{"id":32249,"date":"2022-03-29T17:00:17","date_gmt":"2022-03-29T11:30:17","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32249"},"modified":"2022-03-29T17:08:07","modified_gmt":"2022-03-29T11:38:07","slug":"ncert-solutions-for-class-12-maths-chapter-8-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise"},"content":{"rendered":"

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

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

\"NCERT<\/p>\n

Question 1.
\nFind the area under the given curves and given lines:
\ni. y = x\u00b2, x = 1, x = 2 and x – axis
\nii. y = x4<\/sup>, x = 1, x = 5 and x-axis
\nSolution:
\n\"NCERT
\ni. The given curves are y = x\u00b2 , x = 1 and x = 2. The required area lies between the lines x = 1 and x = 2. Hence the limits of integration are 1 and 2.
\n\"NCERT<\/p>\n

ii. The given curves are y = x4<\/sup>, x = 1 and x = 5. The required area lies between the lines x = 1 and x = 5. Hence the limits of integration are 1 and 5.
\n\"NCERT<\/p>\n

Question 2.
\nFind the area between the curves y = x and y = x\u00b2.
\nSolution:
\n\"NCERT
\ny = x … (1)
\ny = x\u00b2 … (2)
\n(2) \u2192 x = x\u00b2
\n\u21d2 x\u00b2 – x = 0 \u21d2 x(x – 1) = 0
\n\u21d2 x = 0 and x = 1
\nHence the limits of integration are 0 and 1. Required area = (Area under the line y = x) – (Area under the parabola y = x\u00b2)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the area of the region lying in the first quadrant and bounded by y = 4x\u00b2, x = 0, y = 1 and y = 4.
\nSolution:
\nThe given curves are y = 4x\u00b2
\n\u21d2 x\u00b2 = \\(\\frac { y }{ 4 }\\) \u21d2 x = \\(\\frac{1}{2} \\sqrt{y}\\)
\nx = 0, y= 1, y = 4
\nThe required area lies in the first quadrant between the lines y= 1 and y = 4
\n\"NCERT
\nHence the limits of integration are 1 and 4.
\nRequired area = Shaded area
\n\"NCERT<\/p>\n

Question 4.
\nSketch the graph of y = |x + 3| and evaluate \\(\\int_{-6}^{0}|x+3| d x\\).
\nSolution:
\ny = |x + 3| is redefined as follows,
\n\"NCERT<\/p>\n

Question 5.
\nFind the area bounded by the curve y = sin x between x = 0 and x = 2\u03c0.
\nSolution:
\nThe area bounded by the curve y = sinx between x = 0 and x = 2\u03c0 is shaded in the figure.1
\n\"NCERT
\nThe area bounded by the curve between x = 0 and x = \u03c0 is same as the area bounded by it between x = \u03c0 and x = 2\u03c0
\nRequired area 2 x Area of the shaded
\nregion above x – axis between x = 0
\nand \u03c0 = 2\\(2 \\int_{0}^{\\pi} y d x=2 \\int_{0}^{\\pi} \\sin x d x\\)
\n= 2\\([-\\cos x]_{0}^{\\pi}=-2[\\cos x]_{0}^{\\pi}\\)
\n= – 2(cos \u03c0 – cos0) = – 2(- 1 – 1) = 4 sq.unit<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFind the area enclosed between the parabola y\u00b2 = 4ax and the line y = mx.
\nSolution:
\n\"NCERT
\nThe curves are y\u00b2 = 4ax … (1)
\nand y = mx … (2)
\nSolving (1) and (2), we get
\nm\u00b2x\u00b2 = 4ax \u21d2 m\u00b2x\u00b2 – 4ax = 0
\ni.e., x(m\u00b2x – 4a) = 0 \u21d2 x = 0 or x = \\(\\frac{4 a}{m^{2}}\\)
\nThe x coordinate of the point of intersection
\nare x = 0 and x = \\(\\frac{4 a}{m^{2}}\\)
\nRequired area = (Area under the parabola y\u00b2 = 4ax) – (Area under the line y = mx)
\n\"NCERT<\/p>\n

Question 7.
\nFind the area enclosed by the parabola 4y = 3x\u00b2 and the line 2y = 3x+ 12.
\nSolution:
\n\"NCERT
\nThe curves are 4y = 3x\u00b2
\n\"NCERT
\n\u2234 The points of intersection are (4, 12) and (-2, 3)
\nRequired area is shaded in the figure
\n\"NCERT<\/p>\n

Question 8.
\nFind the area of the smaller region bounded by the ellipse \\(\\frac{x^{2}}{9}+\\frac{y^{2}}{4}\\) = 1 and the line \\(\\frac{x}{3}+\\frac{y}{2}=1\\)
\nSolution:
\n\"NCERT
\n\\(\\frac{x^{2}}{9}+\\frac{y^{2}}{4}\\) = 1 is the ellipse with centre at origin intersecting the positive x-axis at A(3, 0) and positive y-axis at B(0, 2)
\n\"NCERT
\nThe required area is shaded in the figure.
\n\"NCERT<\/p>\n

