To solve the problem, we will first derive the Cumulative Distribution Function (CDF) for the random variable $Y_1 = X^3$ given the Probability Density Function (PDF) of $X$, and then find the PDF of $Y_1$.

Step 1: Derive the CDF of $Y_1$ The CDF of $Y_1$, denoted as $F_{Y_1}(y)$, is the probability that $Y_1 \leq y$. Given that $Y_1 = X^3$, this is equivalent to calculating $P(X^3 \leq y)$. We start by considering $y$ values and their corresponding $x$ intervals.

For $Y_1 = X^3$ where $X \in [-1, 2]$:

• If $y < -1$, then there are no $x$ values such that $x^3 \leq y$ because the minimum cube value $x$ can take is $-1^3 = -1$.
• If $-1 \leq y \leq 1$, then $-\sqrt[3]{y} \leq x \leq \sqrt[3]{y}$ (considering $x$ values between -1 and 1).
• If $y > 1$, then $-1 \leq x \leq \sqrt[3]{y}$ (since $x \leq 2$ and $2^3 = 8$).

CDF Calculation: $F_{Y_1}(y) = P(X^3 \leq y)$ [ = \int_{-\sqrt[3]{y}}^{\sqrt[3]{y}} f_X(x) , dx ] when $-1 \leq y \leq 1$ $= \int_{-1}^{\sqrt[3]{y}} f_X(x) \, dx$ when $1 < y \leq 8$

Given $f_X(x) = \frac{x^2}{3}$ for $x \in [-1, 2]$, we compute the integrals:

For $-1 \leq y \leq 1$: [ F_{Y_1}(y) = \int_{-\sqrt[3]{y}}^{\sqrt[3]{y}} \frac{x^2}{3} , dx ] $= \frac{2}{3} \int_{0}^{\sqrt[3]{y}} x^2 \, dx$ $= \frac{2}{3} \cdot \frac{x^3}{3} \Big|_{0}^{\sqrt[3]{y}}$ $= \frac{2}{9} y$

For $1 < y \leq 8$: $F_{Y_1}(y) = \int_{-1}^{\sqrt[3]{y}} \frac{x^2}{3} \, dx$ $= \frac{1}{3} \left( \frac{x^3}{3} \right) \Big|_{-1}^{\sqrt[3]{y}}$ $= \frac{1}{9} \left( (\sqrt[3]{y})^3 + 1 \right)$ $= \frac{y + 1}{9}$

Step 2: Derive the PDF of $Y_1$ The PDF, $f_{Y_1}(y)$, is the derivative of the CDF: $f_{Y_1}(y) = \frac{d}{dy} F_{Y_1}(y)$

For $-1 \leq y \leq 1$: $f_{Y_1}(y) = \frac{d}{dy} \left( \frac{2}{9} y \right) = \frac{2}{9}$

For $1 < y \leq 8$: $f_{Y_1}(y) = \frac{d}{dy} \left( \frac{y + 1}{9} \right) = \frac{1}{9}$

Summary: