NCERT Solutions for Class 12 Science Maths Chapter 2 – Application Of Integrals
Explore comprehensive NCERT Solutions for Class 12 Science Maths Chapter 2: Application Of Integrals. These solutions offer straightforward, step-by-step explanations, making them a favored resource among Class 12 Science students. Whether you’re tackling homework assignments or gearing up for exams, these Maths Application Of Integrals Solutions prove invaluable for efficient preparation. Access free answers to all questions from Chapter 2 of the NCERT Book for Class 12 Science Maths, ensuring a reliable aid in your academic journey. Page No 365: 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 No 365: 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 No 366: 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 No 366: 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 Therefore, area bounded by the ellipse = 4 × 3π = 12π units Page No 366: 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 Therefore, area bounded by the ellipse = Page No 366: 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, , and the x-axis is the area OAB. The point of intersection of the line and the circle in the first quadrant is . Area OAB = Area ΔOCA + Area ACB Area of OAC Area of ABC Therefore, required area enclosed = 3√2 + π3 − 3√2 = π3 square units32 + π3 – 32 = π3 square units Page No 366: Question 7: Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line ANSWER: The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is the area ABCDA. It can be observed that the area ABCD is symmetrical about x-axis. ∴ Area ABCD = 2 × Area ABC Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, , is units. Page No 366: 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 It can be observed that the given area is symmetrical about x-axis. ⇒ Area OED = Area EFCD From (1) and (2), we obtain Therefore, the value of a is . Page No 366: 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,, can be represented as 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 Area of OMACO ⇒ Area of OACO = Area of ΔOAM – Area of OMACO Therefore, required area = units Page No 366: 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 Similarly, Area OACO = Area OLAC – Area OLAO Therefore, required area = Page No 366: 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. The area OACO is symmetrical about x-axis. ∴ Area of OACO = 2 (Area of OAB) Therefore, the required area is units. Page No 366: 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 A. π B. C. 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 No 366: Question 13: Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is A. 2 B. C. ANSWER: The area bounded by the curve, y2 = 4x, y-axis, and y = 3 is represented as Thus, the correct answer is B. Page No 371: Question 1: Find the area of the circle 4×2 + 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, 4×2 + 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, the required area OBCDO is units Page No 371: 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 On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, …
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