NCERT Solutions Class 12 Mathematics Chapter 8

Integral Calculus is related to Eudoxus' (440 B.C.) and Archimedes' (440 B.C.) exhaustion approach (300 B.C.). This method emerged from the need to solve problems estimating the areas of plane figures, surface areas, and solid body volumes. Newton began working on Calculus in 1665, which led to the formulation of concepts as we know them. NCERT solutions class 12 mathematics chapter 8 provides various hints, tactics, and intriguing facts to help students break up the monotony of studying and develop a mathematical base that will help them throughout their studies.

NCERT solutions class 12 mathematics chapter 8 - Application of Integrals is a continuation of the previous chapter, in which students learned about integration and how to solve problems using various methods. Students have learnt formulae for calculating the areas of triangles, rectangles, trapeziums, and circles, among other geometrical figures. These fundamental formulae are crucial in applying mathematics to a wide range of real-world issues. However, this method fails when estimating the areas encompassed by curves. Integral calculus is required to calculate the areas under a curve. NCERT solutions class 12 mathematics chapter 8 use integrals to get the area under curves. Curves such as circles, parabolas, and hyperbolas can be simple or complex.

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 8

All the topics covered in NCERT Solutions Class 12 Mathematics Chapter 8 are equally important. They go over many aspects of integration and how to calculate the area under a curve throughout multiple scenarios. Before attempting the practice questions, students must first grasp the underlying concepts. They should also be familiar with the preceding chapter and the formulas listed. The  NCERT Solutions Class 12 Mathematics Chapter 8 broadly covers two main concepts of the chapter 'Application of Integrals' and  'Integral Calculus'.

List of NCERT Solutions Class 12 Mathematics Chapter 8 Exercises

Students learned about definite integrals as the limit of a sum in Chapter 7 and how to evaluate them using the Fundamental Theorem of Calculus. This concept is used in NCERT solutions for class 12 mathematics chapter 8 to give students an easy approach to computing the area under curves. The area beneath the curve can be thought of as a collection of several very thin vertical stripes. We may now determine the total area necessary by integrating this region between some specific limitations. NCERT Solutions for class 12 mathematics chapter 8 Application of Integrals gives students a better grasp of integration using real-life examples. We'll look at a range of examples, formulas, and theories in depth in this article to help students create a strong mathematical foundation.

Several formulas and theorems are used in this lesson. There is also a blend of current topics with those that the students learned in the previous chapter. If not approached systematically, integration can become a difficult and perplexing concept. Students must regularly review this lesson for all these reasons, particularly the formulas, to learn the material. With the help of examples and practice problems, all the tasks in NCERT Solutions Class 12 Mathematics Chapter 8 - Application of Integrals are well explained. Here are some NCERT solutions for students to practise –

 mathematics Chapter 8 Ex 8.1 - 13 Questions

Class 12 Mathematics Chapter 8 Ex 8.2 - 7 Questions

Class 12 Mathematics Chapter 8 Miscellaneous Ex - 19 Questions

The area bounded by the curve of a function, the area between two curves, and the area under simple curves are all covered in NCERT solutions class 12 mathematics chapter 8.

NCERT Solutions Class 12 Mathematics Chapter 8 Formula List

NCERT solutions class 12 mathematics chapter 8 requires using formulas and a constant focus. Students must comprehend what the question requires and carry out the procedures stepwise. They should also divide the problem into smaller parts to make it easier to understand and avoid ambiguity, hence lowering the risk of errors. More significantly, children must understand the processes in the technique since they are the keys to answering a variety of questions.

The following are some formulas that are regularly used in NCERT solutions for class 12 mathematics chapter 8 –

  • Area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) : a∫by dx = a∫bf(x) dx.
  • The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d : c∫d x dy = c∫d φ (y) dy.
  • The area of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b : b∫a[f(x) - g(x)] dx.

Class 12 NCERT Mathematics Syllabus

Term - 1

Unit Name

 

Chapter Name
 

Relations and Function

Relations and Functions

Inverse Trigonometric Functions

Algebra

 

Matrices

Determinants

Calculus

 

Continuity and Differentiability

Application of Derivatives 

Linear Programming Linear Programming

 Term - 2

Unit Name Chapter Name
 

Calculus

 

Integrals

Application of Integrals

Differential Equations

Vectors and Three-Dimensional Geometry  Vector Algebra

Three Dimensional Geometry 

Probability Probability 

Experts at Extramarks create NCERT Solutions to assist students to understand ideas more quickly and correctly. NCERT Solutions provide extensive, step-by-step explanations of textbook problems. All classes can benefit from such solutions –

  • NCERT Solutions class 1
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  • NCERT Solutions class 8
  • NCERT Solutions class 9
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  • NCERT Solutions class 11
  • NCERT Solutions class 12

NCERT Mathematics Exam Pattern

Duration of Marks 3 hours 15 minutes
Marks for Internal 20 marks
Marks for Theory 80 marks
Total Number of Questions 38 Questions
Very short answer question 20 Questions
Short answer questions 7 Questions
Long Answer Questions (4 marks each) 7 Questions
Long Answer Questions (6 marks each) 4 Questions

Key Features of NCERT Solutions Class 12 Mathematics Chapter 8

To do well in CBSE and pass various competitive exams such as NEET and IIT, one must have a strong mathematical foundation. The NCERT books present you with difficult problems that improve your analytical ability and expose you to many questions that could appear in all of these exams. Our integrals class 12 solutions adhere to the most recent CBSE curriculum and are incredibly valuable to your preparation for the following reasons:

  • The answers are exceptional because teachers with deep subject knowledge create them.
  • You can revise all the significant points of a chapter in a short amount of time.
  • The solutions also prepare you for managing your stress and time, giving you a better chance of finishing your examinations on time.
  • NCERT solutions class 12 mathematics chapter 8 will assist you in grasping the fundamental concepts quickly and easily.

NCERT Exemplar Class 12 Mathematics 

Here are some solutions and problems to help students prepare for their final exams. These example questions are a little more complex, and they cover a variety of concepts in each chapter of Class 12 Mathematics subject. Students will fully understand all the concepts covered in each chapter by practising these NCERT Exemplars for mathematics Class 12. Each question in these materials is connected to concepts covered in the CBSE Class 12 syllabus (2022-2023). Our experts provide the best solutions to challenges that students confront. To match the ideas taught in each class and provide the greatest practising materials or worksheets for students, all these questions reflect the question pattern found in NCERT books.

Q.1 Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.

Ans.

The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

Area of ABCD= 1 4 ydx = 1 4 x dx = [ x 3 2 3 2 ] 1 4 = 2 3 [ 4 3 2 1 3 2 ] = 2 3 ( 81 ) = 14 3 units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8012@

Q.2 Find the area of the region bounded by y2= 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Ans.

The area of the region bounded by the curve, y2 = 9x, the lines, x = 2 and x = 4, and the x-axis is the area ABCD.

Area of ABCD= 2 4 ydx = 2 4 3 x dx =3 [ x 3 2 3 2 ] 2 4 =3× 2 3 [ 4 3 2 2 3 2 ] =2( 82 2 ) =( 164 2 )units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@87C3@

Q.3 Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Ans.

The area of the region bounded by the curve, x2 = 4y, the lines, y = 2 and y = 4, and the y-axis is the area ABCD.

Area of ABCD= 2 4 xdy = 2 4 2 y dx =2 [ y 3 2 3 2 ] 2 4 = 4 3 [ 4 3 2 2 3 2 ] = 4 3 ( 82 2 ) =( 328 2 3 )units

Q.4 Find the area of the region bounded by the ellipse (x2/16) + (y2/9) = 1.

Ans.

The given ellipse (x2/16) + (y2/9)=1 can be represented as given below:

Ellipse is symmetrical about x-axis and y-axis.
So, area bounded by ellipse = 4 x Area of OAB

Area of OAB= 0 4 ydx = 0 4 3 1 x 2 16 dx = 3 4 0 4 16 x 2 dx = 3 4 [ x 2 16 x 2 + 16 2 sin 1 x 4 ] 0 4 = 3 4 [ 4 2 16 4 2 + 16 2 sin 1 4 4 ] 3 4 [ 0 2 16 0 2 + 16 2 sin 1 0 4 ] = 3 4 ×8 π 2 =3πsquareunits Therefore, area bounded by ellipse =4×Area of OAB =4×3πsquareunits =12πsquareunits MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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b8aWjaaykW7caWGZbGaamyCaiaadwhacaWGHbGaamOCaiaadwgacaaMc8UaamyDaiaad6gacaWGPbGaamiDaiaadohaaeaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGymaiaaikdacaaMc8Uae8hWdaNaaGPaVlaadohacaWGXbGaamyDaiaadggacaWGYbGaamyzaiaaykW7caWG1bGaamOBaiaadMgacaWG0bGaam4Caaaaaa@5BAF@

Q.5 Find the area of the region bounded by the ellipse (x2/4) + (y2/9) = 1.

Ans.

The given ellipse (x2/4) + (y2/9) =1 can be represented as given below:

The given equation of ellipse is x 2 4 + y 2 9 =1 y=3 1 x 2 4 It can be observed that ellipse is symmetrical about x-axis and y-axis. Area bounded by ellipse =4×Area OAB Area of OAB= 0 2 ydx = 0 2 3 1 x 2 4 dx = 3 2 0 2 ( 4 x 2 ) dx = 3 2 [ x 2 ( 4 x 2 ) + 4 2 sin 1 x 2 ] 0 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@0E74@ = 3 2 [ 2 2 ( 4 2 2 ) + 4 2 sin 1 2 2 ] 3 2 [ 0 2 ( 4 0 2 ) + 4 2 sin 1 0 2 ] = 3 2 ×2 π 2 = 3 2 πsquare units Therefore, area bounded by the ellipse =4× 3 2 πsquare units =6square unitMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@CA92@

Q.6

Findtheareaoftheregioninthefirst  quadrantenclosedbyxaxis,linex=3yandthecirclex2+y2=4.

Ans.

The area of the smaller part of the circle, x2+y2=a2,cutoffbythe line, x=a2, is the area ABCDA.

It is observed that the area ABCD is symmetrical about x-axis. area ABCD=2×Area ABC Area ofABC= a 2 a ydx = a 2 a a 2 x 2 dx = [ x 2 a 2 x 2 + a 2 2 sin 1 x a ] a 2 a =[ a 2 a 2 a 2 + a 2 2 sin 1 a a ][ ( a 2 ) 2 a 2 ( a 2 ) 2 + a 2 2 sin 1 ( a 2 ) a ] = a 2 2 × π 2 a 2 2 × a 2 a 2 2 ( π 4 ) = a 2 2 × π 2 a 2 4 a 2 2 ( π 4 ) = a 2 4 ( π1 π 2 ) = a 2 4 ( π 2 1 ) Area ABCD=2{ a 2 4 ( π 2 1 ) } = a 2 2 ( π 2 1 )squareunits Therefore, the area of the 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 )squareunits. 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Q.7 The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Ans.

