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>\nNCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1<\/h2>\n <\/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 <\/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 <\/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 <\/p>\n
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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 <\/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 <\/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 \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 <\/p>\n
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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 \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 <\/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 \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 <\/p>\n
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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 \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 <\/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 <\/p>\n
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Question 10. \nFind the area bounded by the curve x\u00b2 = 4y and the line x = 4y – 2 \nSolution: \n \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 <\/p>\n
Question 11. \nFind the area of the region bounded by the curve y\u00b2 = 4x and the line x = 3. \nSolution: \n \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 <\/p>\n
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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 <\/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 <\/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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n