Question 9.
\nFind the area of the smaller region bounded by the ellipse \\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}\\) and the line \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1.
\nSolution:
\n\"NCERT
\nThe ellipse intersects the positive x-axis at A (a, 0) and y-axis at B(0, b).
\n\"NCERT
\nRequired area is shaded in the figure.
\nRequired area = Area under the ellipse in the first quadrant – Area under the line AB
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nFind the area of the region enclosed by the parabola x\u00b2 = y, the line y = x + 2 and the x-axis.
\nSolution:
\nThe given curves are x\u00b2 = y … (1)
\nand y = x + 2 … (2)
\n\"NCERT
\n(1) \u2192 x\u00b2 = x + 2 \u21d2 x\u00b2 – x – 2 = 0
\n\u21d2 (x – 2)(x + 1) = 0 \u21d2 x = 2, x = – 1
\nWhen x = 2, y = 4
\nWhen x = – 1, y = 1
\nThe points of intersection are (-1, 1) and (2,4)
\nRequired area is shaded in the figure
\n\"NCERT<\/p>\n

Question 11.
\nUsing the method of integration find the area bounded by the curve |x| + |y| = 1.
\nSolution:
\n\"NCERT
\nThe given curve is |x| + |y| = 1
\nx + y = 1 … (1)
\nx – y = 1 … (2)
\n– x + y = 1 … (3)
\n– x – y = 1 … (4)
\nRequired area is symmetric w.r.t. both the axes.
\nRequired area = 4 x Area of shaded region
\n= \\(4 \\int_{0}^{1}(1-x) d x=4\\left[x-\\frac{x^{2}}{2}\\right]_{0}^{1}\\)
\n= \\(4\\left(1-\\frac{1}{2}\\right)=4\\left(\\frac{1}{2}\\right)\\)
\n= 2 sq.units<\/p>\n

Question 12.
\nFind the area bounded by curves {(x, y) : y \u2265 x\u00b2 and y = |x|}
\nSolution:
\nThe area bounded by the curves y \u2265 x\u00b2 and y = |x| is same as the area bounded by y = x\u00b2, y = |x| are same
\n\"NCERT
\nThe point B is obtained by solving y = x
\nand y = x\u00b2
\ni.e., x = x\u00b2 \u21d2 x\u00b2 – x = 0
\nx(x – 1) = 0 \u21d2 x = 0 or x = 1
\nWhen x = 0, y = 0 and when x = 1, y = 1
\n\u2234 B is(1, 1)
\nRequired area = 2 x (Area of the shaded portion in the first quadrant)
\n= 2[(Area under the line y = x) – (Area under the parabola y = x\u00b2)]
\n= \\(2\\left[\\int_{0}^{1} x d x-\\int_{0}^{1} x^{2} d x\\right]=2\\left[\\left[\\frac{x^{2}}{2}\\right]_{0}^{1}-\\left[\\frac{x^{3}}{3}\\right]_{0}^{1}\\right]\\)
\n= \\(2\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=2\\left(\\frac{1}{6}\\right)=\\frac{1}{3} \\text { sq.unit }\\)<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nUsing the 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).
\nSolution:
\n\"NCERT
\nEquation of BC: y – 5 = \\(\\left(\\frac{3-5}{6-4}\\right)\\)(x – 4)
\ny – 5 = – (x – 4) \u21d2 y = 9 – x
\nEquation of AC : y – 0 = \\(\\left(\\frac{3-0}{6-2}\\right)\\)(x – 2)
\ny = \\(\\frac{3}{4}\\)(x – 2)
\nArea of \u2206ABC = Area under AB + Area under BC – Area under AC
\n\"NCERT
\n= 5 + 8 – 6 = 7 sq. unit<\/p>\n

Question 14.
\nUsing the method of integration find the area of the region bounded by lines:
\n2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0
\nSolution:
\n\"NCERT
\nThe given lines are 2x + y = 4 … (1)
\n3x – 2y = 6 … (2)
\nx – 3y + 5 = 0 … (3)
\nSolving (1) and (2), we get x = 2, y = 0
\nLet A be the point (2, 0)
\nSolving (2) and (3), we get x = 4, y = 3
\nLet B be the point (4, 3)
\nSolving (1) and (3), we get x = 1, y = 2
\nLet C be the point (1, 2)
\nEquation of AB is y = \\(\\frac{3x-6}{2}\\)
\nEquation of BC is y = \\(\\frac{x+5}{3}\\)
\nEquation of AC is y = 4 – 2x
\nArea of \u2206ABC Area of trapezium DCBE – Area of \u2206DCA – Area of \u2206ABE
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 15.
\nFind the area of the region \\(\\left\\{(x, y): y^{2} \\leq 4 x, 4 x^{2}+4 y^{2} \\leq 9\\right\\}\\)
\nSolution:
\n\"NCERT
\ncircle with centre at origin and intersecting positive x-axis at C(\\(\\frac{3}{2}\\), 0)
\nSolving 4x\u00b2 + 4y\u00b2 = 9 and y\u00b2 = 4x, we get 4x\u00b2 + 16x = 9 \u21d2 4x\u00b2 + 16x – 9 = 0
\n\"NCERT
\nArea of the shaded region = Area under OA + Area under AC
\n\"NCERT
\nThe required region is symmetric with respect to jt-axis.
\n\u2234 Required area = 2 x shaded area
\n= \\(2\\left(\\frac{\\sqrt{2}}{12}+\\frac{9 \\pi}{16}-\\frac{9}{8} \\sin ^{-1}\\left(\\frac{1}{3}\\right)\\right)\\)
\n= \\(\\frac{\\sqrt{2}}{6}+\\frac{9 \\pi}{8}-\\frac{9}{4} \\sin ^{-1}\\left(\\frac{1}{3}\\right) \\text { sq.unit }\\)<\/p>\n