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

AreaOED= 0 a ydx = 0 a x dx = [ x 3 2 3 2 ] 0 a = 2 3 a 3 2 Area of CDEF = a 4 ydx = a 4 x dx = [ x 3 2 3 2 ] a 4 = 2 3 ( 4 3 2 a 3 2 ) = 2 3 ( 8 a 3 2 ) According to equation ( i ), we have 2 3 a 3 2 = 2 3 ( 8 a 3 2 ) a 3 2 =( 8 a 3 2 ) 2 a 3 2 =8 a 3 2 =4 a= 4 2 3 Therefore, the value of a is 4 2 3 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1583@

Q.8 Find the area of the region bounded by the parabola y = x2 and y = |x|.

Ans.

The area bounded by the parabola y = x2 and
y = |x| is shown as below:

Since the required area is symmetrical about y-axis.

So, area OACO = area OBDO

The points of intersection of parabola and line are A(1,1) and B(–1, 1).

Area of OACO=Area ΔAMOareaOMACO = 1 2 ×OM×AM 0 1 ydx = 1 2 ×1×1 0 1 x 2 dx = 1 2 [ x 3 3 ] 0 1 = 1 2 [ 1 3 3 0 3 3 ] = 1 2 1 3 = 1 6 Required area =2( Area of OACO ) =2× 1 6 = 1 3 square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EE87@

Q.9 Find the area of the region bounded by the curve x2 = 4y and the line x = 4y – 2.

Ans.

The area bounded by the parabola x2 = 4y and x = 4y – 2 is shown in figure.
The intersection points of line and parabola are A(–1, 1/4) and B(2,1).

HereALXaxis and BMXaxis.Then AreaOBAO=AreaOBCO+AreaOACO =( AreaOMBCOAreaOMBO )+( AreaOLACAreaOLAO ) =( 0 2 x+2 4 dx 0 2 x 2 4 dx )+( 1 0 x+2 4 dx 1 0 x 2 4 dx ) = 1 4 [ x 2 2 +2x ] 0 2 1 4 [ x 3 3 ] 0 2 + 1 4 [ x 2 2 +2x ] 1 0 1 4 [ x 3 3 ] 1 0 = 1 4 [ 2 2 2 +2( 2 ) ] 1 4 [ 2 3 3 ]+0 1 4 [ ( 1 ) 2 2 +2( 1 ) ]0+ 1 4 [ ( 1 ) 3 3 ] = 1 4 ×6 1 4 × 8 3 1 4 × 3 2 + 1 4 × 1 3 = 3 2 2 3 + 3 8 1 12 = 3616+92 24 = 27 24 = 9 8 squnits Thus, the area of OBAO is 9 8 sq.units. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@F72B@

Q.10 Find the area of the region bounded by the curve y2 = 4x and the line x = 3.

Ans.

The region bounded by the parabola and line is given below:

The area COAC is symmetrical about x-axis. So,
Area OABO = Area OCBO
Then, Area OCAO = 2(area OABO)

AreaOACO=2 0 3 ydx =2 0 3 2 x dx =4 [ x 3 2 3 2 ] 0 3 = 2 3 ×4[ 3 3 2 0 ]=8 3 Therefore, the required area is 8 3 squareunits. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AF9D@

Q.11 Choose the correct answer
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is

(a)π(b) π 2 (c) π 3 (d) π 4 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaGGOaGaamyyaiaacMcacaaMe8UaaGjbVlabec8aWjaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaiikaiaadkgacaGGPaGaaGjbVlaaysW7daWcaaqaaiabec8aWbqaaiaaikdaaaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caGGOaGaam4yaiaacMcacaaMe8+aaSaaaeaaiiaacqWFapaCaeaacaaIZaaaaiaaysW7caaMe8UaaGjbVlaaysW7caGGOaGaamizaiaacMcacaaMe8UaaGjbVpaalaaabaGaeqiWdahabaGaaGinaaaaaaa@6BAF@

Ans.

The area bounded by the circle and the lines x = 0 and x = 2 lies in first quadrant.

AreaOABO= 0 2 ydx = 0 2 4 x 2 dx = [ x 2 4 x 2 + 4 2 sin 1 ( x 2 ) ] 0 2 =[ 2 2 4 2 2 + 4 2 sin 1 ( 2 2 ) ][ 0 2 4 0 2 + 4 2 sin 1 ( 0 2 ) ] =[ 0+2× π 2 ][ 0+0 ]=π Thus,the correct option is A. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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b8aWbqaaiaaikdaaaaacaGLBbGaayzxaaGaeyOeI0YaamWaaeaacaaIWaGaey4kaSIaaGimaaGaay5waiaaw2faaiabg2da9iab=b8aWbqaaiaadsfacaWGObGaamyDaiaadohacaGGSaGaaGzaVlaaykW7caaMc8UaamiDaiaadIgacaWGLbGaaeiiaiaabogacaqGVbGaaeOCaiaabkhacaqGLbGaae4yaiaabshacaqGGaGaae4BaiaabchacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaabgeacaqGUaaaaaa@CF9E@

Q.12 Choose the correct answer

Area of the region bounded by the curve y2 = 4x, and y-axis and the line y = 3 is

(A) 2 (B) 9/4 (C) 9/3 (D) 9/2

Ans.

The area bounded by the parabola and the lines y = 0 and y = 3 lies in first quadrant.

AreaOAB= 0 3 xdy = 0 2 y 2 4 dy = 1 4 [ y 3 3 ] 0 3 = 1 12 [ 3 3 0 3 ] = 27 12 = 9 4 Thus,the correct option is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@96C3@

Q.13 Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.

Ans.

The required area is represented by shaded area OBCDO.

On solving the given equations 4x 2 + 4y 2 = 9 and parabola x 2 =4y, we get the coordinates of point of intersection as B( 2 , 1 2 )and C( 2 , 1 2 ).This area is symmetrical about y-axis. AreaOBCD=2( areaOBCO ) Since,BMXaxis. So,the coordinates of M are ( 2 ,0 ). Therefore,area OBCO=AreaOMBCOAreaOMBO = 0 2 94 x 2 4 dx 0 2 x 2 4 dx = 1 2 0 2 94 x 2 dx 0 2 x 2 4 dx = 1 2 0 2 2 9 t 2 dt 2 1 4 0 2 x 2 dx [ Let t=2x dt dx =2 andt=2× 2 ,( Whenx= 2 ) =2 2 t=2×0( Whenx=0 ) =0 ] = 1 4 [ t 2 9 t 2 + 9 2 sin 1 t 3 ] 0 2 2 1 4 [ x 3 3 ] 0 2 = 1 4 [ 2 2 2 9 ( 2 2 ) 2 + 9 2 sin 1 2 2 3 ] 1 4 [ ( 0 ) 94 ( 0 ) 2 + 9 2 sin 1 2( 0 ) 3 ] 1 4 [ ( 2 ) 3 3 ] = 1 4 ( 2 + 9 2 sin 1 2 2 3 ) 2 6 = 2 4 + 9 8 sin 1 2 2 3 2 6 = 2 12 + 9 8 sin 1 2 2 3 = 1 2 ( 2 6 + 9 4 sin 1 2 2 3 ) Therefore, area of shaded region is 2× 1 2 ( 2 6 + 9 4 sin 1 2 2 3 ) =( 2 6 + 9 4 sin 1 2 2 3 )square units. 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Q.14 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 ( x1 ) 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. AreaOBCAO=2( areaOCAO ) Join AB such thatAMOC. So,the coordinates of M are ( 1 2 ,0 ). Therefore,area OCAO=AreaOMAO+AreaMCAM = 0 1 2 1 ( x1 ) 2 dx + 1 2 1 1 x 2 dx = [ ( x1 ) 2 1 ( x1 ) 2 + 1 2 sin 1 ( x1 ) ] 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 ) ] [ ( 01 ) 2 1 ( 01 ) 2 + 1 2 sin 1 ( 01 ) ]+[ 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 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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Q.15 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,areaOCBAO=areaODBAOareaODCO = 0 3 ( x 2 +2 )dx 0 3 xdx = [ x 3 3 +2x ] 0 3 [ x 2 2 ] 0 3 =[ 3 3 3 +2( 3 )00 ][ 3 2 2 0 ] =15 9 2 = 21 2 units Thus, the required area of shaded region is 21 2 squnits. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E0CF@

Q.16 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 y0= 30 1+1 ( x+1 ) y= 3 2 ( x+1 ) Equation of line BC is y3= 23 31 ( x1 ) y= 1 2 ( x1 )+3 = 1 2 ( x+7 ) Equation of line CA is y2= 20 3+1 ( x3 ) y= 1 2 ( x3 )+2 = 1 2 ( x+1 ) AreaΔACB=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+54

=4square units

Q.17 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).

AreaΔABC=AreaOLBAOAreaOLCAO = 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 +43 ( 0 ) 2 2 0 ][ 2. 4 2 2 +42. 0 2 2 +0 ] =( 24+4 )( 16+4 ) =2820 =8 square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaadgeacaWGYbGaamyzaiaadggacaaMc8UaeyiLdqKaamyqaiaadkeacaWGdbGaeyypa0JaamyqaiaadkhacaWGLbGaamyyaiaaykW7caWGpbGaamitaiaadkeacaWGbbGaam4taiabgkHiTiaadgeacaWGYbGaamyzaiaadggacaaMc8UaaGPaVlaad+eacaWGmbGaam4qaiaadgeacaWGpbaabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0Zaa8qmaeaadaqadaqaaiaaiodacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaWcbaGaaGimaaqaaiaaisdaa0Gaey4kIipakiaadsgacaWG4bGaeyOeI0Yaa8qmaeaadaqadaqaaiaaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaiaadsgacaWG4baaleaacaaIWaaabaGaaGinaaqdcqGHRiI8aaGcbaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0ZaamWaaeaacaaIZaWaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmaaaacqGHRaWkcaWG4baacaGLBbGaayzxaaWaa0baaSqaaiaaicdaaeaacaaI0aaaaOGaeyOeI0YaamWaaeaacaaIYaGaaiOlamaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaey4kaSIaamiEaaGaay5waiaaw2faamaaDaaaleaacaaIWaaabaGaaGinaaaaaOqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maadmaabaGaiGjGiodadGaMaUaaaeacyc4aiGjGbmaabGaMakacyciI0aaacGaMaAjkaiacycOLPaaadGaMaYbaaSqajGjGbGaMakacyciIYaaaaaGcbGaMakacyciIYaaaaiadycOHRaWkcGaMaIinaiabgkHiTiaaiodadaWcaaqaamaabmaabaGaaGimaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaeyOeI0IaaGimaaGaay5waiaaw2faaiabgkHiTmaadmaabaGaiGjGikdacGaMakOlamacyc4caaqaiGjGcGaMaIinamacycihaaWcbKaMagacycOaiGjGikdaaaaakeacycOaiGjGikdaaaGamGjGgUcaRiacyciI0aGaeyOeI0IaaGOmaiaac6cadaWcaaqaaiaaicdadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgUcaRiaaicdaaiaawUfacaGLDbaaaeaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpdaqadaqaaiaaikdacaaI0aGaey4kaSIaaGinaaGaayjkaiaawMcaaiabgkHiTmaabmaabaGaaGymaiaaiAdacqGHRaWkcaaI0aaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGOmaiaaiIdacqGHsislcaaIYaGaaGimaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iaaiIdacaqGGaGaae4CaiaabghacaqG1bGaaeyyaiaabkhacaqGLbGaaeiiaiaabwhacaqGUbGaaeyAaiaabshacaqGZbaaaaa@0D1D@

Q.18 Choose the correct answer

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

(a)2( π2 ) (b)π2 (c)2π1 (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.

