Calculus

Application of Integrals

Medium

common

The area (in sq. units) of the region in the first quadrant enclosed between the line $x + y = 2$ , parabola $y^2 = x$ and the x-axis, is

$\frac{7}{6}$

$\frac{4}{3}$

$\frac{2}{3}$

$\frac{7}{3}$

✅ Correct Option: 1

Solve $y^2 = 2-y$ to get $y=1$ (first quadrant), so the curves meet at $(1,1)$ .

Integrating horizontally is cleanest: for $y$ from $0$ to $1$ , $x$ runs from the parabola $x=y^2$ to the line $x=2-y$ .

$A=\int_0^1 (2-y-y^2)\,dy = 2-\frac{1}{2}-\frac{1}{3}=\frac{7}{6}$ .

2025: 21 May Shift 1

The area (in square units) of the region enclosed between the lines $x + y = 2$ , $x = 0$ , $x = 3$ and $x$ -axis is equal to

2023: 15 June Shift 2

Area of the region bounded by $y = -1$ , $y = 2$ , $x = y^3$ and $x = 0$ is :

2026: 30 May Shift 2

The area (in square units) bounded by the curve $y = x^2|x|$ , x-axis and the lines $x=-2$ and $x=2$ , is

2023: 15 June Shift 2

Area of the region bounded by the curve $|x| + |y| = 1$ and x-axis is :

2026: 21 May Shift 2

The area of region $R = \{(x,y): x^2 + y^2 \le 16, y \le \sqrt{3}x, x \ge 0, y \ge 0\}$ is: (in Sq. units)

2023: 22 May Shift 3

The area enclosed between the curve $x^2 + y^2 = 16$ and the coordinate axes in the first quadrant is :

The area (in sq. units) of the region in the first quadrant enclosed between the line x+y=2x + y = 2x+y=2, parabola y2=xy^2 = xy2=x and the x-axis, is