NCERT Solutions Miscellaneous Exercise: Application of Integrals

# NCERT Solutions Miscellaneous Exercise: Application of Integrals | Mathematics (Maths) for JEE Main & Advanced PDF Download

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``` 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)

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## Mathematics (Maths) for JEE Main & Advanced

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## FAQs on NCERT Solutions Miscellaneous Exercise: Application of Integrals - Mathematics (Maths) for JEE Main & Advanced

 1. What is the concept of integrals in mathematics?
Ans. Integrals in mathematics are used to find the area under a curve or the accumulation of quantities over a certain interval. It is a fundamental concept in calculus that involves finding the antiderivative of a function.
 2. How are integrals applied in real-life situations?
Ans. Integrals are used in various real-life situations such as calculating the area under a curve in physics to find work done, determining the volume of a solid in engineering, and analyzing growth rates in economics.
 3. Can integrals be used to find the total distance traveled by an object?
Ans. Yes, integrals can be used to find the total distance traveled by an object by integrating the absolute value of the velocity function over a specific time interval.
 4. What is the difference between definite andindefinite integrals?
Ans. Definite integrals have specific limits of integration and give a numerical value, representing the area under a curve. Indefinite integrals do not have limits and give a general function, representing the antiderivative of the original function.
 5. How can integrals be used to calculate the average value of a function?
Ans. The average value of a function over an interval can be calculated using integrals by dividing the integral of the function over the interval by the length of the interval. This value represents the average height of the function over that interval.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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