AreaABCA=AreaOACBOAreaΔAOB = 0 2 ( 4 x 2 ) dx 0 2 ( 2x )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 )( 42 ) =( π2 ) square units Therefore, correct option is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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b8aWbqaaiaaikdaaaaacaGLOaGaayzkaaGaeyOeI0YaaeWaaeaacaaI0aGaeyOeI0IaaGOmaaGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maabmaabaGae8hWdaNaeyOeI0IaaGOmaaGaayjkaiaawMcaaiaabccacaqGZbGaaeyCaiaabwhacaqGHbGaaeOCaiaabwgacaqGGaGaaeyDaiaab6gacaqGPbGaaeiDaiaabohaaeaacaqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabogacaqGVbGaaeOCaiaabkhacaqGLbGaae4yaiaabshacaqGGaGaae4BaiaabchacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaabkeacaqGUaaaaaa@017C@

Q.19 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.

AreaOBAO=areaOCABOareaOCAO = 0 1 2 x dx 0 1 2xdx =2 [ x 3 2 3 2 ] 0 1 2 [ x 2 2 ] 0 1 = 4 3 [ 10 ][ 1 2 0 ] = 4 3 1= 1 3 square units

Therefore, correct option is B.

Q.20 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.

AreaABCDA= 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=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@757E@

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

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

Q.21 Find the area between the curves y = x and y = x2.

Ans.

The point of intersection of curve y = x2 and line y = x is (1, 1). Here AC is drawn as perpendicular to x-axis.

AreaOBAO=Area( ΔOCA )Area( OCABO ) = 0 1 xdx 0 1 x 2 dx = [ x 2 2 ] 0 1 [ x 3 3 ] 0 1 =[ 1 2 2 0 ][ 1 3 3 0 ] = 1 2 1 3 = 1 6 square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AAF3@

Q.22 Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4.

Ans.

The area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4 is shown by shaded region ABCDA.

AreaABCD= 1 4 y 2 dy [ y=4 x 2 ] = 1 2 1 4 y 1 2 dy = 1 2 [ y 3 2 3 2 ] 1 4 = 1 3 [ 4 3 2 1 3 2 ] = 1 3 ( 81 )= 7 3 square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C1F5@

Q.23

Sketch the graph of y=| x+3 | and evaluate 6 0 | x+3 |dx. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWHtbGaaC4AaiaahwgacaWH0bGaaC4yaiaahIgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahEgacaWHYbGaaCyyaiaahchacaWHObGaaeiiaiaah+gacaWHMbGaaeiiaiaahMhacqGH9aqpdaabdaqaaiaahIhacqGHRaWkcaWHZaaacaGLhWUaayjcSdGaaeiiaiaahggacaWHUbGaaCizaiaabccacaWHLbGaaCODaiaahggacaWHSbGaaCyDaiaahggacaWH0bGaaCyzaiaaykW7daWdXaqaamaaemaabaacbeGaa8hEaiaa=TcacaWFZaaacaGLhWUaayjcSdGaaGzaVlaa=rgacaWF4bGaaiOlaaWcbaGaeyOeI0IaaGOnaaqaaiaaicdaa0Gaey4kIipaaaa@6D98@

Ans.

The graph of y = |x + 3| is as shown below:

y=| x+3 |={ ( x+3 ),x+3<0x<3 ( x+3 ),x+30x3 6 0 | ( x+3 ) |dx = 6 3 ( x+3 )dx + 3 0 ( x+3 )dx = [ x 2 2 +3x ] 6 3 + [ x 2 2 +3x ] 3 0 =[ ( 3 ) 2 2 +3( 3 ) ( 6 ) 2 2 3( 6 ) ]+[ 0 ( 3 ) 2 2 3( 3 ) ] =( 9 2 918+18 )+( 9 2 +9 ) = 9 2 + 9 2 =9 square units MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@61CE@

Q.24

Findtheareaboundedbythecurvey=sinxbetweenx=0andx=2π.

Ans.

The graph of y = sin x is shown below:

Required area=Area OABO + Area BCDB  =0πsinxdx+|π2πsinxdx|  =[cosx]0π+|[cosx]π2π|  =[cosπ+cos0]+|cos2πcosπ|  =[11]+(1+1)=4square units

Q.25 Find the area enclosed between the parabola y2 = 4ax and the line y = mx.

Ans.

Solving y2 = 4ax and y = mx, we get x = (4a/m2)
and y = (4a/m).

The points of intersection of both the curves are ( 0,0 ) and ( 4a m 2 , 4a m ). AreaOBAO=AreaOCABOArea( ΔOCA ) = 0 4a m 2 2 ax dx 0 4a m 2 mx dx =2 a [ x 3 2 3 2 ] 0 4a m 2 m [ x 2 2 ] 0 4a m 2 = 4 3 a { ( 4a m 2 ) 3 2 0 } m 2 { ( 4a m 2 ) 2 0 } = 4 3 a ( 8a a m 3 ) m 2 ( 16 a 2 m 4 ) = 4 3 × 8 a 2 m 3 8 a 2 m 3 = 8 a 2 3 m 3 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadsfacaWGObGaamyzaiaabccacaqGWbGaae4BaiaabMga caqGUbGaaeiDaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyAai aab6gacaqG0bGaaeyzaiaabkhacaqGZbGaaeyzaiaabogacaqG0bGa aeyAaiaab+gacaqGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaabkgaca qGVbGaaeiDaiaabIgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaa bogacaqG1bGaaeOCaiaabAhacaqGLbGaae4CaiaabccacaqGHbGaae OCaiaabwgacaqGGaWaaeWaaeaacaaIWaGaaiilaiaaicdaaiaawIca caGLPaaacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaaqaamaabmaaba WaaSaaaeaacaaI0aGaamyyaaqaaiaad2gadaahaaWcbeqaaiaaikda aaaaaOGaaiilaiaaykW7daWcaaqaaiaaisdacaWGHbaabaGaamyBaa aaaiaawIcacaGLPaaacaGGUaaabaGaeyinIWLaamyqaiaadkhacaWG LbGaamyyaiaaykW7caWGpbGaamOqaiaadgeacaWGpbGaeyypa0Jaam yqaiaadkhacaWGLbGaamyyaiaaykW7caWGpbGaam4qaiaadgeacaWG cbGaam4taiabgkHiTiaadgeacaWGYbGaamyzaiaadggacaaMc8+aae WaaeaacqGHuoarcaWGpbGaam4qaiaadgeaaiaawIcacaGLPaaaaeaa caWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlabg2da9maapedabaGaaGOmamaakaaabaGa amyyaiaadIhaaSqabaaabaGaaGimaaqaamaalaaabaGaaGinaiaadg gaaeaacaWGTbWaaWbaaWqabeaacaaIYaaaaaaaa0Gaey4kIipakiaa dsgacaWG4bGaeyOeI0Yaa8qmaeaacaWGTbGaamiEaiaaykW7aSqaai aaicdaaeaadaWcaaqaaiaaisdacaWGHbaabaGaamyBamaaCaaameqa baGaaGOmaaaaaaaaniabgUIiYdGccaWGKbGaamiEaaqaaiaaxMaaca WLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8Uaeyypa0JaaGOmamaakaaabaGaamyyaaWcbeaakmaadm aabaWaaSaaaeaacaWG4bWaaWbaaSqabeaadaWcaaqaaiaaiodaaeaa caaIYaaaaaaaaOqaamaalaaabaGaaG4maaqaaiaaikdaaaaaaaGaay 5waiaaw2faamaaDaaaleaacaaIWaaabaWaaSaaaeaacaaI0aGaamyy aaqaaiaad2gadaahaaadbeqaaiaaikdaaaaaaaaakiabgkHiTiaad2 gadaWadaqaamaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaOqa aiaaikdaaaaacaGLBbGaayzxaaWaa0baaSqaaiaaicdaaeaadaWcaa qaaiaaisdacaWGHbaabaGaamyBamaaCaaameqabaGaaGOmaaaaaaaa aaGcbaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaaisdaaeaa caaIZaaaamaakaaabaGaamyyaaWcbeaakmaacmaabaWaaeWaaeaada WcaaqaaiaaisdacaWGHbaabaGaamyBamaaCaaaleqabaGaaGOmaaaa aaaakiaawIcacaGLPaaadaahaaWcbeqaamaalaaabaGaaG4maaqaai aaikdaaaaaaOGaeyOeI0IaaGimaaGaay5Eaiaaw2haaiabgkHiTmaa laaabaGaamyBaaqaaiaaikdaaaWaaiWaaeaadaqadaqaamaalaaaba GaaGinaiaadggaaeaacaWGTbWaaWbaaSqabeaacaaIYaaaaaaaaOGa ayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaicdaai aawUhacaGL9baaaeaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maalaaaba GaaGinaaqaaiaaiodaaaWaaOaaaeaacaWGHbaaleqaaOWaaeWaaeaa daWcaaqaaiaaiIdacaWGHbWaaOaaaeaacaWGHbaaleqaaaGcbaGaam yBamaaCaaaleqabaGaaG4maaaaaaaakiaawIcacaGLPaaacqGHsisl daWcaaqaaiaad2gaaeaacaaIYaaaamaabmaabaWaaSaaaeaacaaIXa GaaGOnaiaadggadaahaaWcbeqaaiaaikdaaaaakeaacaWGTbWaaWba aSqabeaacaaI0aaaaaaaaOGaayjkaiaawMcaaaqaaiaaxMaacaWLja GaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8Uaeyypa0ZaaSaaaeaacaaI0aaabaGaaG4maaaacqGHxdaTda WcaaqaaiaaiIdacaWGHbWaaWbaaSqabeaacaaIYaaaaaGcbaGaamyB amaaCaaaleqabaGaaG4maaaaaaGccqGHsisldaWcaaqaaiaaiIdaca WGHbWaaWbaaSqabeaacaaIYaaaaaGcbaGaamyBamaaCaaaleqabaGa aG4maaaaaaaakeaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maalaaabaGa aGioaiaadggadaahaaWcbeqaaiaaikdaaaaakeaacaaIZaGaamyBam aaCaaaleqabaGaaG4maaaaaaaaaaa@65D4@

Q.26 Find the area enclosed between the parabola 4y = 3x2 and the line 2y = 3x + 12.

