Calculus

Application of Integrals

Easy

common

The area (in sq. units) of the bounded region $\{(x, y): x^2 \le y \le x\}$ is

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{5}{6}$

✅ Correct Option: 1

The curves $y=x^2$ and $y=x$ meet at $x=0$ and $x=1$ , with the line above the parabola between them.

Area $= \int_0^1 (x - x^2)\,dx = 1/2 - 1/3 = 1/6$ .

2026: 14 May Shift 2

The area of the region bounded by the curve $y=x^3, y=8$ and $x=0$ is equal to :

2025: 21 May Shift 1

The area (in square units) bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$ in first quadrant is equal to:

2024: 16 May Shift 1

The area of the region bounded by the lines $\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4, x=0$ and $y=0$ is :

2025: 22 May Shift 1

Area (in sq. units) of the region bounded by the curve $y^2 = 4x$ , y-axis and the line $y = 3$ is

2026: 18 May Shift 1

The area (in sq. units) bounded by a parabola $y^2=4x$ and its latus rectum, is

2025: 16 May Shift 1

The area (sq.units) bounded by the curve y = sinx, π ≤ x ≤ 2π and the x-axis is

The area (in sq. units) of the bounded region {(x,y):x2≤y≤x}\{(x, y): x^2 \le y \le x\}{(x,y):x2≤y≤x} is