Question 16.
\nArea bounded by the curve y = x\u00b3, the x- axis and the ordinates x = – 2 and x = 1 is
\na. – 9
\nb. \\(\\frac{-15}{4}\\)
\nc. \\(\\frac{15}{4}\\)
\nd. \\(\\frac{17}{4}\\)
\nSolution:
\n\"NCERT<\/p>\n

Question 17.
\nThe area bounded by the curve y = x|x|, x-axis and the ordinates x = – 1 and x = 1 is given by
\na. 0
\nb. \\(\\frac{1}{3}\\)
\nc. \\(\\frac{2}{3}\\)
\nd. \\(\\frac{4}{3}\\)
\nSolution:
\ny = x|x| is redefined as
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 18.
\nThe area of the circle x\u00b2 + y\u00b2 = 16 exterior to the parabola y\u00b2 = 6x is
\na. \\(\\frac{4}{3}(4 \\pi-\\sqrt{3})\\)
\nb. \\(\\frac{4}{3}(4 \\pi+\\sqrt{3})\\)
\nc. \\(\\frac{4}{3}(8 \\pi-\\sqrt{3})\\)
\nd. \\(\\frac{4}{3}(8 \\pi+\\sqrt{3})\\)
\nSolution:
\nc. \\(\\frac{4}{3}(8 \\pi-\\sqrt{3})\\)
\n\"
\nx\u00b2 + y\u00b2 = 16 is a circle with centre at origin
\nand intersecting positive x-axis at C(4,0).
\nSolving x\u00b2 + y\u00b2 = 16 and y\u00b2 = 6x, we get
\nx\u00b2 + 6x = 16
\nx\u00b2 + 6x – 16 = 0 \u21d2 (x + 8)(x – 2) = 0
\n\u2234 x = – 8 and x = 2
\nBut x = – 8 is not possible \u2234 x = 2
\nWhen x = 2, y\u00b2 = 12 \u2234 y = \u00b12\\(\\sqrt{3}\\)
\nThe points of intersection are
\nA(2, \\(\\sqrt{3}\\)) and B(2, -2\\(\\sqrt{3}\\))
\nThe required area is shaded in the figure.
\nThe nonshaded region OACBO is symmetric w.r.t. x-axis and lies between x = 0 and x = 4.
\n\u2234 Area of the nonshaded region = 2 x Area of the nonshaded region in the first quadrant.
\n= 2 [Area under OA + Area under AC]
\n\"NCERT<\/p>\n

Question 19.
\nThe area bounded by the y-axis, y = cos x and y = sin x when 0 \u2264 x \u2264 \\(\\frac { \u03c0 }{ 2 }\\) is
\na. 2(\\(\\sqrt{2}\\) – 1)
\nb. \\(\\sqrt{2}\\) – 1
\nc. \\(\\sqrt{2}\\) + 1
\nd. \\(\\sqrt{2}\\)
\nSolution:
\nb. \\(\\sqrt{2}\\) – 1
\n\"NCERT
\nThe curves are y = sinx … (1)
\nand y = cos x … (2)
\nFrom (1) and (2), we get sin x = cos x
\n\u21d2 x = \\(\\frac { \u03c0 }{ 4 }\\)
\n\u2234 y = sin\\(\\frac { \u03c0 }{ 4 }\\) = \\(\\frac{1}{\\sqrt{2}}\\)
\nPoint of intersection of curves is (\\(\\frac { \u03c0 }{ 4 }\\), \\(\\frac{1}{\\sqrt{2}}\\))
\nRequired area = Area of the shaded region = (Area under the curve y = cos x) – (Area above the curve y = sinx)
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Question 1. Find the area under the given curves and given lines: i. y = x\u00b2, …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise<\/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 Miscellaneous Exercise - 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-miscellaneous-exercise\/\" \/>\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 Miscellaneous Exercise - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Miscellaneous Exercise Question 1. 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