Ans.

The area enclosed between parabola 4y = 3x2 and line 2y = 3x + 12 is shown by shaded part AOBA in figure and point of intersection of the given curves are A(–2, 3) and B(4, 12). Two perpendiculars AC and BD are drawn on x-axis.

AreaAOBA=AreaCDBA( AreaACOA+AreaODBO ) = 2 4 3x+12 2 dx 2 4 3 x 2 4 dx = 1 2 [ 3 x 2 2 +12x ] 2 4 3 4 [ x 3 3 ] 2 4 = 1 2 { 3 2 × 4 2 +12×4 3 2 × ( 2 ) 2 12×( 2 ) } 1 4 { 4 3 ( 2 ) 3 } = 1 2 ( 24+486+24 ) 1 4 ( 64+8 ) = 1 2 ×9018 =4518 =27squareunits MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiabgsJiCjaadgeacaWGYbGaamyzaiaadggacaaMc8Uaaeyq aiaab+eacaqGcbGaaeyqaiabg2da9iaadgeacaWGYbGaamyzaiaadg gacaaMc8Uaam4qaiaadseacaWGcbGaamyqaiabgkHiTmaabmaabaGa amyqaiaadkhacaWGLbGaamyyaiaaykW7caWGbbGaam4qaiaad+eaca WGbbGaey4kaSIaamyqaiaadkhacaWGLbGaamyyaiaaykW7caWGpbGa amiraiaadkeacaWGpbaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaaca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7cqGH9aqpdaWdXaqaamaalaaabaGaaG4maiaadIhacqGHRaWkca aIXaGaaGOmaaqaaiaaikdaaaaaleaacqGHsislcaaIYaaabaGaaGin aaqdcqGHRiI8aOGaamizaiaadIhacqGHsisldaWdXaqaamaalaaaba GaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaaakeaacaaI0aaaaaWc baGaeyOeI0IaaGOmaaqaaiaaisdaa0Gaey4kIipakiaadsgacaWG4b aabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaaigdaaeaaca aIYaaaamaadmaabaGaaG4mamaalaaabaGaamiEamaaCaaaleqabaGa aGOmaaaaaOqaaiaaikdaaaGaey4kaSIaaGymaiaaikdacaWG4baaca GLBbGaayzxaaWaa0baaSqaaiabgkHiTiaaikdaaeaacaaI0aaaaOGa eyOeI0YaaSaaaeaacaaIZaaabaGaaGinaaaadaWadaqaamaalaaaba GaamiEamaaCaaaleqabaGaaG4maaaaaOqaaiaaiodaaaaacaGLBbGa ayzxaaWaa0baaSqaaiabgkHiTiaaikdaaeaacaaI0aaaaaGcbaGaaC zcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaam aacmaabaWaaSaaaeaacaaIZaaabaGaaGOmaaaacqGHxdaTcaaI0aWa aWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaiaaikdacqGHxdaTca aI0aGaeyOeI0YaaSaaaeaacaaIZaaabaGaaGOmaaaacqGHxdaTdaqa daqaaiabgkHiTiaaikdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaik daaaGccqGHsislcaaIXaGaaGOmaiabgEna0oaabmaabaGaeyOeI0Ia aGOmaaGaayjkaiaawMcaaaGaay5Eaiaaw2haaaqaaiaaxMaacaWLja GaaCzcaiaaxMaacaWLjaGaaCzcaiabgkHiTmaalaaabaGaaGymaaqa aiaaisdaaaWaaiWaaeaacaaI0aWaaWbaaSqabeaacaaIZaaaaOGaey OeI0YaaeWaaeaacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqa beaacaaIZaaaaaGccaGL7bGaayzFaaaabaGaaCzcaiaaxMaacaaMc8 UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 cqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaamaabmaabaGaaGOmai aaisdacqGHRaWkcaaI0aGaaGioaiabgkHiTiaaiAdacqGHRaWkcaaI YaGaaGinaaGaayjkaiaawMcaaiabgkHiTmaalaaabaGaaGymaaqaai aaisdaaaWaaeWaaeaacaaI2aGaaGinaiabgUcaRiaaiIdaaiaawIca caGLPaaaaeaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maalaaabaGaaGym aaqaaiaaikdaaaGaey41aqRaaGyoaiaaicdacqGHsislcaaIXaGaaG ioaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGinaiaaiwdacqGHsi slcaaIXaGaaGioaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGOmai aaiEdacaaMc8Uaam4CaiaadghacaWG1bGaamyyaiaadkhacaWGLbGa aGPaVlaaykW7caWG1bGaamOBaiaadMgacaWG0bGaam4Caaaaaa@597E@

Q.27

Find 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. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaahAeacaWHPbGaaCOBaiaahsgacaqGGaGaaCiDaiaahIga caWHLbGaaeiiaiaahggacaWHYbGaaCyzaiaahggacaqGGaGaaC4Bai aahAgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahohacaWHTbGa aCyyaiaahYgacaWHSbGaaCyzaiaahkhacaqGGaGaaCOCaiaahwgaca WHNbGaaCyAaiaah+gacaWHUbGaaeiiaiaahkgacaWHVbGaaCyDaiaa h6gacaWHKbGaaCyzaiaahsgacaqGGaGaaCOyaiaahMhacaqGGaGaaC iDaiaahIgacaWHLbGaaeiiaiaahwgacaWHSbGaaCiBaiaahMgacaWH WbGaaC4CaiaahwgaaeaadaWcaaqaaGqabiaa=HhadaahaaWcbeqaai aa=jdaaaaakeaacaWF5aaaaiaa=TcadaWcaaqaaiaa=LhadaahaaWc beqaaiaa=jdaaaaakeaacaWF0aaaaiaa=1dacaWFXaGaaGzaVlaa=b cacaWFGaGaa8xyaiaa=5gacaWFKbGaa8hiaiaa=rhacaWFObGaa8xz aiaa=bcacaWFSbGaa8xAaiaa=5gacaWFLbGaa8hiamaalaaabaGaa8 hEaaqaaiaa=ndaaaGaa83kamaalaaabaGaa8xEaaqaaiaa=jdaaaGa a8xpaiaa=fdacaWFUaaaaaa@883B@

Ans.

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

Area( ABCA )=Area( OBCAO )Area( OBAO ) = 0 3 2 1 x 2 9 dx 0 3 2( 1 x 3 )dx = 2 3 0 3 9 x 2 dx 2 3 0 3 ( 3x )dx = 2 3 [ x 2 9 x 2 + 9 2 sin 1 ( x 3 ) ] 0 3 2 3 [ 3x x 2 2 ] 0 3 = 2 3 { 3 2 9 3 2 + 9 2 sin 1 ( 3 3 ) 0 2 9 0 2 9 2 sin 1 ( 0 3 ) } 2 3 { 3( 3 ) 3 2 2 3( 0 )+ 0 2 2 } = 2 3 { 9 2 × π 2 } 2 3 { 9 9 2 } = 3π 2 3= 3 2 ( π2 )square units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiabgsJiCjacyc4GbbGaiGjGdkhacGaMaoyzaiacyc4GHbWa iGjGbmaabGaMakacyc4GbbGaiGjGdkeacGaMao4qaiacyc4GbbaacG aMaAjkaiacycOLPaaacqGH9aqpcaWGbbGaamOCaiaadwgacaWGHbWa aeWaaeaacaWGpbGaamOqaiaadoeacaWGbbGaam4taaGaayjkaiaawM caaiabgkHiTiaadgeacaWGYbGaamyzaiaadggadaqadaqaaiaad+ea caWGcbGaamyqaiaad+eaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcai aaxMaacaaMc8UaaGPaVlabg2da9maapedabaGaaGOmamaakaaabaGa aGymaiabgkHiTmaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaO qaaiaaiMdaaaaaleqaaOGaamizaiaadIhaaSqaaiaaicdaaeaacaaI ZaaaniabgUIiYdGccqGHsisldaWdXaqaaiaaikdadaqadaqaaiaaig dacqGHsisldaWcaaqaaiaadIhaaeaacaaIZaaaaaGaayjkaiaawMca aiaadsgacaWG4baaleaacaaIWaaabaGaaG4maaqdcqGHRiI8aaGcba GaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaa ikdaaeaacaaIZaaaamaapedabaWaaOaaaeaacaaI5aGaeyOeI0Iaam iEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaamizaiaadIhaaSqaaiaa icdaaeaacaaIZaaaniabgUIiYdGccqGHsisldaWcaaqaaiaaikdaae aacaaIZaaaamaapedabaWaaeWaaeaacaaIZaGaeyOeI0IaamiEaaGa ayjkaiaawMcaaiaadsgacaWG4baaleaacaaIWaaabaGaaG4maaqdcq GHRiI8aaGcbaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqp daWcaaqaaiaaikdaaeaacaaIZaaaamaadmaabaWaaSaaaeaacaWG4b aabaGaaGOmaaaadaGcaaqaaiaaiMdacqGHsislcaWG4bWaaWbaaSqa beaacaaIYaaaaaqabaGccqGHRaWkdaWcaaqaaiaaiMdaaeaacaaIYa aaaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaa kmaabmaabaWaaSaaaeaacaWG4baabaGaaG4maaaaaiaawIcacaGLPa aaaiaawUfacaGLDbaadaqhaaWcbaGaaGimaaqaaiaaiodaaaGccqGH sisldaWcaaqaaiaaikdaaeaacaaIZaaaamaadmaabaGaaG4maiaadI hacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaa caaIYaaaaaGaay5waiaaw2faamaaDaaaleaacaaIWaaabaGaaG4maa aaaOqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaaSaa aeaacaaIYaaabaGaaG4maaaadaGadaqaamaalaaabaGaaG4maaqaai aaikdaaaWaaOaaaeaacaaI5aGaeyOeI0IaaG4mamaaCaaaleqabaGa aGOmaaaaaeqaaOGaey4kaSYaaSaaaeaacaaI5aaabaGaaGOmaaaaci GGZbGaaiyAaiaac6gadaahaaWcbeqaaiabgkHiTiaaigdaaaGcdaqa daqaamaalaaabaGaaG4maaqaaiaaiodaaaaacaGLOaGaayzkaaGaey OeI0YaaSaaaeaacaaIWaaabaGaaGOmaaaadaGcaaqaaiaaiMdacqGH sislcaaIWaWaaWbaaSqabeaacaaIYaaaaaqabaGccqGHsisldaWcaa qaaiaaiMdaaeaacaaIYaaaaiGacohacaGGPbGaaiOBamaaCaaaleqa baGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaacaaIWaaabaGaaG 4maaaaaiaawIcacaGLPaaaaiaawUhacaGL9baaaeaacaWLjaGaaCzc aiaaxMaacaWLjaGaeyOeI0YaaSaaaeaacaaIYaaabaGaaG4maaaada GadaqaaiaaiodadaqadaqaaiaaiodaaiaawIcacaGLPaaacqGHsisl daWcaaqaaiaaiodadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaai abgkHiTiaaiodadaqadaqaaiaaicdaaiaawIcacaGLPaaacqGHRaWk daWcaaqaaiaaicdadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaa Gaay5Eaiaaw2haaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8Ua eyypa0ZaaSaaaeaacaaIYaaabaGaaG4maaaadaGadaqaamaalaaaba GaaGyoaaqaaiaaikdaaaGaey41aq7aaSaaaeaaiiaacqWFapaCaeaa caaIYaaaaaGaay5Eaiaaw2haaiabgkHiTmaalaaabaGaaGOmaaqaai aaiodaaaWaaiWaaeaacaaI5aGaeyOeI0YaaSaaaeaacaaI5aaabaGa aGOmaaaaaiaawUhacaGL9baaaeaacaWLjaGaaCzcaiaaxMaacaaMc8 UaaGPaVlabg2da9maalaaabaGaaG4maiab=b8aWbqaaiaaikdaaaGa eyOeI0IaaG4maiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaWaae WaaeaacqWFapaCcqGHsislcaaIYaaacaGLOaGaayzkaaGaaGPaVlaa dohacaWGXbGaamyDaiaadggacaWGYbGaamyzaiaabccacaqG1bGaae OBaiaabMgacaqG0bGaae4Caaaaaa@3889@

