4<\/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 <\/p>\nQuestion 2. \nFind the area between the curves y = x and y = x\u00b2. \nSolution: \n \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 <\/p>\n
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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 \nHence the limits of integration are 1 and 4. \nRequired area = Shaded area \n <\/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 <\/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 \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
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Question 6. \nFind the area enclosed between the parabola y\u00b2 = 4ax and the line y = mx. \nSolution: \n \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 <\/p>\n
Question 7. \nFind the area enclosed by the parabola 4y = 3x\u00b2 and the line 2y = 3x+ 12. \nSolution: \n \nThe curves are 4y = 3x\u00b2 \n \n\u2234 The points of intersection are (4, 12) and (-2, 3) \nRequired area is shaded in the figure \n <\/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 \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 \nThe required area is shaded in the figure. \n <\/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 \nThe ellipse intersects the positive x-axis at A (a, 0) and y-axis at B(0, b). \n \nRequired area is shaded in the figure. \nRequired area = Area under the ellipse in the first quadrant – Area under the line AB \n <\/p>\n
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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 \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 <\/p>\n
Question 11. \nUsing the method of integration find the area bounded by the curve |x| + |y| = 1. \nSolution: \n \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 \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
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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 \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 \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 \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 <\/p>\n
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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 \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 \nArea of the shaded region = Area under OA + Area under AC \n \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 <\/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 <\/p>\n
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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 <\/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 \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 <\/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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n