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

$\frac{17}{4}$

The correct answer is Option (3) → $\frac{17}{4}$

The curve is $x = y^3$. For $y \in [-1,0]$, $y^3$ is negative, so the region lies to the LEFT of the y–axis. For $y \in [0,2]$, $y^3$ is positive, so the region lies to the RIGHT of the y–axis.

Area must always be taken as positive, so:

Area $=\displaystyle \int_{-1}^{0} |y^3|\,dy + \int_{0}^{2} y^3\,dy$

First part (negative region):

$\int_{-1}^{0} -y^3\,dy = \left[ -\frac{y^4}{4} \right]_{-1}^{0} = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4}$

Second part (positive region):

$\int_{0}^{2} y^3\,dy = \left[\frac{y^4}{4}\right]_{0}^{2} = \frac{16}{4} = 4$

Total area = $\frac{1}{4} + 4 = \frac{17}{4}$

The area of the region is $\frac{17}{4}$ square units.