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Class XII, , Chapter 8 – Application of Integrals, , Maths, , www.ncrtsolutions.blogspot.com, Exercise 8.1, , Question 1:, Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and, the x-axis., Answer, , The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the, x-axis is the area ABCD., , Page 1 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 2:, Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first, quadrant., Answer, , The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis, is the area ABCD., , Page 2 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 3:, Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first, quadrant., Answer, , The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis, is the area ABCD., , Page 3 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Question 4:, , Find the area of the region bounded by the ellipse, Answer, , The given equation of the ellipse,, , , can be represented as, , It can be observed that the ellipse is symmetrical about x-axis and y-axis., , ∴ Area bounded by ellipse = 4 × Area of OAB, , Page 4 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Therefore, area bounded by the ellipse = 4 × 3π = 12π units, , Question 5:, , Find the area of the region bounded by the ellipse, Answer, The given equation of the ellipse can be represented as, , It can be observed that the ellipse is symmetrical about x-axis and y-axis., , ∴ Area bounded by ellipse = 4 × Area OAB, , Page 5 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Therefore, area bounded by the ellipse =, , Question 6:, Find the area of the region in the first quadrant enclosed by x-axis, line, , and the, , circle, Answer, The area of the region bounded by the circle,, area OAB., , Page 6 of 53, , , and the x-axis is the
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Class XII, , Chapter 8 – Application of Integrals, , The point of intersection of the line and the circle in the first quadrant is, , Maths, , ., , Area OAB = Area ∆OCA + Area ACB, Area of OAC, , Area of ABC, , Therefore, area enclosed by x-axis, the line, , , and the circle, , quadrant =, , Question 7:, , Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line, Answer, , Page 7 of 53, , in the first
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Class XII, , Chapter 8 – Application of Integrals, , The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line,, area ABCDA., , It can be observed that the area ABCD is symmetrical about x-axis., , ∴ Area ABCD = 2 × Area ABC, , Page 8 of 53, , Maths, , , is the
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line,, , is, , units., , Question 8:, The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find, the value of a., Answer, The line, x = a, divides the area bounded by the parabola and x = 4 into two equal, parts., , Page 9 of 53, , ,
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Class XII, , Chapter 8 – Application of Integrals, , ∴ Area OAD = Area ABCD, , It can be observed that the given area is symmetrical about x-axis., , ⇒ Area OED = Area EFCD, , Page 10 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , From (1) and (2), we obtain, , Therefore, the value of a is, , ., , Question 9:, Find the area of the region bounded by the parabola y = x2 and, Answer, The area bounded by the parabola, x2 = y,and the line,, , Page 11 of 53, , , can be represented as
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Class XII, , Chapter 8 – Application of Integrals, , The given area is symmetrical about y-axis., , ∴ Area OACO = Area ODBO, , The point of intersection of parabola, x2 = y, and line, y = x, is A (1, 1)., Area of OACO = Area ∆OAB – Area OBACO, , ⇒ Area of OACO = Area of ∆OAB – Area of OBACO, , Therefore, required area =, , units, , Page 12 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 10:, Find the area bounded by the curve x2 = 4y and the line x = 4y – 2, Answer, The area bounded by the curve, x2 = 4y, and line, x = 4y – 2, is represented by the, shaded area OBAO., , Let A and B be the points of intersection of the line and parabola., Coordinates of point, ., Coordinates of point B are (2, 1)., We draw AL and BM perpendicular to x-axis., It can be observed that,, Area OBAO = Area OBCO + Area OACO … (1), Then, Area OBCO = Area OMBC – Area OMBO, , Page 13 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Similarly, Area OACO = Area OLAC – Area OLAO, , Therefore, required area =, , Question 11:, Find the area of the region bounded by the curve y2 = 4x and the line x = 3, Answer, The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO., , Page 14 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , The area OACO is symmetrical about x-axis., , ∴ Area of OACO = 2 (Area of OAB), , Therefore, the required area is, , units., , Question 12:, Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0, and x = 2 is, , Page 15 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , A. π, B., , C., , D., Answer, The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is, represented as, , Thus, the correct answer is A., , Page 16 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Question 13:, Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is, A. 2, B., , C., , D., Answer, The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as, , Thus, the correct answer is B., , Page 17 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Exercise 8.2, , Question 1:, Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y, Answer, The required area is represented by