The area bounded by the curves y2 = 4a2 (x – 1) and lines x = 1 and y = 4a is 1) 4a2 sq units 2) 16a/3 sq units 3) 18a/3 sq units 4) None of these Solution: (2) 16a/3 sq units y2 = 4a2 (x - 1) ⇒ (y -... View Article
The area enclosed between the curve y = 1 + x2, the y-axis and the straight line y = 5 is given by 1) 14/3 sq units 2) 7/3 sq units 3) 5 sq units 4) 16/3 sq units Solution: (4) 16/3 sq units... View Article
The area bounded by y = sin-1 x, x = 1 / √2 and x-axis is 1) [(1 / √2) + 1] square units 2) [1 - (1 / √2)] square units 3) (π / 4√2) square units 4) [(π / 4 √2) + (1 / √2) - 1] square units... View Article
The area of the region bounded by y = 2x – x2 and the x-axis is 1) 8/3 sq units 2) 4/3 sq units 3) 7/3 sq units 4) 2/3 sq units Solution: (2) 4/3 sq units The given curve is y = 2x - x2... View Article
The area bounded by the curve y = 1/2 x2, the X-axis and the line x = 2 is 1) 1/3 sq units 2) 2/3 sq units 3) 1 sq units 4) 4/3 sq units Solution: (4) 4/3 sq units... View Article
Area bounded by the lines y = x, x = -1, x = 2 and x-axis is 1) 5/2 sq units 2) 3/2 sq units 3) 1/7 sq units 4) None of the above Solution: (1) 5/2 sq units... View Article
The area bounded by the curve x = 4 – y2 and the y-axis is 1) 16 sq units 2) 32 sq units 3) 32/3 sq units 4) 16/3 sq units Solution: (3) 32/3 sq units... View Article
Area enclosed by the curve Ï€[4(x – √2)2 + y2] = 8 is 1) Ï€ sq units 2) 2 sq units 3) 3Ï€ sq units 4) 4 sq units Solution: (4) 4 sq units... View Article
The area ( in sq units ) of the region bounded by the curves 2x = y2 – 1 and x = 0 is 1) 1/3 sq units 2) 2/3 sq units 3) 1 sq units 4) 2 sq units Solution: (2) 2/3 sq units... View Article
The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to 1) 4/3 sq units 2) 5/3 sq units 3) 1/3 sq units 4) 2/3 sq units Solution: (1) 4/3 sq units... View Article
The area of the region between the curves y = √(1 + sin x) / (cos x) and y = √(1 – sin x) / cos x bounded by the lines x = 0 and x = Ï€ / 4 is 1) ∫0√2-1 {t / [(1 + t2) (√1 - t2)]} dt 2) ∫0√2-1 {4t / [(1 + t2) (√1 - t2)]} dt 3) ∫0√2+1 {4t / [(1 + t2) (√1 - t2)]} dt 4) ∫0√2+1 {t /... View Article
The area bounded by the curve y = x|x|, x-axis and the ordinates x = 1, x = -1 is given by 1) 0 2) 1/3 3) 2/3 4) None of these Solution: (3) â…”... View Article
The area of the region bounded by the curves y = ex, y = loge x and lines x = 1, x = 2 is 1) (e - 1)2 2) e2 - e + 1 3) e2 - e + 1 - 2 loge 2 4) e2 + e - 2 loge 2 Solution: (3) e2 - e + 1 - 2 loge 2... View Article
The line x = π/4 divides the area of the region bounded by y = sin x, y = cos x and x-axis (0 ≤ x ≤ π/2) into two regions of areas A1 and A2. Then, A1 : A2 equals to 1) 4 : 1 2) 3 : 1 3) 2 : 1 4) 1 : 1 Solution: (4) 1 : 1... View Article
If f (x) is continuous function such that the area bounded by the curve y = f (x), the x-axis and the lines x = a and x = 0 is (a2 / 2) + (a / 2) sin a + (π / 2) cos a. Then the value of f (π / 2) is 1) 1/2 2) a/2 3) a2/2 4) π/2 Solution: (1) ½... View Article
The area of the region bounded by curves y = x log x and y = 2x – 2x2 is 1) 1/2 sq units 2) 3/12 sq units 3) 7/12 sq units 4) None of these Solution: (3) 7/12 sq units... View Article
The area bounded by = |sin x|, x-axis and the lines |x| = π is 1) 2 sq units 2) 3 sq units 3) 4 sq units 4) None of these Solution: (3) 4 sq units |x| = π x = - π, π Area = 2 ∫0π sin x dx = 2 [- cos x]0π = - 2 [cos x]0π = - 2 (- 1 - 1) = 4... View Article
The given figure shows a ∆AOB and the parabola y = x2. The ratio of the area of the ∆AOB to the area of the region AOB of the parabola y = x2 is equal to 1) 3/5 2) 3/4 3) 7/8 4) 5/6 5) 2/3 Solution: (2) ¾... View Article
The area bounded between the parabola y2 = 4x and the line y = 2x – 4 is equal to 1) 17/3 sq units 2) 19/3 sq units 3) 9 sq units 4) 15 sq units Solution: (3) 9 sq units... View Article
The area bounded by the curves y = √5 – x2 and y = |x – 1| is 1) [(5π / 4) - 2] square units 2) [(5Ï€ - 2) / 4] square units 3) [(5Ï€ - 2) / 2] square units 4) [(π / 2) - 5] square units Solution:... View Article
