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

$\frac{17}{4}$

The correct answer is Option (2) → $\frac{17}{4}$

Region bounded by: $y=-1$, $y=2$, $x=y^3$, $x=0$

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

Since $y^3$ is negative for $y < 0$ and positive for $y > 0$, split the integral:

$A = \int_{-1}^{0} (-y^3) \, dy + \int_{0}^{2} y^3 \, dy$

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

$\int_{0}^{2} y^3 \, dy = \left[ \frac{y^4}{4} \right]_0^2 = \frac{16}{4} - 0 = 4$

Total area: $A = \frac{1}{4} + 4 = \frac{17}{4} \, \text{sq. units}$