The area of the region \(x,y):y^2≤4x,x0,x≠3\ is

(a) : The area of region xy((x-1)(x-2) / (x-3)(x-4)), \ y^2 - 4x le 0 \ => yx(x-1)(x-2)(x-3)(x-4) 0 So, x(x-1)(x-2)(x-3)(x-4) 0 \ => x in (1, 2) union (3, 4) Area = integral _0^4 2sqrt(x)dx = (2 / 3) x 16 => (32 / 3)

The area of the region (x,y):y^2≤4x,x<4,\frac{xy(x-1)(x... | Area of Bounded Region (Mathematics)

(a) : The area of region $xy\frac{(x-1)(x-2)}{(x-3)(x-4)}$ , $\\ y^2 - 4x \le 0$ $\\ \Rightarrow yx(x-1)(x-2)(x-3)(x-4) < 0 \\$ Case 1: $y > 0 \\$ So, $x(x-1)(x-2)(x-3)(x-4) < 0$ $\\ \Rightarrow x \in (0, 1) \cup (2, 3)\\$ Case 2: $y < 0 \\$ So, $x(x-1)(x-2)(x-3)(x-4) > 0 \\ \Rightarrow x \in (1, 2) \cup (3, 4)\\$ Area $= \int_{0}^{4} 2\sqrt{x}dx = \frac{2}{3} \times 16 \Rightarrow \frac{32}{3}$

Explanation

The area of the region enclosed by the parabolas y=4x-x^2 and 3y=(x-4)^2 is equal to

Area of the region (x,y):x^2+(y-2)^2≤4,x^2≥2y is

The area of the region {(x,y):x^2 \leq y \leq \mid x^2-4 \mid, y ? 1} is

The area enclosed by the curves y^2+4x=4 \ and \ y-2x=2 is

The area of the region enclosed by the curve f(x)=max\{sin⁡x,cos⁡x\},- π ≤x ≤ π and the x-axis is

The area enclosed by the curves xy+4y=16 and x+y=6 is equal to :

The parabola y^2=4x divides the area of the circle x^2+y^2=5 in two parts. The area of the smaller part is equal to:

Let the area of the region enclosed by the curves y=3x,2y=27-3x and y=3x-x \sqrt x be A. Then 10 A is equal to

The area of the region in the first quadrant inside the circle x^2+y^2=8 and outside the parabola y^2=2x is equal to:

The area of the region enclosed by the curve y=x^3 and its tangent at the point (-1,-1) is

Download Divine JEE App

Quick Links

Exam Categories

Contact Info

Divine JEE