{"id":32099,"date":"2022-03-29T16:30:07","date_gmt":"2022-03-29T11:00:07","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32099"},"modified":"2022-03-29T16:55:03","modified_gmt":"2022-03-29T11:25:03","slug":"ncert-solutions-for-class-12-maths-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-1\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 8 Application of Integrals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-1\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the area of the region bounded by the curve y\u00b2 = x and the lines x = 1, x = 4, and the x-axis.
\nSolution:
\nThe required area is shaded in the figure and lies between the lines x = 1 and x = 4. Hence the limits of integration are 1 and 4.
\n\"NCERT<\/p>\n

Question 2.
\nFind the area of the region bounded by y\u00b2 = 9x, x = 2, x = 4 and x-axis in the first quadrant
\nSolution:
\nThe required area is shaded in the figure lies between the lines x = 2 and x = 4. Hence the limits of integration are 2 and 4.
\n\"NCERT<\/p>\n

Question 3.
\nFind the area of the region bounded by x\u00b2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
\nSolution:
\nx\u00b2 = 4y \u2234 x = 2\\(\\sqrt{y}\\)
\nThe required area is shaded in the figure and lies between the lines y = 2 and y = 4. Hence the limits of integration are 2 and 4.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFind the area of the region bounded by the ellipse \\(\\frac { { x }^{ 2 } }{ 16 } +\\frac { { y }^{ 2 } }{ 9 }\\) = 1
\nSolution:
\ni. Put y = o in \\(\\frac { { x }^{ 2 } }{ 16 } +\\frac { { y }^{ 2 } }{ 9 }\\) = 1, we get \\(\\frac{x^{2}}{16}=1\\)
\ni.e., x\u00b2 = 16 \u2234 x = \u00b1 4
\n\u2234 The points of intersection with x-axis are (-4, 0) and (4, 0)
\n\"NCERT<\/p>\n

ii. The shaded area lies between x = 0 and x = 4.
\n\u2234 The limits of integration are 0 and 4.
\nThe curve is symmetrical w.r.t both the axes
\n\u2234 Required area = 4 x area of the shaded region
\n\\(\\frac { { x }^{ 2 } }{ 16 } +\\frac { { y }^{ 2 } }{ 9 }\\) = 1 i.e., \\(\\frac{y^{2}}{9}=1\\) – \\(\\frac{x^{2}}{16}=1\\)
\n\"NCERT<\/p>\n

Question 5.
\nFind the area of the region bounded by the ellipse \\(\\frac { { x }^{ 2 } }{ 4 } +\\frac { { y }^{ 2 } }{ 9 } \\) = 1
\nSolution:
\n\"NCERT
\nThe points of intersection of the curve and the x-axis is obtained by substituting y = 0 in the equation of the curve.
\nWe get \\(\\frac{x^{2}}{4}=1\\), x\u00b2 = 4 \u2234 x = \u00b12
\nThe curve intersects the x-axis at (-2, 0) and (2,0).
\n\\(\\frac { { x }^{ 2 } }{ 4 } +\\frac { { y }^{ 2 } }{ 9 } \\) = 1 \u2234 \\(\\frac{y^{2}}{9}\\) = 1 – \\(\\frac{x^{2}}{4}=1\\)
\n= \\(\\frac{4-x^{2}}{4}\\)
\n\u2234 y\u00b2 = \\(\\frac { 9 }{ 4 }\\)(4 – x\u00b2) \u2234 y = \\(\\frac{3}{2} \\sqrt{4-x^{2}}\\)
\nThe curve is symmetrical about both the axes.
\nArea of the ellipse = 4 x area of the shaded region
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFind the area of the region in the first quadrant enclosed by x-axis, line x = \\(\\sqrt{3}\\)y and the circle x\u00b2 + y\u00b2 = 4.
\nSolution:
\n\"NCERT
\nThe given equations are x = \\(\\sqrt{3}\\)y … (1)
\nx\u00b2 + y = 4 … (2)
\nTo get the point of intersection of the line and the circle, solve (1) and (2)
\nFrom (1), we get y = \\(\\frac{x}{\\sqrt{3}}\\)
\n\u2234 (2) \u2192 x\u00b2 + \\(\\left(\\frac{x}{\\sqrt{3}}\\right)^{2}\\) = 4 \u21d2 x\u00b2 + \\(\\frac{x^{2}}{3}\\) = 4
\n\\(\\frac{4x^{2}}{3}\\) = 4 \u21d2 x\u00b2 = 3 \u21d2 x = \u00b1\\(\\sqrt{3}\\)
\nTo get the point of intersection of the circle and x-axis puty = 0 in (2)
\n(2) \u2192 x\u00b2 = 4 \u2234x = \u00b1 2
\n\u2234 The x coordinate of A and B are \\(\\sqrt{3}\\) and 2 respectively.
\nRequired area = Area of OAC + Area of ACB
\n\"NCERT<\/p>\n

