NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 12 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 12 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 12 Mathematics are an important part of exams for Class 12 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 12 Mathematics and also download more latest study material for all subjects. Chapter 8 Application of Integrals is an important topic in Class 12, please refer to answers provided below to help you score better in exams

## Chapter 8 Application of Integrals Class 12 Mathematics NCERT Solutions

Class 12 Mathematics students should refer to the following NCERT questions with answers for Chapter 8 Application of Integrals in Class 12. These NCERT Solutions with answers for Class 12 Mathematics will come in exams and help you to score good marks

### Chapter 8 Application of Integrals NCERT Solutions Class 12 Mathematics

**Exercise 8.1**

**Question. Find the area of the region bounded by the curve y ^{2} = x and the lines x = 1, x = 4 and the x - axis. **

**Answer :**

**Question. Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.**

**Answer :**

**Question. Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.**

**Answer :**

**Question. Find the area of the region bounded by the ellipse x ^{2}/16 + y^{2}/9 = 1 **

**Answer :**

The given equation of the ellipse, x

^{2}/16 + y

^{2}/9 = 1 , can be represented as

It can be observed that the ellipse is symmetrical about x - axis and y - axis.

∴ Area bounded by ellipse = 4× Area of OAB

Area of OAB = ∫_{2}^{4} y dx

**Question. Find the area of the region bounded by the ellipse x ^{2}/4 + y^{2}/9 = 1 **

**Answer :**

The given equation of the ellipse can be represented as

It can be observed that the ellipse is symmetrical about x - axis and y - axis.

∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse = 4× (3π/2) = 6π units

**Question. Find the area of the region in the first quadrant enclosed by x-axis, line x = √3 y and the circle x ^{2} + y^{2} = 4. **

**Answer :**

The area of the region bounded by the circle, x

^{2}+ y

^{2}= 4, x = √3 y, and the x - axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3, 1) .

Area OAB = Area ΔOCA + Area ACB

Area of OAC = (1/2) × OC × AC = (1/2) × √3 × 1 = √3/2 **...(1)**

Therefore, area enclosed by x - axis, the line x = √3 y, and the circle x^{2} + y^{2} = 4 in the

first quadrant = √3/2 + π/2 - √3/2 = π/3 units

**Question. Find the area of the smaller part of the circle x ^{2} + y^{2} = a^{2} cut off by the line x = a/√2**

**Answer :**

The area of the smaller part of the circle, x

^{2}+ y

^{2}= a

^{2}, cut off by the line, x = a/√2, is the area ABCD.

It can be observed that the area ABCD is symmetrical about x - axis.

∴ Area ABCD = 2 × Area ABC

Therefore, the area of smaller part of the circle , x^{2} + y^{2} = a^{2} , cut off by the line, x = a/√2, is [(a^{2} /2)(π/2 - 1)] units.

**Question. The area between x = y^{2} and x = 4 is divided into two equal parts by the line x = a, find the value of a.**

**Answer :**

The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts.

Area OAD = Area ABCD

It can be observed that the given area is symmetrical about x - axis.

∴ Area OED = Area EFCD

**Question. Find the area of the region bounded by the parabola y = x ^{2} and y = |x|**

**Answer :**

The area bounded by the parabola, x

^{2}= y , and the line, y = |x|, can be represented as

The given area is symmetrical about y - axis,

∴ Area OACO = Area ODBO

The point of intersection of parabola, x^{2} = y, and line, y = x, is A(1, 1).

**Question. Find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2**

**Answer :**

The area bounded by the curve, x

^{2}= 4

*y*and line, x = 4y - 2, is represented by the shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point A are (-1, 1/4).

Coordinates of point B are (2, 1)

We draw AL and BM perpendicular to x - axis.

Similarly, Area OACO = Area OLAC - Area OLAO

**Question. Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3**

**Answer :**

The region bounded by the parabola,

*y*

^{2}= 4

*x,*and the line, x = 3, is the area OACO.

