(x−2)+1(x−2)=0

  or 

For ,from eq. (i),  (2, 1)

For  from eq. (i), 

Therefore, the two points of intersection of parabola (i) and line (ii) are C and D (2, 1).

Step III. Area CMOEDN between parabola (i) and axis =  = 

=  =  =  sq. units……..(iii)

Step IV. Area of trapezium CMND between line (ii) and axis =  = 

=  =  =  =  

=  =  sq. units……….(iv)

 Required shaded area = Area given by eq. (iv) – Area given by eq. (iii)

=  sq. units

Ans. Equation of the (parabola) curve is
 ………(i)
 y=2x−−√y=2x ………(ii)
Here required shaded area OAMB = 2 x Area OAM

=  =  = 
=  =  =  sq. units 

Ans. Equation of the circle is  ……….(i) 

  ……….(ii)

Required area =  = 

= 

= 

=  =  sq. units

Therefore, option (A) is correct.

Ans. Equation of the curve (parabola) is  

Required area = Area OAM =  = ∣∣∣∫03y24dy∣∣∣|∫03y24dy|
= 14∣∣∣(y33)30∣∣∣14|(y33)03| =  sq. units
Therefore, option (B) is correct.