Question 7.
\nFind the area of the smaller part of the circle x\u00b2 + y\u00b2 = a\u00b2 cut off by the line x = \\(\\frac { a }{ \\sqrt { 2 } }\\).
\nSolution:
\n\"NCERT
\nx\u00b2 + y\u00b2 = a\u00b2 is a circle which intersects the positive x axis at (a, 0).
\nx\u00b2 + y\u00b2 = a\u00b2
\n\u2234 y = \\(\\sqrt{a^{2}-x^{2}}\\)
\nThe required area is shaded in the figure and lies between x = \\(\\frac { a }{ \\sqrt { 2 } }\\) and x – a
\n\u2234 Limits of integration are \\(\\frac { a }{ \\sqrt { 2 } }\\) and a
\nArea of the shaded region = 2 x Area of the shaded region in the first quadrant
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nThe area between x = y\u00b2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
\nSolution:
\n\"NCERT
\nThe given curves are x = y\u00b2 … (1)
\nx = 4 ….. (2) and x = a … (3)
\nSince the line x = a divides the area in two equal parts, we get \\(2 \\int_{0}^{a} y d x=2 \\int_{a}^{4} y d x\\)
\nSince the areas are symmetric w.r.t. x-axis
\n\"NCERT<\/p>\n

Question 9.
\nFind the area of the region bounded by the parabola y = x\u00b2 and y = |x|.
\nSolution:
\nClearly x\u00b2 = y represents a parabola with vertex at (0, 0) positive direction of y-axis as its axis it opens upwards.
\ny = |x| i.e., y = x and y = – x represent two lines passing through the origin and making an angle of 45\u00b0 and 135\u00b0 with the positive direction of the x-axis.
\nThe required region is the shaded region as shown in the figure. Since both the curve are symmetrical about y-axis.
\nSo,
\nRequired area = 2 (shaded area in the first quadrant)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nFind the area bounded by the curve x\u00b2 = 4y and the line x = 4y – 2
\nSolution:
\n\"NCERT
\nThe given curves are x\u00b2 = 4y … (1)
\nand x – 4y – 2 … (2)
\nSubstituting (2) in (1),
\nwe get (4y – 2)\u00b2 = 4y
\n\u21d2 16y\u00b2 – 16y + 4 = 4y
\n\u21d2 16y\u00b2 – 20y + 4 = 0
\n\u21d2 4y\u00b2 – 5y + 1 = 0
\n\u21d2 4y\u00b2 – 4y – y + 1 = 0
\n\u21d2 4y(y – 1) – 1(y – 1) = 0
\n\u21d2 (y – 1)(4y – 1) = 0
\n\u21d2 y = 1, y = \\(\\frac { 1 }{ 4 }\\)
\nWhen y = 1, x = 2
\nWhen y = \\(\\frac { 1 }{ 4 }\\), x = – 1
\n\u2234 Points of intersection are (-1, \\(\\frac { 1 }{ 4 }\\)) and (2, 1)
\nThe required area is shaded in the figure. Required area = (Area under the line x = 4y – 2) – (Area under the parabola x\u00b2 = 4y) :
\n\"NCERT<\/p>\n

Question 11.
\nFind the area of the region bounded by the curve y\u00b2 = 4x and the line x = 3.
\nSolution:
\n\"NCERT
\nThe given curves are y\u00b2 = 4x … (1)
\nand x = 3
\nThe shaded region is symmetric w.r.t. x-axis.
\n\u2234 Required area = 2 x Shaded area in the first quadrant
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 12.
\nArea lying in the first quadrant and bounded by the circle x\u00b2 + y\u00b2 = 4 and the lines x = 0 and x = 2 is
\n(a) \u03c0
\n(b) \\(\\frac { \\pi }{ 2 } \\)
\n(c) \\(\\frac { \\pi }{ 3 } \\)
\n(d) \\(\\frac { \\pi }{ 4 } \\)
\nSolution:
\n(a) \u03c0
\n\"NCERT<\/p>\n

Question 13.
\nArea of the region bounded by the curve y\u00b2 = 4x, y-axis and the line y = 3 is
\n(a) 2
\n(b) \\(\\frac { 9 }{ 4 }\\)
\n(c) \\(\\frac { 9 }{ 3 }\\)
\n(d) \\(\\frac { 9 }{ 2 }\\)
\nSolution:
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-1\/ NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1 Question 1. Find the area of the region bounded by the curve y\u00b2 = x and …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1<\/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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-1\/\" \/>\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 8.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-1\/ NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1 Question 1. 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