**Question. 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) π/2

(C) π/3

(D) π/4

**Answer :**

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as

**Question. Area of the region bounded by the curve y^{2} = 4x , y - axis and the line y = 3 is **

(A) 2

(B) 9/4

(C) 9/3

(D) 9/2

**Answer :**

The area bounded by the curve,

*y*

^{2}= 4x, y - axis, and y = 3 is represented as

**Exercise 8.2**

**Question. Find the area of the circle 4 x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y **

**Answer :**

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4*x*^{2} + 4*y*^{2} = 9, and parabola, *x*^{2} = 4*y* , we obtain the point of intersection as B(√2, 1/2) and D(-√2, 1/2).

It can be observed that the required area is symmetrical about y - axis.

Area OBCDO = 2 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are (√2, 0).

Therefore, Area OBCO = Area OMBCO - Area OMBO

**Question. Find the area bounded by curves ( x – 1)^{2} + y^{2} = 1 and x^{2} + y^{ 2} = 1**

**Answer :**

The area bounded by the curves, (x - 1)

^{2}+ y

^{2}= 1 and x

^{2}+ y

^{2}= 1, is represented by the shaded area as

On solving the equations, (x - 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1, we obtain the point of intersections as A (1/2, √3/2) and B(1/2, √3/2)

It can be observed that the required area is symmetrical about x - axis.

∴ Area OBCAO = 2 × Area OCAO

We join AB, which intersects OC at M, such that AM is perpendicular to OC.

The coordinates of M are (1/2, 0)

⇒ Area OCAO = Area OMAO + Area MCAM

**Question. Find the area of the region bounded by the curves y = x^{2 }+ 2, y = x, x = 0 and x = 3**

**Answer :**

The area bounded by the curves, y =

*x*

^{2 }+ 2,

*y*=

*x*,

*x*= 0 and x = 3, is represented b the shaded area OCBAO as

**Question. Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).**

**Answer : **

BL and CM are drawn perpendicular to x - axis.

It can be observed in the following figure that,

Area (Δ ACB) = Area (ALBA) + Area (BLMCB) - Area (AMCA) **...(1)**

**Question. Using integration find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and x = 4.**

**Answer :**

The equations of sides of the triangle are y = 2x + 1, y = 3x + 1, and x = 4

On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C(4, 9)

**Question. Smaller area enclosed by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is**

A. 2 (π – 2)

B. π – 2

C. 2π – 1

D. 2 (π + 2)

**Answer :**

The smaller area enclosed by the circle,

*x*

^{2}+

*y*

^{2}= 4 and the line

*x*+

*y*= 2 , is represented by the shaded area ACBA as

**Question. Area lying between the curve y^{2} = 4x and y = 2x is**

A. 2/3

B. 1/3

C. 1/4

D. 3/4

**Answer :**

The area lying between the curve, y

^{2}= 4

*x*and

*y*= 2

*x*, is represented by the shaded area OBAO as

**Miscellaneous Solutions**

**Question. Find the area under the given curves and given lines:
(i) y = x^{2}, x = 1, x = 2 and x-axis
(ii) y = x^{4}, x = 1, x = 5 and x –axis**

**Answer :**

(i) The required area is represented by the shaded area ADCBA as

(ii) The required area is represented by the shaded area ADCBA as

**Question. Find the area between the curves y = x and y = x^{2}.**

**Answer :**

The required area is represented by the shaded area OBAO as

**Question. Find the area of the region lying in the first quadrant and bounded by y = 4x^{2}, x = 0, y = 1 and y = 4.**

**Answer :**

The area in the first quadrant bounded by y = 4x

^{2}, x = 0, y = 1 and y = 4 is represented by the shaded area ABCDA as

**Question. Sketch the graph of y = |x + 3| and evaluate ∫ _{-6}^{0} |x + 3| dx.**

**Answer :**

The given equation is y = |x + 3|

The corresponding values of x and y are given in the following table :

**Question. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.**

**Answer :**

The graph of y = sin x can be drawn as

= [ - cosπ + cos 0] + |- cos 2π + cos π|

= 1 + 1+ |(-1-1)|

= 2 + |-2|

= 2 + 2 = 4 units

**Question. Find the area enclosed between the parabola y^{2} = 4ax and the line y = mx.**

**Answer :**

The area enclosed between the parabola, y

^{2}= 4

*ax*and the line

*y*=

*mx*, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and (4a/m^{2}, 4a/m)

We draw AC perpendicular to x - axis.

