Chapter 8 Application of Integrals

NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals

Chapter 8 of Class 12 Maths, Application of Integrals focuses on different applications of of integration, mainly, to find areas related to curves. This chapter explains how definite integrals are applied to calculate the area under simple curves, between two curves, and bounded by axes or lines. It helps students understand the practical use of integrals beyond formulas and calculations.

This page provides the NCERT Solutions which are prepared in accordance with the latest NCERT syllabus. The solutions are explained step by step in clear and simple language, helping Class 12 students visualise problems, apply correct limits, and solve questions accurately for board exams.

1.0Key Concepts of Class 12 Maths Chapter 8 Applications of Integrals

Class 12 Maths Chapter 8, Applications of Integrals, explains how definite integrals can be used to calculate the areas of different regions. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 8 include:

Area Under a Curve: Learn how definite integrals are used to find the area between a curve and the x-axis.
Area Between Two Curves: Understand how to calculate the area of a region enclosed between two curves using suitable limits.
Choice of Axis and Limits: Identify the appropriate axis of integration and set correct limits to obtain accurate results.
Graphical Representation: Use rough graphs to identify bounded regions and understand the area that needs to be calculated.
Standard Curves and Lines: Solve problems involving straight lines, parabolas, and other simple algebraic curves using integration.

2.0NCERT Class 12 Maths Chapter 8 Applications of Integrals : Detailed Solutions

EXERCISE - 8.1

Find the area of the region bounded by the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ . Sol. The given curve is an ellipse with centre at $(0, 0)$ and symmetrical about both X-axis and Y-axis ( ∵ the power of x and y both are even) Area bounded by the ellipse $= 4 \times$ (Area of shaded region in the first quadrant only) ( ∵ By symmetry)

$= 4 \times \int_{a}^{b} y dx = 4 \int_{0}^{4} y dx = 4 \int_{0}^{4} \frac{3}{4} 16 - x^{2} dx$

$= 12 π$ sq. units. Therefore, area bounded by the ellipse is $12 π$ sq. units.

Find the area of the region bounded by the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ . Sol. The given curve is an ellipse with centre at $(0, 0)$ and symmetrical about both X-axis and Y-axis.

Area bounded by the ellipse $= 4 \times$ (Area of shaded region in the first quadrant only) ( ∵ By symmetry )

$= 4 \times \int_{a}^{b} ydx = 4 \int_{0}^{2} ydx = 4 \int_{0}^{2} \frac{3}{2} 4 - x^{2} dx (∵ \frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1, ∴ y = \frac{3}{2} 4 - x^{2}) = 6 \int_{0}^{2} 2^{2} - x^{2} dx = 6 [\frac{x}{2} 4 - x^{2} + \frac{2 ^{2}}{2} sin^{- 1} (\frac{x}{2})]_{0}^{2} [∵ \int a^{2} - x^{2} dx = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} (\frac{x}{a}) + C] = 6 {0 + 2 sin^{- 1} (1) - 0} = 6 \times 2 \times (\frac{π}{2}) = 6 π sq. units.$

Choose the correct answer in the following Q. 3 and 4

Area lying in the first quadrant and bounded by the circle $x^{2} + y^{2} = 4$ and the lines $x = 0$ and $x = 2$ is. (A) $π$ (B) $\frac{π}{2}$ (C) $\frac{π}{3}$ (D) $\frac{π}{4}$ Sol. (A) The area bounded by the circle and the lines $x = 0$ and $x = 2$ , in the first quadrant is represented in the figure by shaded region.

$Required area = \int_{0}^{2} ydx = \int_{0}^{2} 4 - x^{2} dx = [\frac{x}{2} 4 - x^{2} + \frac{4}{2} sin^{- 1} (\frac{x}{2})]_{0}^{2} = 0 + 2 sin^{- 1} (1) - 0 = 2 \times \frac{π}{2} = π sq units.$

Thus the correct option is (A)

Area of the region bounded by the curve $y^{2} = 4 x, Y$ -axis and the line $y = 3$ is. (A) 2 (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$ Sol. (B) The area bounded by the curve, $y^{2} = 4 x$ , Y-axis and $y = 3$ is represented in the figure by shaded region.

$Required area = \int_{0}^{3} xdy = \int_{0}^{3} \frac{y ^{2}}{4} dy = \frac{1}{4} [\frac{y ^{3}}{3}]_{0}^{3} = \frac{1}{12} (3^{3} - 0) = \frac{1}{12} (27) = \frac{9}{4} sq units.$

Thus, the correct answer is (B).

