The area of the region bounded by $y = -1$, $y = 2$, $x = y^3$ and $x = 0$ is $\frac{m}{n}$ sq. units, where $\gcd(m, n) = 1$, then $m - n$ is equal to:

The area $A$ of a region bounded by a curve $x = f(y)$, the $y$-axis ($x = 0$), and the horizontal lines $y = c$ and $y = d$ is given by the integral:

$A = \int_{c}^{d} |f(y)| \, dy$

Identify the integral limits and function

The boundaries are $y = -1$ and $y = 2$ with the curve $x = y^3$. Since the curve $x = y^3$ crosses the $y$-axis at $y = 0$ (where it changes from negative to positive), we must split the integral to find the total area:

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

Evaluate the integrals

For the first part ($y$ is negative):

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

For the second part ($y$ is positive):

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

Calculate the total area

$A = \frac{1}{4} + 4 = \frac{1 + 16}{4} = \frac{17}{4}$

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Find $m - n$:

We are given that the area is $\frac{m}{n}$ where $\gcd(m, n) = 1$.

Comparing $\frac{17}{4}$ with $\frac{m}{n}$:

$m = 17$

$n = 4$

$\gcd(17, 4) = 1$ (This condition is satisfied).

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The value of $m - n$ is 13.