Calculus

Application of Integrals

Hard

core

The area of the shaded portion is :

$\left(\pi - \frac{8}{3}\right)$ sq. units

$\left(\frac{8}{3} + \pi\right)$ sq. units

$\left(\frac{\pi}{2} - \frac{4}{3}\right)$ sq. units

$\left(\frac{\pi}{2} + \frac{4}{3}\right)$ sq. units

✅ Correct Option: 1

Curves intersect at $(0,0)$ and $(2, 2)$ . Integrating w.r.t. $y$ from 0 to 2:

Area $= \int_0^2 \left[\frac{y^2}{2} - (2 - \sqrt{4-y^2})\right]dy$

$= \frac{4}{3} - 4 + \pi = \pi - \frac{8}{3}$ .

2026: 19 May Shift 2

The area bounded by the curve $|-x|+|-y|=2$ is

2022: 4 Aug Shift 1

The area of the region bounded by the curve $x^2 = 4y$ and the straight line $x = 4y - 2$ is:

2025: 3 June Shift 1

The area of the smaller region of the circle $x^2 + y^2 = 8$ cut off by the line $x = 2$ is

2026: 31 May Shift 2

The area enclosed between $y^2 = 4x$ , $x = 1$ and $x = 4$ in first quadrant is :

2026: 29 May Shift 1

The area (in sq. units) of the region bounded by $x^2=4y, y=1, y=9$ and $x=0$ in the first quadrant is :

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 :

Solution