the shaded area OBCDO., , Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we obtain the, point of intersection as, ., It can be observed that the required area is symmetrical about y-axis., , ∴ Area OBCDO = 2 × Area OBCO, , We draw BM perpendicular to OA., Therefore, the coordinates of M are, , ., , Therefore, Area OBCO = Area OMBCO – Area OMBO, , Page 18 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Therefore, the required area OBCDO is, units, , Question 2:, Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1, Answer, The area bounded by the curves, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, is represented by, the shaded area as, , Page 19 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of, intersection as A, , and B, , It can be observed that the required area is symmetrical about x-axis., , ∴ Area OBCAO = 2 × Area OCAO, , We join AB, which intersects OC at M, such that AM is perpendicular to OC., , The coordinates of M are, , ., , Page 20 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Therefore, required area OBCAO =, , Maths, , units, , Question 3:, Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3, Answer, The area bounded by the curves, y = x2 + 2, y = x, x = 0, and x = 3, is represented by, the shaded area OCBAO as, , Page 21 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Then, Area OCBAO = Area ODBAO – Area ODCO, , Question 4:, Using integration finds the area of the region bounded by the triangle whose vertices are, (–1, 0), (1, 3) and (3, 2)., Answer, BL and CM are drawn perpendicular to x-axis., It can be observed in the following figure that,, Area (∆ACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1), , Page 22 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Equation of line segment AB is, , Equation of line segment BC is, , Equation of line segment AC is, , Therefore, from equation (1), we obtain, , Page 23 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Area (∆ABC) = (3 + 5 – 4) = 4 units, , Question 5:, Using integration find the area of the triangular region whose sides have the equations y, = 2x +1, y = 3x + 1 and x = 4., Answer, The equations of sides of the triangle are y = 2x +1, y = 3x + 1, and x = 4., On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C, (4, 9)., , It can be observed that,, Area (∆ACB) = Area (OLBAO) –Area (OLCAO), , Question 6:, Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is, A. 2 (π – 2), , Page 24 of 53
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Class XII, , Chapter 8 – Application of Integrals, , B. π – 2, C. 2π – 1, D. 2 (π + 2), Answer, The smaller area enclosed by the circle, x2 + y2 = 4, and the line, x + y = 2, is, represented by the shaded area ACBA as, , It can be observed that,, Area ACBA = Area OACBO – Area (∆OAB), , Thus, the correct answer is B., , Question 7:, Area lying between the curve y2 = 4x and y = 2x is, , A., , Page 25 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , B., , C., , D., Answer, The area lying between the curve, y2 = 4x and y = 2x, is represented by the shaded, area OBAO as, , The points of intersection of these curves are O (0, 0) and A (1, 2)., We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0)., , ∴ Area OBAO = Area (∆OCA) – Area (OCABO), , Page 26 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Thus, the correct answer is B., , Page 27 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Miscellaneous Solutions, , Question 1:, Find the area under the given curves and given lines:, (i) y = x2, x = 1, x = 2 and x-axis, (ii) y = x4, x = 1, x = 5 and x –axis, Answer, i., , The required area is represented by the shaded area ADCBA as, , ii., , The required area is represented by the shaded area ADCBA as, , Page 28 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Question 2:, Find the area between the curves y = x and y = x2, Answer, The required area is represented by the shaded area OBAO as, , Page 29 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , The points of intersection of the curves, y = x and y = x2, is A (1, 1)., We draw AC perpendicular to x-axis., , ∴ Area (OBAO) = Area (∆OCA) – Area (OCABO) … (1), , Question 3:, Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y, = 1 and y = 4, , Page 30 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Answer, The area in the first quadrant bounded by y = 4x2, x = 0, y = 1, and y = 4 is, represented by the shaded area ABCDA as, , Question 4:, Sketch the graph of, , and evaluate, , Answer, , Page 31 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , The given equation is, The corresponding values of x and y are given in the following table., x, , –6, , –5, , –4, , –3, , –2, , –1, , 0, , y, , 3, , 2, , 1, , 0, , 1, , 2, , 3, , On plotting these points, we obtain the graph of, , It is known that,, , Page 32 of 53, , as follows., , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 5:, Find the area bounded by the curve y = sin x between x = 0 and x = 2π, Answer, The graph of y = sin x can be drawn as, , ∴ Required area = Area OABO + Area BCDB, , Question 6:, Find the area enclosed between the parabola y2 = 4ax and the line y = mx, Answer, The area enclosed between the parabola, y2 = 4ax, and the line, y = mx, is represented, by the shaded area OABO as, , Page 33 of 53
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Class XII, , Chapter 8 – Application of Integrals, , The points of intersection of both the curves are (0, 0) and, We draw AC perpendicular to x-axis., , ∴ Area OABO = Area OCABO – Area (∆OCA), , Page 34 of 53, , Maths, , .
