NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.2) Exercise 8.2

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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.2) Exercise 8.2 

The Central Board of Secondary Education (CSBE) conducts the Class 12 Board Examination annually, in which over 16 lakh candidates appear. It could be an intimidating time for students. Strengths and weaknesses are part of preparation. Mathematics is a very important subject in Class 12. Every chapter carries significant marks value in the examination, and it should be revised thoroughly. With the help of the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2, students can easily understand the exercise. It helps students prepare for their final exam. Extramarks provides NCERT Solutions to help candidates with all the relevant resources. If students can deal with these subjects tactfully then they will not face many problems. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 can be used during the time of exam preparation for revision purposes. To fetch more marks in the class, it is very important to have conceptual clarity of the topic. Students should not delay taking references from the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2.

Exercise 8.2 Class 12 Maths contains several questions. Sometimes these questions can get tricky and confusing. Therefore, students need to verify their answers from the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2. All these questions are solved by the Extramarks experts.

Taking references from the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 is considered the best solution for CBSE students. This chapter contains exercises. Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2. Students can download these solutions from the Extramarks website.

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NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 is inclusive of the process of calculating the areas bounded by the curve. Students are also introduced to the Application of Integrals such as the Area under Simple Curves, Definite Integral as the Limit of a Sum, between Lines, and Ellipses, and Parabolas. By the Integration Method, we can calculate the Average Value of a Function.

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 helps students  strategise their learning of this topic.

The Area under Simple Curves:

The definition of the area under simple curves is the area bounded by the curve y = f(x) using the principles. It helps to calculate the area of the space by using a line and a parabola, a line and an ellipse, a line and a circle. A few examples accompany the method of calculation in the chapter.

The Area between Two Curves:

This part explains the method of calculating the area between two curves with a solution. It denotes that the area can be calculated by dividing the region into a number of pieces of small area and then adding up the area of those tiny pieces. Calculating the area of these tiny pieces is vertical is easy.

Candidate can align their notes with the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 for taking quick references and understanding difficult topics.

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 deals with multiple Integrals applications. Students should not delay any more to take benefit from the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2.

Q.1 Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1.

Ans

The required area bounded by (x – 1)2 + y2 = 1 and x2 + y2 = 1 is represented by shaded area ACBOA.

On solving the given equations ( x−1 ) 2 + y 2 = 1 and x 2 +y 2 =1, we get the coordinates of point of intersection as A( 1 2 ,  3 2 ) and  B( 1 2 ,−  3 2 ).This  area is symmetrical about x-axis. ∴​ Area  OBCAO=2( area OCAO ) Join​  AB such that AM⊥OC. So, the  coordinates of M are ( 1 2 ,0 ). Therefore, area  OCAO=Area OMAO+Area MCAM          = ∫ 0 1 2 1− ( x−1 ) 2 dx + ∫ 1 2 1 1− x 2 dx          = [ ( x−1 ) 2 1− ( x−1 ) 2 + 1 2 sin −1 ( x−1 ) ] 0 1 2 + [ x 2 1− x 2 + 1 2 sin −1 x ] 1 2 1          =[ ( 1 2 −1 ) 2 1− ( 1 2 −1 ) 2 + 1 2 sin −1 ( 1 2 −1 ) ] −[ ( 0−1 ) 2 1− ( 0−1 ) 2 + 1 2 sin −1 ( 0−1 ) ]+[ 1 2 1− 1 2 + 1 2 sin −1 1 ] −[ 1 2 2 1− ( 1 2 ) 2 + 1 2 sin −1 ( 1 2 ) ]    =[ − 3 8 + 1 2 ( − π 6 )− 1 2 ( − π 2 ) ]+ 1 2 ( π 2 )− 3 8 − 1 2 ( π 6 )    =− 3 8 − π 12 + π 4 + π 4 − 3 8 − π 12    =− 3 4 − π 6 + π 2   = 2π 6 − 3 4 Therefore, required area of OBCAO    =2( 2π 6 − 3 4 )  =( 2π 3 − 3 2 ) square units 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Q.2 Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3.

