Calculus

Application of Integrals

Easy

common

Area of the region bounded by the curves $y = x^2$ , $x=0$ , $x=2$ and $x-axis$ is

$\frac{16}{3}$ sq. units

$\frac{8}{3}$ sq. units

$\frac{13}{6}$ sq. units

$\frac{13}{3}$ sq. units

✅ Correct Option: 2

The curve lies above the $x$ -axis on $[0,2]$ , so the area is $\int_0^2 x^2dx=\left[\frac{x^3}{3}\right]_0^2=\frac{8}{3}$ square units.

2025: 14 May Shift 2

Area of the region bounded by the curve $y^2 = 4x$ , $y$ -axis and the line $y = 3$ is equal to

2026: 21 May Shift 1

The area in the first quadrant bounded by $y = 9x^2$ , $x = 0$ , $y = 1$ and $y = 4$ is :

2025: 19 May Shift 1

The area (in square units) of the region bounded by the curves $3y^2 = ax$ , $y = a$ , $a > 0$ and $y$ -axis is:

2026: 15 May Shift 1

The area of the region bounded by the curves $y = 1+|x-1|, x=-1, x=2, y=0$ is

2025: 22 May Shift 2

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

2025: 29 May Shift 2

The area of the region bounded by the curves $y = x^2 + 2$ , $y = x$ , $x = 0$ and $x = 2$ is

Area of the region bounded by the curves y=x2y = x^2y=x2, x=0x=0x=0, x=2x=2x=2 and x−axisx-axisx−axis is