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 7:, Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12, Answer, The area enclosed between the parabola, 4y = 3x2, and the line, 2y = 3x + 12, is, represented by the shaded area OBAO as, , The points of intersection of the given curves are A (–2, 3) and (4, 12)., We draw AC and BD perpendicular to x-axis., , ∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO), , Page 35 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 8:, , Find the area of the smaller region bounded by the ellipse, , and the line, , Answer, , The area of the smaller region bounded by the ellipse,, , , is represented by the shaded region BCAB as, , Page 36 of 53, , , and the line,
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 9:, , Find the area of the smaller region bounded by the ellipse, , and the line, , Answer, , The area of the smaller region bounded by the ellipse,, , , is represented by the shaded region BCAB as, , ∴ Area BCAB = Area (OBCAO) – Area (OBAO), , Page 38 of 53, , , and the line,
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 10:, Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and xaxis, Answer, The area of the region enclosed by the parabola, x2 = y, the line, y = x + 2, and x-axis, is represented by the shaded region OABCO as, , Page 39 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1)., , ∴ Area OABCO = Area (BCA) + Area COAC, , Page 40 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 11:, Using the method of integration find the area bounded by the curve, [Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x, – y = 11], Answer, The area bounded by the curve,, , , is represented by the shaded region ADCB, , as, , The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0)., It can be observed that the given curve is symmetrical about x-axis and y-axis., , ∴ Area ADCB = 4 × Area OBAO, , Page 41 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Question 12:, Find the area bounded by curves, Answer, The area bounded by the curves,, , , is represented by the, , shaded region as, , It can be observed that the required area is symmetrical about y-axis., , Question 13:, Using 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), , Page 42 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Answer, The vertices of ∆ABC are A (2, 0), B (4, 5), and C (6, 3)., , Equation of line segment AB is, , Equation of line segment BC is, , Equation of line segment CA is, , Page 43 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Area (∆ABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA), , Question 14:, Using the method of integration find the area of the region bounded by lines:, 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0, Answer, The given equations of lines are, 2x + y = 4 … (1), 3x – 2y = 6 … (2), And, x – 3y + 5 = 0 … (3), , Page 44 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , The area of the region bounded by the lines is the area of ∆ABC. AL and CM are the, perpendiculars on x-axis., Area (∆ABC) = Area (ALMCA) – Area (ALB) – Area (CMB), , Question 15:, Find the area of the region, Answer, The area bounded by the curves,, , , is represented as, , Page 45 of 53
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Class XII, , Chapter 8 – Application of Integrals, , The points of intersection of both the curves are, , Maths, , ., , The required area is given by OABCO., It can be observed that area OABCO is symmetrical about x-axis., , ∴ Area OABCO = 2 × Area OBC, , Area OBCO = Area OMC + Area MBC, , Question 16:, Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is, A. – 9, B., , Page 46 of 53
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , C., , D., Answer, , Solve it yourself., The correct option is D., , Question 17:, The area bounded by the curve, , , x-axis and the ordinates x = –1 and x = 1 is, , given by, [Hint: y = x2 if x > 0 and y = –x2 if x < 0], , Page 47 of 53
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Class XII, , Chapter 8 – Application of Integrals, , A. 0, B., , C., , D., Answer, , Thus, the correct answer is C., , Page 48 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Question 18:, The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is, A., , B., , C., , D., Answer, The given equations are, x2 + y2 = 16 … (1), y2 = 6x … (2), , Area bounded by the circle and parabola, , Page 49 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Area of circle = π (r)2, = π (4)2, = 16π units, , Thus, the correct answer is C., , Page 50 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Question 19:, , The area bounded by the y-axis, y = cos x and y = sin x when, A., B., C., D., Answer, The given equations are, y = cos x … (1), And, y = sin x … (2), , Required area = Area (ABLA) + area (OBLO), , Integrating by parts, we obtain, , Page 51 of 53, , Maths
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Class XII, , Chapter 8 – Application of Integrals, , Maths, , Thus, the correct answer is B., , units, , Therefore, the required area is, , Page 52 of 53