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Maths for all chapters, with exercise-wise answers, important formulas, and step-by-step solutions to help students understand concepts and practise questions effectively.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 8 Application of Integrals Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 8.1	4 Questions & Solutions	Finding the area of regions bounded by curves using integration

5.0Key Features and Benefits of Class 12 Maths Chapter 8 Applications of Integrals

Step-by-Step Solutions: Each solution explains the calculation clearly, helping students understand how areas are determined between curves.
Graph and Limits Guidance: The solutions help students understand how to draw suitable graphs and choose the correct limits before applying integration.
NCERT-Based Coverage: The solutions follow the latest NCERT syllabus and cover the questions given in the chapter exercises.
Improved Accuracy: Regular practice helps students solve application-based calculus problems with greater accuracy and fewer calculation mistakes.
Useful for Competitive Exams: A strong understanding of applications of integrals can support preparation for Mathematics Olympiads and competitive entrance examinations.
Geometrical Understanding: The chapter helps students connect integration concepts with the geometric interpretation of areas and bounded regions.

NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals

Chapter 8 of Class 12 Maths, Application of Integrals focuses on different applications of of integration, mainly, to find areas related to curves. This chapter explains how definite integrals are applied to calculate the area under simple curves, between two curves, and bounded by axes or lines. It helps students understand the practical use of integrals beyond formulas and calculations.

This page provides the NCERT Solutions which are prepared in accordance with the latest NCERT syllabus. The solutions are explained step by step in clear and simple language, helping Class 12 students visualise problems, apply correct limits, and solve questions accurately for board exams.

1.0Key Concepts of Class 12 Maths Chapter 8 Applications of Integrals

Class 12 Maths Chapter 8, Applications of Integrals, explains how definite integrals can be used to calculate the areas of different regions. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 8 include:

Area Under a Curve: Learn how definite integrals are used to find the area between a curve and the x-axis.
Area Between Two Curves: Understand how to calculate the area of a region enclosed between two curves using suitable limits.
Choice of Axis and Limits: Identify the appropriate axis of integration and set correct limits to obtain accurate results.
Graphical Representation: Use rough graphs to identify bounded regions and understand the area that needs to be calculated.
Standard Curves and Lines: Solve problems involving straight lines, parabolas, and other simple algebraic curves using integration.

2.0NCERT Class 12 Maths Chapter 8 Applications of Integrals : Detailed Solutions

EXERCISE - 8.1

Find the area of the region bounded by the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ . Sol. The given curve is an ellipse with centre at $(0, 0)$ and symmetrical about both X-axis and Y-axis ( ∵ the power of x and y both are even) Area bounded by the ellipse $= 4 \times$ (Area of shaded region in the first quadrant only) ( ∵ By symmetry)

$= 4 \times \int_{a}^{b} y dx = 4 \int_{0}^{4} y dx = 4 \int_{0}^{4} \frac{3}{4} 16 - x^{2} dx$

$= 12 π$ sq. units. Therefore, area bounded by the ellipse is $12 π$ sq. units.

Find the area of the region bounded by the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ . Sol. The given curve is an ellipse with centre at $(0, 0)$ and symmetrical about both X-axis and Y-axis.

Area bounded by the ellipse $= 4 \times$ (Area of shaded region in the first quadrant only) ( ∵ By symmetry )

$= 4 \times \int_{a}^{b} ydx = 4 \int_{0}^{2} ydx = 4 \int_{0}^{2} \frac{3}{2} 4 - x^{2} dx (∵ \frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1, ∴ y = \frac{3}{2} 4 - x^{2}) = 6 \int_{0}^{2} 2^{2} - x^{2} dx = 6 [\frac{x}{2} 4 - x^{2} + \frac{2 ^{2}}{2} sin^{- 1} (\frac{x}{2})]_{0}^{2} [∵ \int a^{2} - x^{2} dx = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} (\frac{x}{a}) + C] = 6 {0 + 2 sin^{- 1} (1) - 0} = 6 \times 2 \times (\frac{π}{2}) = 6 π sq. units.$

Choose the correct answer in the following Q. 3 and 4

Area lying in the first quadrant and bounded by the circle $x^{2} + y^{2} = 4$ and the lines $x = 0$ and $x = 2$ is. (A) $π$ (B) $\frac{π}{2}$ (C) $\frac{π}{3}$ (D) $\frac{π}{4}$ Sol. (A) The area bounded by the circle and the lines $x = 0$ and $x = 2$ , in the first quadrant is represented in the figure by shaded region.