Area OABO = Area OCABO - Area (ΔOCA)

**Question. Find the area enclosed by the parabola 4 y = 3x^{2} and the line 2y = 3x + 12**

**Answer :**

The area enclosed between the parabola, 4y = 3x

^{2}, and the line, 2y = 3x + 12, is represented by the shaded area OBAO as

The points of intersection of the given curves are A(-2, 3) and (4, 12).

We draw AC and BD perpendicular to x - axis.

Area OBAO = Area CDBA - (Area ODBO + Area OACO)

**Question. Find the area of the smaller region bounded by the ellipse x ^{2} /9 + y^{2} /4 and the line x/3 + y/2 = 1.**

**Answer :**

The area of the smaller region bounded by the ellipse, x

^{2}/9 + y

^{2}/4 = 1, and the line x/3 + y/2 = 1 , is represented by the shaded region BCAB as

**Question. Find the area of the smaller region bounded by the ellipse x ^{2} /a^{2} + y^{2} /b^{2} = 1 and the line x/a + y/b = 1.**

**Answer :**

The area of the smaller region bounded by the ellipse, x

^{2}/a

^{2}+ y

^{2}/b

^{2}= 1 and the line, x/a + y/b = 1, is represented by the the shaded region BCAB as

**Question. Find the area of the region enclosed by the parabola x^{2} = y, the line y = x + 2 and x-axis.**

**Answer :**

The area of the region enclosed by the parabola,

*x*

^{2}=

*y*, the line, y = x + 2 and x - axis is represented by the shaded region OABCO as

**Question. Using the method of integration find the area bounded by the curve |x| + |y| = 1.**

[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and – x – y = 1].

**Answer : **

The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as

**Question. Find the area bounded by curves {(x, y) : y ≥ x ^{2} and y = |x|}.**

**Answer :**

The area bounded by the curves, {(x, y) : y ≥ x

^{2}and y = |x|}, is represented by the shaded region as

**Question. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).**

**Answer : **

The Vertices of ΔABC are A(2, 0), B(4, 5) and C(6, 3).

Equation of line segment AB is

y - 0 = [(5 - 0)/(4 - 2)] (x - 2)

⇒ 2y = 5x - 10

⇒ y = (5/2) (x - 2) **...(1) **

Equation of line segment BC is

y - 5 = [(3 - 5)(6 - 4)] (x - 4)

⇒ 2y - 10 = -2x + 8

⇒ 2y = -2x + 18

⇒ y = -x + 9 **...(2) **

Equation of line segment CA is

y - 3 = 0 - 3 / 2 - 6 (x-6)

⇒ -4y + 12 = -3x + 18

⇒ 4y = 3x - 6

⇒ y = (3/4)(x - 2) **...(3) **

Area (ΔABC) = Area (ABLA) + Area (BLMCB) - Area(ACMA)

**Question. Using the method of integration find the area of the region bounded by lines: 2 x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0**

**Answer :**

The given equations of lines are

2x + y= 4

**...(1)**

3x -2 y = 6

**...(2)**

And, x - 3y + 5 = 0

**...(3)**

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x - axis.

Area (ΔABC) = Area (ALMCA) - Area (ALB) - Area(CMB)

**Question. Find the area of the region {(x, y) : y ^{2} ≤ 4x, 4x^{2} + 4y^{2} ≤ 9}**

**Answer :**

The area bounded by the curves, {(x, y) : y

^{2}≤ 4x, 4x

^{2}+ 4y

^{2}≤ 9}, is represented as

**Question. Choose the correct answer Area bounded by the curve y = x^{3}, the x-axis and the ordinates x = –2 and x = 1 is**

A. – 9

B. -15/4

C. 15/4

D. 17/4

**Answer :**

**Question. Choose the correct answer The area bounded by the curve y = x | x| ,, x-axis and the ordinates x = –1 and x = 1 is given by**

[Hint: y = x^{2} if x > 0 and y = –x^{2} if x < 0]

