Area of region bounded by the curves $x=y^3, x=0$ between $y = -1$ and $y = 2$ is:

$\frac{17}{4}$ sq. units

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

$Area = \int_{-1}^{2} ( \,x_{\text{right}} - x_{\text{left}}\, )\,dy$

$For -1\le y\le0: x_{\text{right}}=0,\;x_{\text{left}}=y^3. \quad$ For $0\le y\le2: x_{\text{right}}=y^3,\;x_{\text{left}}=0.$

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

$= -\left[\frac{y^4}{4}\right]_{-1}^{0} + \left[\frac{y^4}{4}\right]_{0}^{2}$

$= -\left(0 - \frac{1}{4}\right) + \left(\frac{16}{4} - 0\right)$

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