Ans

The area bounded by y = x2 +2, y = x, x = 0 and x = 3 is represented by the shaded area

Then, area OCBAO=area ODBAO−area ODCO = ∫ 0 3 ( x 2 +2 )dx − ∫ 0 3 x dx = [ x 3 3 +2x ] 0 3 − [ x 2 2 ] 0 3 =[ 3 3 3 +2( 3 )−0−0 ]−[ 3 2 2 −0 ] =15− 9 2 = 21 2  units Thus, the required area of shaded region is 21 2  sq units. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E0CF@

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

Ans

Here, BL and CM are perpendicular to x-axis.

Equation of line AB is    y−0= 3−0 1+1 ( x+1 ) y= 3 2 ( x+1 ) Equation of line BC is y−3= 2−3 3−1 ( x−1 ) y= −1 2 ( x−1 )+3     = 1 2 ( −x+7 ) Equation of line CA is    y−2= 2−0 3+1 ( x−3 ) y= 1 2 ( x−3 )+2     = 1 2 ( x+1 )    Area ΔACB=Area( ALBA )+Area( BLMCB )−Area( ΔAMCA )     = ∫ −1 1 3 2 ( x+1 )dx + ∫ 1 3 1 2 ( −x+7 )dx − ∫ −1 3 1 2 ( x+1 )dx     = 3 2 [ x 2 2 +x ] −1 1 + 1 2 [ − x 2 2 +7x ] 1 3 − 1 2 [ x 2 2 +x ] −1 3     = 3 2 [ 1 2 2 +1− ( −1 ) 2 2 −( −1 ) ]+ 1 2 [ − 3 2 2 +7( 3 )+ 1 2 2 −7( 1 ) ] − 1 2 [ 3 2 2 +3− ( −1 ) 2 2 −( −1 ) ]     = 3 2 ( 2 )+ 1 2 ( − 9 2 +21+ 1 2 −7 )− 1 2 ( 9 2 +3− 1 2 +1 )     =3+5−4

=4​​ square units

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

Ans

The coordinates of intersection points of the given lines are A(0, 1), B(4, 13) and C(4, 9).

Area ΔABC=Area OLBAO−Area  OLCAO     = ∫ 0 4 ( 3x+1 ) dx− ∫ 0 4 ( 2x+1 )dx     = [ 3 x 2 2 +x ] 0 4 − [ 2. x 2 2 +x ] 0 4     =[ 3 ( 4 ) 2 2 +4−3 ( 0 ) 2 2 −0 ]−[ 2. 4 2 2 +4−2. 0 2 2 +0 ]     =( 24+4 )−( 16+4 )     =28−20     =8 square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0D1D@

Q.5 Choose the correct answer

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y =2 is

(a)  2( π−2 ) (b)  π−2 (c)  2π−1 (d)  2( π+2 ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaacIcacaWGHbGaaiykaiaaysW7caaMe8UaaCOmamaabmaabaaccaGae8hWdaNaeyOeI0IaaCOmaaGaayjkaiaawMcaaaqaaiaacIcacaWGIbGaaiykaiaaysW7caaMe8Uae8hWdaNaeyOeI0IaaCOmaaqaaiaacIcacaWGJbGaaiykaiaaysW7caaMe8UaaCOmaiab=b8aWjabgkHiTiaaigdaaeaacaGGOaGaamizaiaacMcacaaMe8UaaGjbVlaahkdadaqadaqaaiab=b8aWjabgUcaRiaahkdaaiaawIcacaGLPaaaaaaa@60F1@

Ans

The smaller area enclosed by the circle, x2 + y2 = 4 and line x + y = 2 is given by shaded region ABCA.

Area ABCA=Area OACBO−AreaΔAOB     = ∫ 0 2 ( 4− x 2 ) dx− ∫ 0 2 ( 2−x )dx     = [ x 2 4− x 2 + 4 2 sin −1 ( x 2 ) ] 0 2 − [ 2x− x 2 2 ] 0 2     =[ 2 2 4− 2 2 +2 sin −1 ( 2 2 ) ]−[ 2( 2 )− 2 2 2 ]     =( 2× π 2 )−( 4−2 )     =( π−2 ) square units Therefore, correct option is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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b8aWbqaaiaaikdaaaaacaGLOaGaayzkaaGaeyOeI0YaaeWaaeaacaaI0aGaeyOeI0IaaGOmaaGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maabmaabaGae8hWdaNaeyOeI0IaaGOmaaGaayjkaiaawMcaaiaabccacaqGZbGaaeyCaiaabwhacaqGHbGaaeOCaiaabwgacaqGGaGaaeyDaiaab6gacaqGPbGaaeiDaiaabohaaeaacaqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabogacaqGVbGaaeOCaiaabkhacaqGLbGaae4yaiaabshacaqGGaGaae4BaiaabchacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaabkeacaqGUaaaaaa@017C@

Q.6 Area lying between the curves y2 = 4x and y = 2x is
(A) 2/3 (B) 1/3 (C) 1/4 (D) 3/4

Ans

The area lying between y2 = 4x and y = 2x is shown by shaded area given graph.

