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

The area is enclosed by:

• $y = -1$ (bottom horizontal line)
• $y = 2$ (top horizontal line)
• $x = y^3$ (curved line)
• $x = 0$ (the y-axis)

Since the boundaries are given as functions of $y$, integrate with respect to $y$.

---

The curve $x = y^3$ position relative to the y-axis:

At $y = -1$: $x = (-1)^3 = -1$ (left of y-axis)

At $y = 0$: $x = 0^3 = 0$ (on the y-axis)

At $y = 2$: $x = 2^3 = 8$ (right of y-axis)

The curve crosses from left to right at $y = 0$, so the region must be split at this point.

---

For $y = -1$ to $y = 0$:

Right boundary: $x = 0$

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

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

For $y = 0$ to $y = 2$:

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

Left boundary: $x = 0$

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

---

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

---

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

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

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

$= \frac{1}{4}$

---

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

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

$= \frac{16}{4}$

$= 4$

---

$\text{Total Area} = \frac{1}{4} + 4$

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

$= \frac{17}{4}$

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