Calculus

Application of Integrals

Medium

core

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

18 Sq.units

8 Sq.units

12 Sq.units

10 Sq.units

✅ Correct Option: 3

Integrate along the $y$ -axis, where $x=y^{1/3}$ and $y$ runs from $0$ to $8$ .

Area $=\int_0^8 y^{1/3}dy=\left[\frac{3}{4}y^{4/3}\right]_0^8=\frac{3}{4}(16)=12$ square units.

2026: 15 May Shift 1

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

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Area bounded by the curves $y = \cos^2 x, x = 0, y = 0$ and $x = \pi$ , is:

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The area of the region bounded by the curve y=x3,y=8y=x^3, y=8y=x3,y=8 and x=0x=0x=0 is equal to :