NCERT Solutions Miscellaneous Exercise: Application of Integrals | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
Miscellaneous Examples                                                  Page No: 371 
 
1. Find the area under the given curves and given lines: 
(i) and x-axis. 
(ii) and x-axis. 
 
Solution:  
(i)Equation of the curve is  
 
y= x
2
 ……..(1) 
 
Required area bounded by curve (1), vertical line x=1, x=2 and x-axis 
=  
=  
=  sq. units 
(ii) Equation of the curve 
 
y= x
4
 …..(1) 
 
 
Page 2


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
Miscellaneous Examples                                                  Page No: 371 
 
1. Find the area under the given curves and given lines: 
(i) and x-axis. 
(ii) and x-axis. 
 
Solution:  
(i)Equation of the curve is  
 
y= x
2
 ……..(1) 
 
Required area bounded by curve (1), vertical line x=1, x=2 and x-axis 
=  
=  
=  sq. units 
(ii) Equation of the curve 
 
y= x
4
 …..(1) 
 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
It is clear that curve (1) passes through the origin because x=0 from (1) y=0. 
Table of values for curve   
x 1 2 3 4 5 
y 1 16 81 256 625 
 
Required shaded area between the curve , vertical lines  and axis 
=  =  
=  =  
=  
= 624.8 sq. units 
2. Find the area between the curves y=x and y=x
2
 
Solution: Equation of one curve (straight line) is y=x …..(i) 
 
Page 3


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
Miscellaneous Examples                                                  Page No: 371 
 
1. Find the area under the given curves and given lines: 
(i) and x-axis. 
(ii) and x-axis. 
 
Solution:  
(i)Equation of the curve is  
 
y= x
2
 ……..(1) 
 
Required area bounded by curve (1), vertical line x=1, x=2 and x-axis 
=  
=  
=  sq. units 
(ii) Equation of the curve 
 
y= x
4
 …..(1) 
 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
It is clear that curve (1) passes through the origin because x=0 from (1) y=0. 
Table of values for curve   
x 1 2 3 4 5 
y 1 16 81 256 625 
 
Required shaded area between the curve , vertical lines  and axis 
=  =  
=  =  
=  
= 624.8 sq. units 
2. Find the area between the curves y=x and y=x
2
 
Solution: Equation of one curve (straight line) is y=x …..(i) 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
Equation of second curve (parabola) is  …..(ii) 
Solving equation (i) and (ii), we get  or  and  or  
 Points of intersection of line (i) and parabola (ii) are O (0, 0) and A (1, 1). 
 
Now Area of triangle OAM 
= Area bounded by line (i) and x-axis 
=  =  
 =  
=  sq. units 
Also Area OBAM = Area bounded by parabola (ii) and x-axis 
=  =  
 =  
Page 4


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
Miscellaneous Examples                                                  Page No: 371 
 
1. Find the area under the given curves and given lines: 
(i) and x-axis. 
(ii) and x-axis. 
 
Solution:  
(i)Equation of the curve is  
 
y= x
2
 ……..(1) 
 
Required area bounded by curve (1), vertical line x=1, x=2 and x-axis 
=  
=  
=  sq. units 
(ii) Equation of the curve 
 
y= x
4
 …..(1) 
 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
It is clear that curve (1) passes through the origin because x=0 from (1) y=0. 
Table of values for curve   
x 1 2 3 4 5 
y 1 16 81 256 625 
 
Required shaded area between the curve , vertical lines  and axis 
=  =  
=  =  
=  
= 624.8 sq. units 
2. Find the area between the curves y=x and y=x
2
 
Solution: Equation of one curve (straight line) is y=x …..(i) 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
Equation of second curve (parabola) is  …..(ii) 
Solving equation (i) and (ii), we get  or  and  or  
 Points of intersection of line (i) and parabola (ii) are O (0, 0) and A (1, 1). 
 
Now Area of triangle OAM 
= Area bounded by line (i) and x-axis 
=  =  
 =  
=  sq. units 
Also Area OBAM = Area bounded by parabola (ii) and x-axis 
=  =  
 =  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
=  sq. units 
Required area OBA between line (i) and parabola (ii) 
= Area of triangle OAM – Area of OBAM 
=  =  sq. units 
3. Find the area of the region lying in the first quadrant and bounded 
by  and y=4. 
Solution: Equation of the curve is  
 
  ……….(i) 
or ……….(ii) 
Here required shaded area of the region lying in first quadrant bounded by parabola (i),  
and the horizontal lines  and  is 
 
Page 5


 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
Miscellaneous Examples                                                  Page No: 371 
 
1. Find the area under the given curves and given lines: 
(i) and x-axis. 
(ii) and x-axis. 
 
Solution:  
(i)Equation of the curve is  
 
y= x
2
 ……..(1) 
 
Required area bounded by curve (1), vertical line x=1, x=2 and x-axis 
=  
=  
=  sq. units 
(ii) Equation of the curve 
 
y= x
4
 …..(1) 
 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
It is clear that curve (1) passes through the origin because x=0 from (1) y=0. 
Table of values for curve   
x 1 2 3 4 5 
y 1 16 81 256 625 
 
Required shaded area between the curve , vertical lines  and axis 
=  =  
=  =  
=  
= 624.8 sq. units 
2. Find the area between the curves y=x and y=x
2
 
Solution: Equation of one curve (straight line) is y=x …..(i) 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
Equation of second curve (parabola) is  …..(ii) 
Solving equation (i) and (ii), we get  or  and  or  
 Points of intersection of line (i) and parabola (ii) are O (0, 0) and A (1, 1). 
 
Now Area of triangle OAM 
= Area bounded by line (i) and x-axis 
=  =  
 =  
=  sq. units 
Also Area OBAM = Area bounded by parabola (ii) and x-axis 
=  =  
 =  
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
 
=  sq. units 
Required area OBA between line (i) and parabola (ii) 
= Area of triangle OAM – Area of OBAM 
=  =  sq. units 
3. Find the area of the region lying in the first quadrant and bounded 
by  and y=4. 
Solution: Equation of the curve is  
 
  ……….(i) 
or ……….(ii) 
Here required shaded area of the region lying in first quadrant bounded by parabola (i),  
and the horizontal lines  and  is 
 
 
 
 
 
 
NCERT Solutions for Class 12 Maths Chapter 8-Application of 
integrals 
=  
=  
=  
=  =  sq. units 
4. Sketch the graph of  and evaluate  
 
 
Solution: Equation of the given curve is  ……….(i)