$Required area = \int_{0}^{2} ydx = \int_{0}^{2} 4 - x^{2} dx = [\frac{x}{2} 4 - x^{2} + \frac{4}{2} sin^{- 1} (\frac{x}{2})]_{0}^{2} = 0 + 2 sin^{- 1} (1) - 0 = 2 \times \frac{π}{2} = π sq units.$

Thus the correct option is (A)

Area of the region bounded by the curve $y^{2} = 4 x, Y$ -axis and the line $y = 3$ is. (A) 2 (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$ Sol. (B) The area bounded by the curve, $y^{2} = 4 x$ , Y-axis and $y = 3$ is represented in the figure by shaded region.

$Required area = \int_{0}^{3} xdy = \int_{0}^{3} \frac{y ^{2}}{4} dy = \frac{1}{4} [\frac{y ^{3}}{3}]_{0}^{3} = \frac{1}{12} (3^{3} - 0) = \frac{1}{12} (27) = \frac{9}{4} sq units.$

Thus, the correct answer is (B).

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Maths for all chapters, with exercise-wise answers, important formulas, and step-by-step solutions to help students understand concepts and practise questions effectively.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 8 Application of Integrals Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 8.1	4 Questions & Solutions	Finding the area of regions bounded by curves using integration

5.0Key Features and Benefits of Class 12 Maths Chapter 8 Applications of Integrals

Step-by-Step Solutions: Each solution explains the calculation clearly, helping students understand how areas are determined between curves.
Graph and Limits Guidance: The solutions help students understand how to draw suitable graphs and choose the correct limits before applying integration.
NCERT-Based Coverage: The solutions follow the latest NCERT syllabus and cover the questions given in the chapter exercises.
Improved Accuracy: Regular practice helps students solve application-based calculus problems with greater accuracy and fewer calculation mistakes.
Useful for Competitive Exams: A strong understanding of applications of integrals can support preparation for Mathematics Olympiads and competitive entrance examinations.
Geometrical Understanding: The chapter helps students connect integration concepts with the geometric interpretation of areas and bounded regions.

Frequently Asked Questions

How are areas under curves calculated in NCERT Class 12 Maths Chapter 8?

What is the role of graphs in NCERT Solutions for Class 12 Maths Chapter 8?

Can NCERT Class 12 Maths Chapter 8 questions involve areas between two curves?

How do I choose the correct limits for an area problem in NCERT Class 12 Maths Chapter 8?

What types of curves are used in NCERT Class 12 Maths Chapter 8 Applications of Integrals?

What can I practise from the Miscellaneous Exercise of NCERT Class 12 Maths Chapter 8?

Can NCERT Solutions for Class 12 Maths Chapter 8 Applications of Integrals help with competitive exams like JEE?

Join ALLEN!

NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals

1.0Key Concepts of Class 12 Maths Chapter 8 Applications of Integrals

2.0NCERT Class 12 Maths Chapter 8 Applications of Integrals : Detailed Solutions

EXERCISE - 8.1

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 8 Application of Integrals Exercise-wise Solutions

5.0Key Features and Benefits of Class 12 Maths Chapter 8 Applications of Integrals

Table of Contents

Frequently Asked Questions

How are areas under curves calculated in NCERT Class 12 Maths Chapter 8?

What is the role of graphs in NCERT Solutions for Class 12 Maths Chapter 8?

Can NCERT Class 12 Maths Chapter 8 questions involve areas between two curves?

How do I choose the correct limits for an area problem in NCERT Class 12 Maths Chapter 8?

What types of curves are used in NCERT Class 12 Maths Chapter 8 Applications of Integrals?

What can I practise from the Miscellaneous Exercise of NCERT Class 12 Maths Chapter 8?

Can NCERT Solutions for Class 12 Maths Chapter 8 Applications of Integrals help with competitive exams like JEE?

Join ALLEN!

NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals

1.0Key Concepts of Class 12 Maths Chapter 8 Applications of Integrals

2.0NCERT Class 12 Maths Chapter 8 Applications of Integrals : Detailed Solutions

EXERCISE - 8.1

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 8 Application of Integrals Exercise-wise Solutions

5.0Key Features and Benefits of Class 12 Maths Chapter 8 Applications of Integrals

Table of Contents