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

The region is bounded by the horizontal lines $y = -1$ and $y = 2$, the y-axis $x = 0$, and the curve $x = y^3$.

Since the boundaries are horizontal lines, integrate with respect to $y$ from $y = -1$ to $y = 2$.

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The curve $x = y^3$ behaves as follows:

• When $y = -1$: $x = (-1)^3 = -1$
• When $y = 0$: $x = 0^3 = 0$
• When $y = 2$: $x = 2^3 = 8$

The curve crosses the y-axis at $y = 0$, switching from left to right.

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For $y \in [-1, 0]$:

Left boundary: $x = y^3$ (negative values)

Right boundary: $x = 0$

Width: $0 - y^3 = -y^3$

$A_1 = \int_{-1}^{0} (-y^3) \, dy$

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

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

$= 0 + \frac{1}{4}$

$= \frac{1}{4}$

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For $y \in [0, 2]$:

Left boundary: $x = 0$

Right boundary: $x = y^3$ (positive values)

Width: $y^3 - 0 = y^3$

$A_2 = \int_{0}^{2} y^3 \, dy$

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

$= \frac{2^4}{4} - \frac{0^4}{4}$

$= \frac{16}{4}$

$= 4$

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Total Area $= A_1 + A_2$

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

$= \frac{1}{4} + \frac{16}{4}$

$= \frac{17}{4}$

Therefore, the area of the region is $\dfrac{17}{4}$ square units.