Q.28

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. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaahAeacaWHPbGaaCOBaiaahsgacaqGGaGaaCiDaiaahIga caWHLbGaaeiiaiaahggacaWHYbGaaCyzaiaahggacaqGGaGaaC4Bai aahAgacaqGGaGaaCiDaiaahIgacaWHLbGaaeiiaiaahohacaWHTbGa aCyyaiaahYgacaWHSbGaaCyzaiaahkhacaqGGaGaaCOCaiaahwgaca WHNbGaaCyAaiaah+gacaWHUbGaaeiiaiaahkgacaWHVbGaaCyDaiaa h6gacaWHKbGaaCyzaiaahsgacaqGGaGaaCOyaiaahMhacaqGGaGaaC iDaiaahIgacaWHLbGaaeiiaiaahwgacaWHSbGaaCiBaiaahMgacaWH WbGaaC4CaiaahwgaaeaadaWcaaqaaGqabiaa=HhadaahaaWcbeqaai aa=jdaaaaakeaacaWFHbWaaWbaaSqabeaacaWFYaaaaaaakiaa=Tca daWcaaqaaiaa=LhadaahaaWcbeqaaiaa=jdaaaaakeaacaWFIbWaaW baaSqabeaacaWFYaaaaaaakiaa=1dacaWFXaGaaGzaVlaa=bcacaWF GaGaa8xyaiaa=5gacaWFKbGaa8hiaiaa=rhacaWFObGaa8xzaiaa=b cacaWFSbGaa8xAaiaa=5gacaWFLbGaa8hiamaalaaabaGaa8hEaaqa aiaa=fgaaaGaa83kamaalaaabaGaa8xEaaqaaiaa=jgaaaGaa8xpai aa=fdacaWFUaaaaaa@8AC3@

Ans.

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

Area( ABCA )=Area( OBCAO )Area( OBAO ) = 0 a b 1 x 2 a 2 dx 0 a b( 1 x a )dx = b a 0 a a 2 x 2 dx b a 0 a ( ax )dx = b a [ x 2 a 2 x 2 + a 2 2 sin 1 ( x a ) ] 0 a b a [ ax x 2 2 ] 0 a = b a { a 2 a 2 a 2 + a 2 2 sin 1 ( a a ) 0 2 a 2 0 2 a 2 2 sin 1 ( 0 a ) } b a { a( a ) a 2 2 3( 0 )+ 0 2 2 } = b a { a 2 2 × π 2 } b a { a 2 a 2 2 } = b a ( π a 2 4 a 2 2 ) = b a 2 2a ( π 2 1 ) = ab 4 ( π2 )square units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiabgsJiCjacyc4GbbGaiGjGdkhacGaMaoyzaiacyc4GHbWa iGjGbmaabGaMakacyc4GbbGaiGjGdkeacGaMao4qaiacyc4GbbaacG aMaAjkaiacycOLPaaacqGH9aqpcaWGbbGaamOCaiaadwgacaWGHbWa aeWaaeaacaWGpbGaamOqaiaadoeacaWGbbGaam4taaGaayjkaiaawM caaiabgkHiTiaadgeacaWGYbGaamyzaiaadggadaqadaqaaiaad+ea caWGcbGaamyqaiaad+eaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcai aaxMaacaaMc8UaaGPaVlabg2da9maapedabaGaamOyamaakaaabaGa aGymaiabgkHiTmaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaO qaaiaadggadaahaaWcbeqaaiaaikdaaaaaaaqabaGccaWGKbGaamiE aaWcbaGaaGimaaqaaiaadggaa0Gaey4kIipakiabgkHiTmaapedaba GaamOyamaabmaabaGaaGymaiabgkHiTmaalaaabaGaamiEaaqaaiaa dggaaaaacaGLOaGaayzkaaGaamizaiaadIhaaSqaaiaaicdaaeaaca WGHbaaniabgUIiYdaakeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPa Vlabg2da9maalaaabaGaamOyaaqaaiaadggaaaWaa8qmaeaadaGcaa qaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWG4bWaaWba aSqabeaacaaIYaaaaaqabaGccaWGKbGaamiEaaWcbaGaaGimaaqaai aadggaa0Gaey4kIipakiabgkHiTmaalaaabaGaamOyaaqaaiaadgga aaWaa8qmaeaadaqadaqaaiaadggacqGHsislcaWG4baacaGLOaGaay zkaaGaamizaiaadIhaaSqaaiaaicdaaeaacaWGHbaaniabgUIiYdaa keaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlabg2da9maalaaaba GaamOyaaqaaiaadggaaaWaamWaaeaadaWcaaqaaiaadIhaaeaacaaI YaaaamaakaaabaGaamyyamaaCaaaleqabaGaaGOmaaaakiabgkHiTi aadIhadaahaaWcbeqaaiaaikdaaaaabeaakiabgUcaRmaalaaabaGa amyyamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaci4CaiaacM gacaGGUbWaaWbaaSqabeaacqGHsislcaaIXaaaaOWaaeWaaeaadaWc aaqaaiaadIhaaeaacaWGHbaaaaGaayjkaiaawMcaaaGaay5waiaaw2 faamaaDaaaleaacaaIWaaabaGaamyyaaaakiabgkHiTmaalaaabaGa amOyaaqaaiaadggaaaWaamWaaeaacaWGHbGaamiEaiabgkHiTmaala aabaGaamiEamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaaacaGL BbGaayzxaaWaa0baaSqaaiaaicdaaeaacaWGHbaaaaGcbaGaaCzcai aaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaadkgaaeaa caWGHbaaamaacmaabaWaaSaaaeaacaWGHbaabaGaaGOmaaaadaGcaa qaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWGHbWaaWba aSqabeaacaaIYaaaaaqabaGccqGHRaWkdaWcaaqaaiaadggadaahaa WcbeqaaiaaikdaaaaakeaacaaIYaaaaiGacohacaGGPbGaaiOBamaa CaaaleqabaGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaacaWGHb aabaGaamyyaaaaaiaawIcacaGLPaaacqGHsisldaWcaaqaaiaaicda aeaacaaIYaaaamaakaaabaGaamyyamaaCaaaleqabaGaaGOmaaaaki abgkHiTiaaicdadaahaaWcbeqaaiaaikdaaaaabeaakiabgkHiTmaa laaabaGaamyyamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaci 4CaiaacMgacaGGUbWaaWbaaSqabeaacqGHsislcaaIXaaaaOWaaeWa aeaadaWcaaqaaiaaicdaaeaacaWGHbaaaaGaayjkaiaawMcaaaGaay 5Eaiaaw2haaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacqGHsisldaWc aaqaaiaadkgaaeaacaWGHbaaamaacmaabaGaamyyamaabmaabaGaam yyaaGaayjkaiaawMcaaiabgkHiTmaalaaabaGaamyyamaaCaaaleqa baGaaGOmaaaaaOqaaiaaikdaaaGaeyOeI0IaaG4mamaabmaabaGaaG imaaGaayjkaiaawMcaaiabgUcaRmaalaaabaGaaGimamaaCaaaleqa baGaaGOmaaaaaOqaaiaaikdaaaaacaGL7bGaayzFaaaabaGaaCzcai aaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaadkgaaeaa caWGHbaaamaacmaabaWaaSaaaeaacaWGHbWaaWbaaSqabeaacaaIYa aaaaGcbaGaaGOmaaaacqGHxdaTdaWcaaqaaGGaaiab=b8aWbqaaiaa ikdaaaaacaGL7bGaayzFaaGaeyOeI0YaaSaaaeaacaWGIbaabaGaam yyaaaadaGadaqaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHsisl daWcaaqaaiaadggadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaa Gaay5Eaiaaw2haaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8Ua eyypa0ZaaSaaaeaacaWGIbaabaGaamyyaaaadaqadaqaamaalaaaba Gae8hWdaNaamyyamaaCaaaleqabaGaaGOmaaaaaOqaaiaaisdaaaGa eyOeI0YaaSaaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaG OmaaaaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcaiaaxMaacaaMc8Ua aGPaVlabg2da9maalaaabaGaamOyaiaadggadaahaaWcbeqaaiaaik daaaaakeaacaaIYaGaamyyaaaadaqadaqaamaalaaabaGae8hWdaha baGaaGOmaaaacqGHsislcaaIXaaacaGLOaGaayzkaaaabaGaaCzcai aaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaadggacaWG IbaabaGaaGinaaaadaqadaqaaiab=b8aWjabgkHiTiaaikdaaiaawI cacaGLPaaacaaMc8Uaam4CaiaadghacaWG1bGaamyyaiaadkhacaWG LbGaaeiiaiaabwhacaqGUbGaaeyAaiaabshacaqGZbaaaaa@66E3@

Q.29 Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2 and x-axis.

