{"id":32188,"date":"2021-08-16T14:50:27","date_gmt":"2021-08-16T09:20:27","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32188"},"modified":"2021-08-16T14:50:27","modified_gmt":"2021-08-16T09:20:27","slug":"ncert-solutions-for-class-12-maths-chapter-8-ex-8-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-2\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 8 Application of Integrals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the area of the circle 4x\u00b2 + 4y\u00b2 = 9 which is interior to the parabola x\u00b2 = 4y.
\nSolution:
\n\"NCERT
\n4x\u00b2 + 4y\u00b2 = 9
\n\u2234 x\u00b2 + y\u00b2 = \\(\\frac { 9 }{ 4 }\\) … (1) is a circle with
\ncentre (0,0) and radius = \\(\\frac { 3 }{ 2 }\\)
\nx\u00b2 = 4y … (2)
\nSubstituting (2) in (1), we get 4y + y\u00b2
\n\u21d2 4y\u00b2 + 16y – 9 = 0
\n\u21d2 4y\u00b2 – 2y + 18y – 9 = 0
\n\u21d2 2y(2y – 1) + 9(2y – 1) = 0
\n\u21d2 (2y – 1)(2y + 9) = 0
\n\u21d2 y = \\(\\frac { 1 }{ 2 }\\), y = \\(\\frac { – 9 }{ 2 }\\)
\n\u2234 y = \\(\\frac { 1 }{ 2 }\\)
\nSubstitute y = \\(\\frac { 1 }{ 2 }\\) in (2) we get x = \u00b1 \\(\\sqrt{2}\\)
\n\u2234 The point of intersection of the curves are (- \\(\\sqrt{2}\\), \\(\\frac { 1 }{ 2 }\\)) and (\\(\\sqrt{2}\\), \\(\\frac { 1 }{ 2 }\\))
\nSince the shaded region is symmetric with respect to y-axis,
\nRequired Area = 2[Area under the circle – Area under the parabola] lying in the first quadrant between the lines x = 0 and x = \\(\\sqrt{2}\\)
\n\"NCERT<\/p>\n

Question 2.
\nFind the area bounded by curves (x – 1)\u00b2 + y\u00b2 = 1 and x\u00b2 + y\u00b2 = 1.
\nSolution:
\n\"NCERT
\nx\u00b2 + y\u00b2 = 1 is a circle with centre at origin and intersecting positive x-axis at (1, 0).
\n(x – 1)\u00b2 + y = 1 is a circle with centre at the origin and passing through the origin.
\nSolving x\u00b2 + y\u00b2 = 1 and (x – 1)\u00b2 + y\u00b2 = 1,
\nwe get x\u00b2 – (x – 1)\u00b2 = 0
\n(x – x + 1)(x + x – 1) = 0
\n2x – 1 = 0
\n\u2234 x = \\(\\frac { 1 }{ 2 }\\)
\nWhen x = \\(\\frac { 1 }{ 2 }\\), we get (\\(\\frac { 1 }{ 2 }\\))\u00b2 + y\u00b2 = 1
\ny\u00b2 = \\(\\frac { 3 }{ 4 }\\)
\n\u2234 y = \\(\\frac{\\pm \\sqrt{3}}{2}\\)
\nHence the points of intersection of the two circles are (\\(\\frac { 1 }{ 2 }\\), \\(\\frac{\\pm \\sqrt{3}}{2}\\)) and (\\(\\frac { 1 }{ 2 }\\), \\(\\frac{\\pm \\sqrt{- 3}}{2}\\)).
\nThe area bounded by the two curves is sym-metric w.r.t. x-axis.
\n\u2234 Required area = 2 x shaded area … (1)
\nArea of shaded region = Area under arc OA + Area under arc AC
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the area of the region bounded by the curves y = x\u00b2 + 2, y = x, x = 0 and x = 3.
\nSolution:
\nEquation of the parabola is y = x\u00b2 + 2 or x\u00b2 = (y – 2)
\nIts vertex is (0,2) axis is y-axis.
\nBoundary lines are y = x, x = 0, x = 3.
\nGraphs of the curve and lines have been shown in the figure.
\nArea of the region PQRO = Area of the region OAQR – Area of region OAP
\n\"NCERT<\/p>\n

Question 4.
\nUsing integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).
\nSolution:
\nLet the vertices of the triangle are A(- 1, 0), B(1, 3) and C(3, 2).
\n\"NCERT
\nEquation of AB is y – 0 = \\(\\frac { 3 – 0 }{ 1+1 }\\)(x+1)
\ni.e, y = \\(\\frac { 3 }{ 2 }\\)(x + 1)
\nEquation of BC is y – 3 = \\(\\frac { 2 – 3 }{ 3 – 1 }\\)(x – 1)
\ni.e, y – 3 = \\(\\frac { – 1 }{ 2 }\\)(x – 1) \u2234 y = \\(\\frac { 7 – x }{ 2 }\\)
\nEquation of AC is y – 0 = \\(\\frac { 2 – 0 }{ 3 + 1 }\\) (x + 1)
\ni.e., y = \\(\\frac { 1 }{ 2 }\\)(x + 1)
\nArea of \u2206 ABC = Area of \u2206 ADB + Area of trapezium BDEC – Area of \u2206 AEC
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nUsing integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4.
\nSolution:
\nThe equations of sides of the triangle are
\ny = 2x+ 1 … (1), y = 3x + 1 … (2), x = 4 … (3)
\nSolving equations (1) and (2), we get x = 0 and y = 1
\nLet A be the point (0,1)
\nSolving equations (1) and (3), we get x = 4 and y = 9.
\nLet B be the point (4, 9) Solving equations (2) and (3) we get x = 4 and y = 13
\nLet C be the point (4, 13)
\n\"NCERT<\/p>\n

Question 6.
\nSmaller area bounded by the circle x\u00b2 + y\u00b2 = 4 and the line x + y = 2
\n(a) 2 (\u03c0 – 2)
\n(b) \u03c0 – 2
\n(c) 2\u03c0 – 1
\n(d) 2(\u03c0 + 2)
\nSolution:
\nx\u00b2 + y\u00b2 = 4 intersects the positive x-axis at (2, 0) and the positive y-axis at (0, 2). The line x + y = 2 passes through (2, 0) and (0, 2).
\n\"NCERT
\nRequired area is shaded in the figure. Required Area = (Area of circle – Area under the line) lying in the first quadrant
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nArea lying between the curves y\u00b2 = 4x and y = 2x.
\n(a) \\(\\frac { 2 }{ 3 }\\)
\n(b) \\(\\frac { 1 }{ 3 }\\)
\n(c) \\(\\frac { 1 }{ 4 }\\)
\n(d) \\(\\frac { 3 }{ 4 }\\)
\nSolution:
\n(b) \\(\\frac { 1 }{ 3 }\\)
\n\"NCERT
\nThe given curves are y\u00b2 = 4x … (1)
\ny = 2x … (2)
\n4x(x – 1) = 0 \u21d2 x = 0, x = 1
\nWhen x = 0, y = 0
\nWhen x = 1, y = 2
\nThe points of intersection are (0, 0) and (1, 2)
\nRequired area is shaded in the figure.
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2 Question 1. Find the area of the circle 4x\u00b2 + 4y\u00b2 = 9 which is interior to …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. 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