Area OBAO=area OCABO−area OCAO      = ∫ 0 1 2 x dx − ∫ 0 1 2xdx      =2 [ x 3 2 3 2 ] 0 1 −2 [ x 2 2 ] 0 1      = 4 3 [ 1−0 ]−[ 1 2 −0 ]      = 4 3 −1 = 1 3  square units

Therefore, correct option is B.

Q.7 Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x- axis

(ii) y = x4, x = 1, x = 5 and x- axis

Ans

(i) The required area ADCBA is represented by shaded region in graph.

   Area ABCDA= ∫ 1 2 x 2 dx = [ x 3 3 ] 1 2 = 1 3 [ 2 3 − 1 3 ] = 7 3   square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaykW7caaMc8UaaGPaVlaadgeacaWGYbGaamyzaiaadggacaaMc8UaamyqaiaadkeacaWGdbGaamiraiaadgeacqGH9aqpdaWdXaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccaWGKbGaamiEaaWcbaGaaGymaaqaaiaaikdaa0Gaey4kIipaaOqaaiaaxMaacaWLjaGaaCzcaiabg2da9maadmaabaWaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maaaaaiaawUfacaGLDbaadaqhaaWcbaGaaGymaaqaaiaaikdaaaaakeaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIZaaaamaadmaabaGaaGOmamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaigdadaahaaWcbeqaaiaaiodaaaaakiaawUfacaGLDbaaaeaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiaaiEdaaeaacaaIZaaaaiaaykW7caaMc8Uaam4CaiaabghacaqG1bGaaeyyaiaabkhacaqGLbGaaeiiaiaabwhacaqGUbGaaeyAaiaabshacaqGZbaaaaa@757E@

(ii) The required area ADCBA is represented by shaded region in graph.

   Area ABCDA= ∫ 1 5 x 4 dx = [ x 5 5 ] 1 5 = 1 5 [ 5 5 − 1 5 ] = 3124 5 =624.8  square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7E61@

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FAQs (Frequently Asked Questions)

1. Que 1.What are the Mathematical Applications of Integrals?

Lots of applications of integrals are used in many subjects like Mathematics, Physics

some of them are the following:

It is used in Physics to know the Centre of Gravity, to calculate the Mass and Momentum of Towers, Satellites, Inertia of Vehicles, and calculate the Centre of Mass, etc.

In Mathematics to find the Area Under a Curve, the Average Value of a Curve or Between Two Curves, etc. These definitions and concepts can also be found in the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2

 

2. Que 2.What are the benefits of having NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2?

To solve problems in Ex 8.2 Class 12 students should use the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 provided by Extramarks. To verify the final answer or to rectify the process students can refer to the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2. Using the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2, candidates can easily identify and correct their errors and simple mistakes.It is uploaded on the Extramarks’ website for the students.

3. How can the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 help with board exam preparation?

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 have been compiled for students to understand the topics easily. Extramarks ensures that students can clarify their doubts and obtain maximum marks in the examination. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 includes questions relating to the Areas Bounded by the Curve, Definite Integral as the Limit of a Sum, between Lines, and Ellipses, Parabolas and introduces the Application of Integrals such as the Area under Simple Curves.

4. What are the key features of NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2?

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 are prepared with the help of subject experts. These solutions help students to firm their core concepts on Applications of Integrals. These are well-structured and organised solutions. It will help students with their assignments and ace their competitive exams.

5. Are the NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 sufficient to score well in the CBSE exams?

These solutions have been created by the professional faculty at Extramarks. These solutions have been curated in consideration of the current CBSE Syllabus and its guidelines. Candidates need to measure their calibre and manage their study routine accordingly. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 are exercise-wise solutions to help students  obtain an overall idea about the concepts for the examination. Practising these questions regularly will improve efficiency and problem-solving techniques.