Calculus

Application of Integrals

Medium

core

The area (in square units) of the region bounded by the curve $x = y^3, y = -1, y = 2$ and $x = 0$ is:

17/4

15/4

13/4

11/4

✅ Correct Option: 1

Integrate with respect to $y$ using absolute values since $x=y^3$ is negative for $y<0$ .

Area $=\int_{-1}^{0}(-y^3)dy+\int_0^2 y^3dy = \dfrac14 + 4 = \dfrac{17}{4}$ square units.

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

2025: 27 May Shift 1

The area of the region bounded by the curves $y = x$ and $y = x^3$ is:

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:

2022: 6 Aug Shift 2

The maximum height (in meters) achieved in the first jump is

2025: 26 May Shift 2

Area (in sq. units) of the region bounded by the curves $y = -1$ , $y = 2$ , $x = y^3$ and $x = 0$ is

2026: 19 May Shift 2

Area bounded by the curves $y = 3^x, y = x$ and $x = 1, x = 2$ is:

The area (in square units) of the region bounded by the curve x=y3,y=−1,y=2x = y^3, y = -1, y = 2x=y3,y=−1,y=2 and x=0x = 0x=0 is: