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CHAPTER 8:, , APPLICATION OF INTEGRALS, 3 mark questions, Question 1:, , Question 2:, , Find the area of the region bounded by the, curve y2 = x and the lines x = 1, x = 4 and the, x-axis., Answer :, , 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 = x, the lines, x = 1 and x = 4, and the xaxis is the area ABCD., , The area of the region bounded by the curve,, y2 = 9x, x = 2, and x = 4, and the x-axis is the, area ABCD., , 1

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Question 3:, , Question 4:, , Find the area of the region bounded by x2 =, 4y, y = 2, y = 4 and the y-axis in the first, quadrant., , Find the area of the region bounded by the, curve y2 = 4x, y-axis and the line y = 3 is, , Answer :, , Answer :, The area bounded by the curve, y2 = 4x, yaxis, and y = 3 is represented as, , The area of the region bounded by the curve,, x2 = 4y, y = 2, and y = 4, and the y-axis is the, area ABCD., , Question 5:, Find the area lying between the curve y2 =, 4x and y = 2x is, Answer :, The area lying between the curve, y2 = 4x, and y = 2x, is represented by the shaded area, OBAO as, 2

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2, , 2, , From (2) and (1) y =4a, y = 2a, Required area A = 2 [area of OSP, =2∫, = 2 ∫ √ . √ . dx, =4√, , ( ) =, , √, , [ a√ ] =, , Sq. units, , 5 MARK QUESTIONS:, , 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 (OCABO) – Area, (ΔOCA), , Question 1:, Find the area of the region bounded by the, ellipse, Answer :, The given equation of the ellipse,, , can be represented as, , square units, Question 5., 2, , Find area enclosed by the Parabola y =4ax, and its latus rectum by integration, 2, Solution: y = 4ax ---- (1) and, the equation of the Latus rectum is given by, x = a ……… (2), , It can be observed that the ellipse is, symmetrical about x-axis and y-axis., 3

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∴ Area bounded by ellipse = 4 × Area of, OAB, , It can be observed that the ellipse is, symmetrical about x-axis and y-axis., ∴ Area bounded by ellipse = 4 × Area OAB, , Therefore, area bounded by the ellipse = 4 ×, 3π = 12π units, , Question 2:, Therefore, area bounded by the ellipse =, , Find the area of the region bounded by the, ellipse, , Question 3:, , Answer :, The given equation of the ellipse can be, represented as, , 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,, , and the x-axis is the, area OAB., , 4

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The area of the smaller part of the circle, x2, + y2 = a2, cut off by the line,, area ABCDA., , , is the, , The point of intersection of the line and the, circle in the first quadrant is, ., Area OAB = Area ΔOCA + Area ACB, Area of OAC, It can be observed that the area ABCD is, symmetrical about x-axis., ∴ Area ABCD = 2 × Area ABC, , Area of ABC, , Therefore, area enclosed by x-axis, the line, , and the circle, , in the, , first quadrant =, Question 7:, Find the area of the smaller part of the circle, x2 + y2 = a2 cut off by the line, Answer :, , 5

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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., ∴ Area OAD = Area ABCD, , From (1) and (2), we obtain, It can be observed that the given area is, symmetrical about x-axis., ⇒ Area OED = Area EFCD, , 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,, 6, , , can be represented as

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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 ΔOAM – Area, OMACO, Area of ΔOAM, , 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 xaxis., It can be observed that,, Area OBAO = Area OBCO + Area OACO, … (1), Then, Area OBCO = Area OMBC – Area, OMBO, , Area of OMACO, , ⇒ Area of OACO = Area of ΔOAM – Area, of OMACO, , Therefore, required area =, , units, , Question 10:, Find the area bounded by the curve x2 = 4y, and the line x = 4y – 2, , Similarly, Area OACO, = Area OLAC – Area OLAO, , Answer :, The area bounded by the curve, x2 = 4y, and, line, x = 4y – 2, is represented by the shaded, area OBAO., 7

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Therefore, the required area is, Therefore, required area =, , units., , Question 12:, Find the area of the circle 4x2 + 4y2 = 9, which is interior to the parabola x2 = 4y, , Question 11:, , Answer :, The required area is represented by the, shaded area OBCDO., , 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., , Solving the given equation of circle, 4x2 +, 4y2 = 9, and parabola, x2 = 4y, we obtain the, point of intersection as, The area OACO is symmetrical about xaxis., ∴ Area of OACO = 2 (Area of OAB), , ., It can be observed that the required area is, symmetrical about y-axis., 8

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∴ Area OBCDO = 2 × Area OBCO, We draw BM perpendicular to OA., Therefore, the coordinates of M are, ., Therefore, Area OBCO = Area OMBCO –, Area OMBO, Therefore, the required area OBCDO is, Equation of line segment AC is, units, Question :13, Using integration finds the area of the region, bounded by the triangle whose vertices are, (–1, 0), (1, 3) and (3, 2)., , Therefore, from equation (1), we obtain, Area (ΔABC) = (3 + 5 – 4) = 4 units, , Answer :, BL and CM are drawn perpendicular to xaxis., It can be observed in the following figure, that,, Area (ΔACB) = Area (ALBA) + Area, (BLMCB) – Area (AMCA) … (1), , Question 14:, 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)., , Equation of line segment AB is, , It can be observed that,, Equation of line segment BC is, 9

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Area (ΔACB) = Area (OLBAO) –Area, (OLCAO), , Question 15:, Find the smaller area enclosed by the circle, x2 + y2 = 4 and the line x + y = 2 is, 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, (ΔOAB), , 10