Probability distribution function: $Y=X^2$ The probability distribution function of a random variable $X$ is given by $f(x).$ Another random variable $Y$ is related to $X$ by $Y=X^2.$
What is the probability of $Y$ being less than or equal to $b$ where $b\in[0,\infty)$
What I have done/ what I know,
EDIT
Using method of transformation suggested by @SeanRoberson,
\begin{align}P(X\le x) &= f(x) \\ 
P(Y\le b )&= P(X^2 \le b)\\
&= P(X \le b^{1/2}) \\
&=\int_0^{b^{1/2}}f(x)~ dx
\end{align}
Is this answer correct.?
and as the value of b is not given , numerical answer is not possible.
If I am missing something please help me .
 A: $P(X^2\le b)=P(-\sqrt{b}\le X\le\sqrt{b})=\int\limits_{-\sqrt{b}}^{\sqrt{b}}f(x)dx$.
You have to include case where $X$ is negative.
A: Thanks for your edits, showing your thoughts. This is not a 1 to 1 transformation.
Practical example:
Let $X\sim\mathsf{Norm}(0,1),$ $Y \sim\mathsf{Chisq}(\nu = 1).$
Then $P(Y \le 3.8416)$ $= P(|X| \le 1.96)$ $= P(-1.96 \le X \le 1.96) \approx 0.95.$
In cae you're familiar with R statistical software, see below. If not look at printed tables of the standard normal CDF and of chi-squared distributions.
pchisq(3.8416, 1)
[1] 0.9500042
pnorm(1.96) - pnorm(-1.96)
[1] 0.9500042

A: I don't know why people are answering this without pointing out conspicuous problems with the way the question is phrased. You wrote:

The probability distribution function of a random variable $X$ is given by $f(x).$

Usually "probability distribution function", if that term is used, means the cumulative distribution function (c.d.f.), often denoted by (capital) $F,$ so that you have $\Pr(X\le x) = F(x),$ and indeed you wrote $\Pr(X\le x) = f(x),$ suggesting that you intended $f$ to be the c.d.f. But then you wrote

$$P(Y\le b )= \cdots =\int_0^{b^{1/2}}f(x)~ dx$$

suggesting you intended $f$ to be the probability density function (p.d.f.) instead.
Then there is also the fact that $\big[X^2\le b\big]$ is equivalent to $\big[ -\sqrt b \le X\le b\big].$