Ans.

The area of the region enclosed by the hyperbola and line is represented by shaded region OABCO.
The point of intersection of the parabola, x2 = y and the line, y = x + 2, is A(–1, –1).

Area( OCAO )= 1 2 ( x+2 )dx 1 2 x 2 dx = [ x 2 2 +2x ] 1 2 [ x 3 3 ] 1 2 =[ { ( 2 ) 2 2 +2( 2 ) }{ ( 1 ) 2 2 +2( 1 ) } ]{ 2 3 3 ( 1 ) 3 3 } =( 2+4 1 2 +2 )( 8 3 + 1 3 ) = 15 2 3 = 9 2 square units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiabgsJiCjaadgeacaWGYbGaamyzaiaadggadaqadaqaaiaa d+eacaWGdbGaamyqaiaad+eaaiaawIcacaGLPaaacqGH9aqpdaWdXa qaamaabmaabaGaamiEaiabgUcaRiaaikdaaiaawIcacaGLPaaacaWG KbGaamiEaaWcbaGaeyOeI0IaaGymaaqaaiaaikdaa0Gaey4kIipaki abgkHiTmaapedabaGaamiEamaaCaaaleqabaGaaGOmaaaakiaadsga caWG4baaleaacqGHsislcaaIXaaabaGaaGOmaaqdcqGHRiI8aaGcba GaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 cqGH9aqpdaWadaqaamaalaaabaGaamiEamaaCaaaleqabaGaaGOmaa aaaOqaaiaaikdaaaGaey4kaSIaaGOmaiaadIhaaiaawUfacaGLDbaa daqhaaWcbaGaeyOeI0IaaGymaaqaaiaaikdaaaGccqGHsisldaWada qaamaalaaabaGaamiEamaaCaaaleqabaGaaG4maaaaaOqaaiaaioda aaaacaGLBbGaayzxaaWaa0baaSqaaiabgkHiTiaaigdaaeaacaaIYa aaaaGcbaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7cqGH9aqpdaWadaqaamaacmqabaWaaSaaaeaadaqadaqaai aaikdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakeaacaaI YaaaaiabgUcaRiaaikdadaqadaqaaiaaikdaaiaawIcacaGLPaaaai aawUhacaGL9baacqGHsisldaGadeqaamaalaaabaWaaeWaaeaacqGH sislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGcba GaaGOmaaaacqGHRaWkcaaIYaWaaeWaaeaacqGHsislcaaIXaaacaGL OaGaayzkaaaacaGL7bGaayzFaaaacaGLBbGaayzxaaGaeyOeI0Yaai WaaeaadaWcaaqaaiaaikdadaahaaWcbeqaaiaaiodaaaaakeaacaaI ZaaaaiabgkHiTmaalaaabaWaaeWaaeaacqGHsislcaaIXaaacaGLOa GaayzkaaWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maaaaaiaawUha caGL9baaaeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlabg2da9maabmaabaGaaGOmaiabgUcaRiaaisdacqGH sisldaWcaaqaaiaaigdaaeaacaaIYaaaaiabgUcaRiaaikdaaiaawI cacaGLPaaacqGHsisldaqadaqaamaalaaabaGaaGioaaqaaiaaioda aaGaey4kaSYaaSaaaeaacaaIXaaabaGaaG4maaaaaiaawIcacaGLPa aaaeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlabg2da9maalaaabaGaaGymaiaaiwdaaeaacaaIYaaaaiabgk HiTiaaiodaaeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlabg2da9maalaaabaGaaGyoaaqaaiaaikdaaaGaaG PaVlaadohacaWGXbGaamyDaiaadggacaWGYbGaamyzaiaabccacaqG 1bGaaeOBaiaabMgacaqG0bGaae4Caaaaaa@E388@

Q.30 Using the method of integration find the area bounded by the curve |x| + |y| = 1.

Ans.

The area bounded by the curve |x| + |y| = 1 can be obtained by tracing curves of the lines x + y = 1,
x – y =1, –x – y= 1 and –x + y = 1. The shaded portion ABCD is showing the required area in the graph.

The curves are intersecting the axes at the points
A (0, 1), B (1, 0), C (0, –1) and D (–1, 0).
The given curve is symmetrical about x-axis and y-axis.

So, Area ABCD = 4( Area AOB ) =4 0 1 ( 1x )dx =4 [ x x 2 2 ] 0 1 =4( 1 1 2 2 0+ 0 2 2 ) =4( 1 2 )=2square units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabofacaqGVbGaaiilaiaabccacaqGbbGaaeOCaiaabwga caqGHbGaaeiiaiaabgeacaqGcbGaae4qaiaabseacaqGGaGaeyypa0 JaaeiiaiaabsdadaqadaqaaiaabgeacaqGYbGaaeyzaiaabggacaqG GaGaaeyqaiaab+eacaqGcbaacaGLOaGaayzkaaaabaGaaCzcaiaaxM aacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyyp a0JaaGinamaapedabaWaaeWaaeaacaaIXaGaeyOeI0IaamiEaaGaay jkaiaawMcaaiaadsgacaWG4baaleaacaaIWaaabaGaaGymaaqdcqGH RiI8aaGcbaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8Uaeyypa0JaaGinamaadmaabaGaamiEaiabgkHi TmaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaa aacaGLBbGaayzxaaWaa0baaSqaaiaaicdaaeaacaaIXaaaaaGcbaGa aCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8Uaeyypa0JaaGinamaabmaabaGaaGymaiabgkHiTmaalaaabaGa aGymamaaCaaaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaeyOeI0IaaG imaiabgUcaRmaalaaabaGaaGimamaaCaaaleqabaGaaGOmaaaaaOqa aiaaikdaaaaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaWLjaGaaG PaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGinamaa bmaabaWaaSaaaeaacaaIXaaabaGaaGOmaaaaaiaawIcacaGLPaaacq GH9aqpcaaIYaGaaGPaVlaadohacaWGXbGaamyDaiaadggacaWGYbGa amyzaiaabccacaqG1bGaaeOBaiaabMgacaqG0bGaae4Caaaaaa@B000@

Q.31

Find the area bounded by curves{ ( x,y ):y x 2 and y=| x | }. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aacaWHgbGaaCyAaiaah6gacaWHKbGaaeiiaiaahshacaWHObGaaCyz aiaabccacaWHHbGaaCOCaiaahwgacaWHHbGaaeiiaiaahkgacaWHVb GaaCyDaiaah6gacaWHKbGaaCyzaiaahsgacaqGGaGaaCOyaiaahMha caqGGaGaaC4yaiaahwhacaWHYbGaaCODaiaahwgacaWHZbGaaGPaVp aacmaabaWaaeWaaeaacaWH4bGaaiilaiaahMhaaiaawIcacaGLPaaa caGG6aGaaCyEaiabgwMiZkaahIhadaahaaWcbeqaaiaaikdaaaGcca qGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaahMhacqGH9aqpdaabdaqa aiaahIhaaiaawEa7caGLiWoaaiaawUhacaGL9baacaGGUaaaaa@6CD9@

Ans.

null

The required area is symmetrical about y-axis.

So,required area=2{ Area( OCAO )Area( OCADO ) } =2[ 0 1 xdx 0 1 x 2 dx ] =2 [ x 2 2 ] 0 1 2 [ x 3 3 ] 0 1 =2[ 1 2 2 0 ]2[ 1 3 3 0 ] =2( 1 2 1 3 ) =2( 1 6 )= 1 3 square units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadofacaWGVbGaaiilaiaaykW7caaMc8UaaeOCaiaabwga caqGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaaeyyai aabkhacaqGLbGaaeyyaiabg2da9iaaikdadaGadaqaaiaadgeacaWG YbGaamyzaiaadggacaaMc8+aaeWaaeaacaWGpbGaam4qaiaadgeaca WGpbaacaGLOaGaayzkaaGaeyOeI0IaamyqaiaadkhacaWGLbGaamyy aiaaykW7daqadaqaaiaad+eacaWGdbGaamyqaiaadseacaWGpbaaca GLOaGaayzkaaaacaGL7bGaayzFaaaabaGaaCzcaiaaxMaacaWLjaGa aCzcaiabg2da9iaaikdadaWadaqaamacyc4dXaqaiGjGcGaMaoiEai acyc4GKbGaiGjGdIhaaSqaiGjGcGaMaIimaaqaiGjGcGaMaIymaaqd cWaMaA4kIipakiabgkHiTmaapedabaGaamiEamaaCaaaleqabaGaiG jGikdaaaGccaWGKbGaamiEaaWcbaGaaGimaaqaaiaaigdaa0Gaey4k IipaaOGaay5waiaaw2faaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacq GH9aqpcaaIYaWaamWaaeaadaWcaaqaaiaadIhadaahaaWcbeqaaiaa ikdaaaaakeaacaaIYaaaaaGaay5waiaaw2faamaaDaaaleaacaaIWa aabaGaaGymaaaakiabgkHiTiaaikdadaWadaqaamaalaaabaGaamiE amaaCaaaleqabaGaaG4maaaaaOqaaiaaiodaaaaacaGLBbGaayzxaa Waa0baaSqaaiaaicdaaeaacaaIXaaaaaGcbaGaaCzcaiaaxMaacaWL jaGaaCzcaiabg2da9iaaikdadaWadaqaamaalaaabaGaaGymamaaCa aaleqabaGaaGOmaaaaaOqaaiaaikdaaaGaeyOeI0IaaGimaaGaay5w aiaaw2faaiabgkHiTiaaikdadaWadaqaamaalaaabaGaaGymamaaCa aaleqabaGaaG4maaaaaOqaaiaaiodaaaGaeyOeI0IaaGimaaGaay5w aiaaw2faaaqaaiaaxMaacaWLjaGaaCzcaiaaxMaacqGH9aqpcaaIYa WaaeWaaeaadaWcaaqaaiaaigdaaeaacaaIYaaaaiabgkHiTmaalaaa baGaaGymaaqaaiaaiodaaaaacaGLOaGaayzkaaaabaGaaCzcaiaaxM aacaWLjaGaaCzcaiabg2da9iaaikdadaqadaqaamaalaaabaGaaGym aaqaaiaaiAdaaaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIXa aabaGaaG4maaaacaqGGaGaae4CaiaabghacaqG1bGaaeyyaiaabkha caqGLbGaaeiiaiaabwhacaqGUbGaaeyAaiaabshacaqGZbaaaaa@C931@

Q.32 Using 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).