A. 0

B. 1/3

C. 2/3

D. 4/3

**Answer :**

**Question. Choose the correct answer The area of the circle x^{2} + y^{2} = 16 exterior to the parabola y^{2} = 6x is**

**A. 4/3 (4π - √3)**

B. 4/3 (4π + √3)

C. 4/3 (8π - √3)

D. 4/3 (4π + √3)

B. 4/3 (4π + √3)

C. 4/3 (8π - √3)

D. 4/3 (4π + √3)

**Answer :**

The given equations are

*x*

^{2}+

*y*

^{2}= 16

**...(1)**

y

^{2}= 6x

**...(2)**

**Question. The area bounded by the y-axis, y = cos x and y = sin x when 0 <= x <= π2**

(A) 2 ( 2 −1)

(B) √2-1

(C) √2+1

(D) √2

**Answer :**

The given equation are

y = cos x

**...(1)**

And, y = sin x

**...(2)**

NCERT Solutions Class 12 Mathematics Chapter 1 Relations and Functions |

NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions |

NCERT Solutions Class 12 Mathematics Chapter 3 Matrices |

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants |

NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiability |

NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives |

NCERT Solutions Class 12 Mathematics Chapter 7 Integrals |

NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals |

NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations |

NCERT Solutions Class 12 Mathematics Chapter 10 Vector Algebra |

NCERT Solutions Class 12 Mathematics Chapter 11 Three Dimensional Geometry |

NCERT Solutions Class 12 Mathematics Chapter 12 Linear Programming |

NCERT Solutions Class 12 Mathematics Chapter 13 Probability |

## More Study Material

### NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals

NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 12 Mathematics textbook online or you can easily download them in pdf.

### Chapter 8 Application of Integrals Class 12 Mathematics NCERT Solutions

The Class 12 Mathematics NCERT Solutions Chapter 8 Application of Integrals are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 8 Application of Integrals of Mathematics Class 12 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 8 Application of Integrals Class 12 chapter of Mathematics so that it can be easier for students to understand all answers.

**NCERT Solutions Chapter 8 Application of Integrals Class 12 Mathematics**

Class 12 Mathematics NCERT Solutions Chapter 8 Application of Integrals is a really good source using which the students can get more marks in exams. The same questions will be coming in your Class 12 Mathematics exam. Learn the Chapter 8 Application of Integrals questions and answers daily to get a higher score. Chapter 8 Application of Integrals of your Mathematics textbook has a lot of questions at the end of chapter to test the students understanding of the concepts taught in the chapter. Students have to solve the questions and refer to the step-by-step solutions provided by Mathematics teachers on studiestoday to get better problem-solving skills.

**Chapter 8 Application of Integrals Class 12 NCERT Solution Mathematics**

These solutions of Chapter 8 Application of Integrals NCERT Questions given in your textbook for Class 12 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 12.

**Class 12 NCERT Solution Mathematics Chapter 8 Application of Integrals**

NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals detailed answers are given with the objective of helping students compare their answers with the example. NCERT solutions for Class 12 Mathematics provide a strong foundation for every chapter. They ensure a smooth and easy knowledge of Revision notes for Class 12 Mathematics. As suggested by the HRD ministry, they will perform a major role in JEE. Students can easily download these solutions and use them to prepare for upcoming exams and also go through the Question Papers for Class 12 Mathematics to clarify all doubts

You can download the NCERT Solutions for Class 12 Mathematics Chapter 8 Application of Integrals for latest session from StudiesToday.com

Yes, you can click on the link above and download NCERT Solutions in PDFs for Class 12 for Mathematics Chapter 8 Application of Integrals

Yes, the NCERT Solutions issued for Class 12 Mathematics Chapter 8 Application of Integrals have been made available here for latest academic session

You can easily access the links above and download the Chapter 8 Application of Integrals Class 12 NCERT Solutions Mathematics for each chapter

There is no charge for the NCERT Solutions for Class 12 Mathematics Chapter 8 Application of Integrals you can download everything free

Regular revision of NCERT Solutions given on studiestoday for Class 12 subject Mathematics Chapter 8 Application of Integrals can help you to score better marks in exams

Yes, studiestoday.com provides all latest NCERT Chapter 8 Application of Integrals Class 12 Mathematics solutions based on the latest books for the current academic session

Yes, studiestoday provides NCERT solutions for Chapter 8 Application of Integrals Class 12 Mathematics in mobile-friendly format and can be accessed on smartphones and tablets.

Yes, NCERT solutions for Class 12 Chapter 8 Application of Integrals Mathematics are available in multiple languages, including English, Hindi