Probability Density Function of $Y=X^3$

Question

$X \sim \text{Uniform}(-\frac{1}{2},\frac{1}{2})$, $Y=X^3$. To find the PDF of $Y$, I did the following:

$$F_Y(y) = P(Y\leq y) = P(X^3 \leq y) = P(-y^{1/3} \leq x \leq y^{1/3}) $$

Then $$ = \int_{-y^{1/3}}^{y^{1/3}} f_x(x) dx = 2y^{1/3}$$

Thus I get $f_y(y) = 2/3y^{-2/3}$. However the correct answer is $1/3y^{-2/3}$. It seems my bounds should be from $0$ to $y^{1/3}$ instead of $-y^{1/3}$ to $y^{1/3}$. Why is this the case when $x \in (-\frac{1}{2}, \frac{1}{2})$?

Answer 1

$ P(X^3 \leq y) = P(-y^{1/3} \leq X \leq y^{1/3})$

This is incorrect. $x\mapsto x^3$ is monotone. So:

$ P(X^3 \leq y) = P(X \leq y^{1/3})$