Ans.

The vertices of triangle ABC are A(2, 0), B(4, 5) and C(6, 3).

Equation of line AB is y0= 50 42 ( x2 ) y= 5 2 ( x2 ) Equation of line BC is y5= 35 64 ( x4 ) y= 2 2 ( x4 )+5=x+9 Equation of line CA is y3= 03 26 ( x6 ) y= 3 4 ( x6 )+3 = 3 4 ( x2 ) Area( ΔABC )=Area( ΔABLA )+Area( ΔBLMCB )Area( ΔACMA ) = 2 4 5 2 ( x2 )dx + 4 6 ( x+9 )dx 2 6 3 4 ( x2 )dx = 5 2 [ x 2 2 2x ] 2 4 + [ x 2 2 +9x ] 4 6 3 4 [ x 2 2 2x ] 2 6 = 5 2 [ 4 2 2 2( 4 ) 2 2 2 +2( 2 ) ]+[ 6 2 2 +9( 6 )+ 4 2 2 9( 4 ) ] 3 4 [ 6 2 2 2( 6 ) 2 2 2 +2( 2 ) ] = 5 2 ( 882+4 )+( 18+54+836 ) 3 4 ( 18122+4 ) = 5 2 ( 2 )+8 3 4 ( 8 ) =5+86 =7 squre units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadweacaWGXbGaamyDaiaadggacaWG0bGaamyAaiaad+ga caWGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaabYgacaqGPbGaaeOBai 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Q.33 Using method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Ans.

The equations of given lines are:
2x + y = 4 … (i)
3x – 2y= 6 … (ii)
x – 3y +5 =0 … (iii)
The graphical representation of these lines is given in below. Here AL and CM are perpendicular to x-axis.

Area(ΔABC)=Area(ALMCA)Area(ALB)Area(CMB)      =14x+53dx12(42x)dx243x62dx      =13[x22+5x]14[4xx2]1212[3x226x]24      =13[422+5(4)1225(1)][4(2)224(1)+12]12[34226(4)3222+6(2)]      =13(8+20125)(844+1)12(24246+12)      =13(452)13      =1524      =72 square units

Q.34

Find the area of the region{ ( x,y ): y 2 4x, 4 x 2 +4 y 2 9 }. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aacaWHgbGaaCyAaiaah6gacaWHKbGaaeiiaiaahshacaWHObGaaCyz aiaabccacaWHHbGaaCOCaiaahwgacaWHHbGaaeiiaiaah+gacaWHMb GaaeiiaiaahshacaWHObGaaCyzaiaabccacaWHYbGaaCyzaiaahEga caWHPbGaaC4Baiaah6gacaaMc8UaaGPaVpaacmaabaWaaeWaaeaaie qacaWF4bGaa8hlaiaa=LhaaiaawIcacaGLPaaacaWF6aGaa8xEamaa CaaaleqabaGaiGjG=jdaaaGccqGHLjYScaWF0aGaa8hEaiaa=Xcaca WFGaGaa8hnaiaa=HhadaahaaWcbeqaaiaa=jdaaaGccaWFRaGaa8hn aiaa=LhadaahaaWcbeqaaiaa=jdaaaGccqGHKjYOcaWF5aaacaGL7b GaayzFaaGaaiOlaaaa@6BA3@

Ans.

The intersection points of both the curves are( 1 2 , 2 ) and( 1 2 , 2 ). The required area is given by OABCO. OABCO is symmetrical about x-axis. AreaOABCO=2×AreaOAB AreaOABO=AreaOMAO+AreaAMBA = 0 1 2 2 x + 1 2 3 2 1 2 94 x 2 dx =2 [ x 3 2 3 2 ] 0 1 2 + 1 4 [ 2x 2 94 x 2 + 9 2 sin 1 ( 2x 3 ) ] 1 2 3 2 = 4 3 [ ( 1 2 ) 3 2 0 ]+ 1 4 [ 2( 3 2 ) 2 94 ( 3 2 ) 2 + 9 2 sin 1 ( 2× 3 2 3 ) 2× 1 2 2 94 ( 1 2 ) 2 9 2 sin 1 ( 2× 1 2 3 ) ] = 4 3 × 1 2 2 + 1 4 { 0+ 9 2 × π 2 1 2 ×2 2 9 2 × sin 1 ( 1 3 ) } = 2 3 + 9π 16 2 4 9 8 sin 1 ( 1 3 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGPbGaaeOBaiaabsha caqGLbGaaeOCaiaabohacaqGLbGaae4yaiaabshacaqGPbGaae4Bai aab6gacaqGGaGaaeiCaiaab+gacaqGPbGaaeOBaiaabshacaqGZbGa aeiiaiaab+gacaqGMbGaaeiiaiaabkgacaqGVbGaaeiDaiaabIgaca qGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqG1bGaaeOCaiaa 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= 9π 16 9 8 sin 1 ( 1 3 )+ 2 12 = 9π 16 9 8 sin 1 ( 1 3 )+ 2 12 Thus, the required area of shaded region is 2×( 9π 16 9 8 sin 1 ( 1 3 )+ 2 12 ) square units i.e., ( 9π 8 9 4 sin 1 ( 1 3 )+ 2 6 ) square units. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 cqGH9aqpdaWcaaqaaiaaiMdaiiaacqWFapaCaeaacaaIXaGaaGOnaa aacqGHsisldaWcaaqaaiaaiMdaaeaacaaI4aaaaiGacohacaGGPbGa aiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaae aacaaIXaaabaGaaG4maaaaaiaawIcacaGLPaaacqGHRaWkdaWcaaqa amaakaaabaGaaGOmaaWcbeaaaOqaaiaaigdacaaIYaaaaaqaaiaaxM aacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpdaWc aaqaaiaaiMdacqWFapaCaeaacaaIXaGaaGOnaaaacqGHsisldaWcaa qaaiaaiMdaaeaacaaI4aaaaiGacohacaGGPbGaaiOBamaaCaaaleqa baGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaacaaIXaaabaGaaG 4maaaaaiaawIcacaGLPaaacqGHRaWkdaWcaaqaamaakaaabaGaaGOm aaWcbeaaaOqaaiaaigdacaaIYaaaaaqaaiaadsfacaWGObGaamyDai aadohacaGGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGYbGa aeyzaiaabghacaqG1bGaaeyAaiaabkhacaqGLbGaaeizaiaabccaca qGHbGaaeOCaiaabwgacaqGHbGaaeiiaiaab+gacaqGMbGaaeiiaiaa bohacaqGObGaaeyyaiaabsgacaqGLbGaaeizaiaabccacaqGYbGaae yzaiaabEgacaqGPbGaae4Baiaab6gacaqGGaGaaeyAaiaabohacaqG GaaabaGaaGOmaiabgEna0oaabmaabaWaaSaaaeaacaaI5aGae8hWda habaGaaGymaiaaiAdaaaGaeyOeI0YaaSaaaeaacaaI5aaabaGaaGio aaaaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiabgkHiTiaaigdaaa GcdaqadaqaamaalaaabaGaaGymaaqaaiaaiodaaaaacaGLOaGaayzk aaGaey4kaSYaaSaaaeaadaGcaaqaaiaaikdaaSqabaaakeaacaaIXa GaaGOmaaaaaiaawIcacaGLPaaacaaMc8Uaae4CaiaabghacaqG1bGa aeyyaiaabkhacaqGLbGaaeiiaiaabwhacaqGUbGaaeyAaiaabshaca qGZbaabaGaaeiiaiaabMgacaqGUaGaaeyzaiaab6cacaqGSaGaaeii amaabmaabaWaaSaaaeaacaaI5aGae8hWdahabaGaaGioaaaacqGHsi sldaWcaaqaaiaaiMdaaeaacaaI0aaaaiGacohacaGGPbGaaiOBamaa CaaaleqabaGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaacaaIXa aabaGaaG4maaaaaiaawIcacaGLPaaacqGHRaWkdaWcaaqaamaakaaa baGaaGOmaaWcbeaaaOqaaiaaiAdaaaaacaGLOaGaayzkaaGaae4Cai aabghacaqG1bGaaeyyaiaabkhacaqGLbGaaeiiaiaabwhacaqGUbGa aeyAaiaabshacaqGZbGaaeOlaaaaaa@DB5C@

Q.35 Area bounded by the curve y = x3, the x-axis and the ordinates x= –2 and x = 1 is

(A) – 9 (B) – 15/4 (C) 15/4 (D) 17/4

Ans.

The required area is shaded in the figure.

Required area=| 2 0 x 3 dx |+ 0 1 x 3 dx =| [ x 4 4 ] 2 0 |+ [ x 4 4 ] 0 1 =| 1 4 ( 016 ) |+ 1 4 ( 10 ) = 16 4 + 1 4 = 17 4 Thus, correct option is D. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiGackfacaGGLbGaamyCaiaadwhacaWGPbGaamOCaiaadwga caWGKbGaaeiiaiaabggacaqGYbGaaeyzaiaabggacqGH9aqpdaabda qaamaapedabaGaamiEamaaCaaaleqabaGaaG4maaaakiaadsgacaWG 4baaleaacqGHsislcaaIYaaabaGaaGimaaqdcqGHRiI8aaGccaGLhW UaayjcSdGaey4kaSYaa8qmaeaacaWG4bWaaWbaaSqabeaacaaIZaaa aOGaamizaiaadIhaaSqaaiaaicdaaeaacaaIXaaaniabgUIiYdaake aacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlabg2da9maaemaabaWa amWaaeaadaWcaaqaaiaadIhadaahaaWcbeqaaiaaisdaaaaakeaaca aI0aaaaaGaay5waiaaw2faamaaDaaaleaacqGHsislcaaIYaaabaGa aGimaaaaaOGaay5bSlaawIa7aiabgUcaRmaadmaabaWaaSaaaeaaca WG4bWaaWbaaSqabeaacaaI0aaaaaGcbaGaaGinaaaaaiaawUfacaGL DbaadaqhaaWcbaGaaGimaaqaaiaaigdaaaaakeaacaWLjaGaaCzcai aaxMaacaaMc8UaaGPaVlabg2da9maaemaabaWaaSaaaeaacaaIXaaa baGaaGinaaaadaqadaqaaiaaicdacqGHsislcaaIXaGaaGOnaaGaay jkaiaawMcaaaGaay5bSlaawIa7aiabgUcaRmaalaaabaGaaGymaaqa aiaaisdaaaWaaeWaaeaacaaIXaGaeyOeI0IaaGimaaGaayjkaiaawM caaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaaSaa aeaacaaIXaGaaGOnaaqaaiaaisdaaaGaey4kaSYaaSaaaeaacaaIXa aabaGaaGinaaaaaeaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqa aiaaigdacaaI3aaabaGaaGinaaaaaeaacaWGubGaamiAaiaadwhaca WGZbGaaiilaiaabccacaqGJbGaae4BaiaabkhacaqGYbGaaeyzaiaa bogacaqG0bGaaeiiaiaab+gacaqGWbGaaeiDaiaabMgacaqGVbGaae OBaiaabccacaqGPbGaae4CaiaabccacaqGebGaaeOlaaaaaa@AD3C@

Q.36 The area bounded by the curve y = x |x|, x-axis and the ordinates x = –1 and x = 1 is given by

(A) 0 (B) 1/3 (C) 2/3 (D) 4/3

Ans.

We have, y=x| x |={ x 2 ,x<0 x 2 ,x0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadEfacaWGLbGaaeiiaiaabIgacaqGHbGaaeODaiaabwga caqGSaaabaGaamyEaiabg2da9iaadIhadaabdaqaaiaadIhaaiaawE a7caGLiWoacqGH9aqpdaGabaabaeqabaGaeyOeI0IaamiEamaaCaaa leqabaGaaGOmaaaakiaacYcacaaMc8UaaGPaVlaaykW7caWG4bGaey ipaWJaaGimaaqaaiaaykW7caaMc8UaamiEamaaCaaaleqabaGaaGOm aaaakiaacYcacaaMc8UaaGPaVlaaykW7caWG4bGaeyyzImRaaGimaa aacaGL7baaaaaa@6193@

Intersection points of curve x| x | and lines x=1 and x=1 are ( 1,1 ) and ( 1,1 ) respectively. Required area = 1 1 x| x | dx =| 1 0 x 2 dx |+ 0 1 x 2 dx =| [ x 3 3 ] 1 0 |+ [ x 3 3 ] 0 1 =| ( 0+ 1 3 ) |+( 1 3 0 ) = 1 3 + 1 3 = 2 3 square units Thus, the correct option is C. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMeacaqGUbGaaeiDaiaabwgacaqGYbGaae4Caiaabwga caqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGWbGaae4Bai aabMgacaqGUbGaaeiDaiaabohacaqGGaGaae4BaiaabAgacaqGGaGa ae4yaiaabwhacaqGYbGaaeODaiaabwgacaqGGaGaaeiEamaaemaaba GaamiEaaGaay5bSlaawIa7aiaabccacaqGHbGaaeOBaiaabsgacaqG GaGaaeiBaiaabMgacaqGUbGaaeyzaiaabohacaqGGaGaaeiEaiabg2 da9iaaigdacaqGGaGaamyyaiaad6gacaWGKbGaaeiiaiaadIhacqGH 9aqpcqGHsislcaqGXaGaaeiiaiaabggacaqGYbGaaeyzaaqaamaabm aabaGaaGymaiaacYcacaaMc8UaaGPaVlaaigdaaiaawIcacaGLPaaa caqGGaGaaeyyaiaab6gacaqGKbGaaeiiamaabmaabaGaeyOeI0IaaG ymaiaacYcacqGHsislcaaIXaaacaGLOaGaayzkaaGaaeiiaiaabkha caqGLbGaae4CaiaabchacaqGLbGaae4yaiaabshacaqGPbGaaeODai aabwgacaqGSbGaaeyEaiaab6caaeaacqGH0icxcaqGsbGaaeyzaiaa bghacaqG1bGaaeyAaiaabkhacaqGLbGaaeizaiaabccacaqGHbGaae OCaiaabwgacaqGHbGaaeiiaiabg2da9maapedabaGaamiEamaaemaa baGaamiEaaGaay5bSlaawIa7aaWcbaGaeyOeI0IaaGymaaqaaiaaig daa0Gaey4kIipakiaadsgacaWG4baabaGaaCzcaiaaxMaacaWLjaGa aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cq GH9aqpdaabdaqaamaapedabaGaeyOeI0IaamiEamaaCaaaleqabaGa aGOmaaaakiaadsgacaWG4baaleaacqGHsislcaaIXaaabaGaaGimaa qdcqGHRiI8aaGccaGLhWUaayjcSdGaey4kaSYaa8qmaeaacaWG4bWa aWbaaSqabeaacaaIYaaaaOGaamizaiaadIhaaSqaaiaaicdaaeaaca aIXaaaniabgUIiYdaakeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9maaem aabaWaamWaaeaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaa iodaaaaakeaacaaIZaaaaaGaay5waiaaw2faamaaDaaaleaacqGHsi slcaaIXaaabaGaaGimaaaaaOGaay5bSlaawIa7aiabgUcaRmaadmaa baWaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maa aaaiaawUfacaGLDbaadaqhaaWcbaGaaGimaaqaaiaaigdaaaaakeaa caWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlabg2da9maaemaabaWaaeWaaeaacaaIWaGa ey4kaSYaaSaaaeaacaaIXaaabaGaaG4maaaaaiaawIcacaGLPaaaai aawEa7caGLiWoacqGHRaWkdaqadaqaamaalaaabaGaaGymaaqaaiaa iodaaaGaeyOeI0IaaGimaaGaayjkaiaawMcaaaqaaiaaxMaacaWLja GaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8Uaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacqGHRaWkda WcaaqaaiaaigdaaeaacaaIZaaaaiaaykW7cqGH9aqpdaWcaaqaaiaa ikdaaeaacaaIZaaaaiaabccacaqGZbGaaeyCaiaabwhacaqGHbGaae OCaiaabwgacaqGGaGaaeyDaiaab6gacaqGPbGaaeiDaiaabohaaeaa caqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaae4yaiaab+gacaqGYbGaaeOCaiaabwgacaqGJbGa aeiDaiaabccacaqGVbGaaeiCaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqGGaGaae4qaiaab6caaaaa@4571@

Q.37

The area of the circle x2+y2=16 exterior to  the parabola y2= 6x is(A)43(4π3) (B)43(4π +3)(C)  43(8π3) (D)43(8π +3)

Ans.

The intersection points of parabola and circle are A(2,23)andB(2,23).Area of circle which is exterior to parabola is shownby shaded area.

UnshadedArea bounded by circle and parabola =2( 0 2 6x dx + 2 4 16 x 2 dx ) =2( 6 [ x 3 2 3 2 ] 0 2 + [ x 2 16 x 2 + 16 2 sin 1 ( x 4 ) ] 2 4 ) =2( 2 6 3 [ 2 3 2 0 ]+ [ 0+8 sin 1 ( 1 ) 12 8 sin 1 ( 1 2 ) ] 2 4 ) =2( 8 3 3 +8× π 2 2 3 8× π 6 ) =2( 2 3 3 + 8π 3 )= 4 3 3 + 16π 3 Area of shaded region=Area of circle( 4 3 3 + 16π 3 ) =π ( 4 ) 2 ( 4 3 3 + 16π 3 ) =16π 16π 3 4 3 3 = 32π 3 4 3 3 = 4 3 ( 8π 3 ) square units Thus, correct option is C. 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Q.38

The area bounded by the y-axis, y = cosx and y = sinx when 0x π 2 is ( A ) 2( 2 1 ) ( B ) 2 1 ( C ) 2 +1 ( D ) 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGHbGaaeOCaiaabwga caqGHbGaaeiiaiaabkgacaqGVbGaaeyDaiaab6gacaqGKbGaaeyzai aabsgacaqGGaGaaeOyaiaabMhacaqGGaGaaeiDaiaabIgacaqGLbGa aeiiaiaabMhacaqGTaGaaeyyaiaabIhacaqGPbGaae4CaiaabYcaca qGGaGaaeyEaiaab2dacaqGJbGaae4BaiaabohacaqG4bGaaeiiaiaa bggacaqGUbGaaeizaiaabccacaqG5bGaaeypaiaabccacaqGZbGaae yAaiaab6gacaqG4bGaaeiiaaqaaiaabEhacaqGObGaaeyzaiaab6ga caqGGaGaaeimaiabgsMiJkaabIhacqGHKjYOdaWcaaqaaGGaaiab=b 8aWbqaaiaabkdaaaGaaeiiaiaabMgacaqGZbaabaGaaeiOamaabmaa baGaaeyqaaGaayjkaiaawMcaaGqabiaa+bcacaqGYaWaaeWaaeaada GcaaqaaiaabkdaaSqabaGccqGHsislcaqGXaaacaGLOaGaayzkaaGa a4hOaiaa+bkacaGFGcGaa4hOaiaaxMaadaqadaqaaiaabkeaaiaawI cacaGLPaaacaqGGaWaaOaaaeaacaqGYaaaleqaaOGaeyOeI0Iaaeym aiaa+bkacaGFGcGaaCzcamaabmaabaGaae4qaaGaayjkaiaawMcaai aa+bcadaGcaaqaaiaabkdaaSqabaGccaqGRaGaaeymaiaabckacaGF GcGaa4hOaiaaxMaadaqadaqaaiaabseaaiaawIcacaGLPaaacaGFGa WaaOaaaeaacaqGYaaaleqaaaaaaa@9635@

Ans.

Given curves are y = sinx and y = cosx Intersection point of both given curves is B( π 4 , 1 2 ). Area of shaded region=Area( AMBA )+Area( OMBO ) = 0 1 2 xdy + 1 2 1 xdy = 0 1 2 sin 1 ydy + 1 2 1 cos 1 ydy = 0 1 2 sin 1 y.1dy + 1 2 1 cos 1 y.1dy Integrating by parts, we get = [ y sin 1 y+ 1 y 2 ] 0 1 2 + [ y cos 1 y 1 y 2 ] 1 2 1 =[ 1 2 sin 1 ( 1 2 )+ 1 ( 1 2 ) 2 01 ] +[ 1 cos 1 ( 1 ) 1 ( 1 ) 2 1 2 cos 1 1 2 + 1 ( 1 2 ) 2 ] =[ 1 2 π 4 + 1 2 1 ]+[ 00 1 2 π 4 + 1 2 ] = 1 2 π 4 + 1 2 1 1 2 π 4 + 1 2 = 2 2 1= 2 1 Thus, the correct option is B. 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FAQs (Frequently Asked Questions)
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Questions that get asked in a board test are highly unpredictable. Math professionals have carefully selected all of the sums in the NCERT Solutions Class 12 Mathematics Chapter 8 – Application of Integrals to provide students with a complete picture of the subject. As a result, if students wish to increase their chances of achieving a high grade on their exams, they must thoroughly practise all problems. They should also look at prior test papers when preparing for the boards.

2. How many problems are there in NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals?

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