Suppose the pdf of $X$ is $f(x) = \frac{-3}{4} (x-3)(x-5)$. Find the pdf of $X^2/8$. Suppose that $X$ is a continuous random variable with a pdf $f(x) =
 \frac{-3}{4} (x-3)(x-5)$ with $3\leq x \leq 5$. What is the pdf of the random variable $Y$,
  where $Y = X^2/8$.


My attempt:
First I notice that if $X\in[3,5]$, then $Y\in[9/8, 25/8]$.

I derived the cumulative distribution function as follows:
$$P(Y&lt;y) = P(X^2&lt;8y)=P(-\sqrt{8y}&lt;X&lt;\sqrt{8y})=\int_{-\sqrt{8y}}^{\sqrt{8y}}{f(x)dx}=-\sqrt{2y}(8y+45)$$

Then differentiate this cdf to obtain the pdf of $Y$:$$\frac{d}{dy}(-\sqrt{2y}(8y+45)) = -\frac{3(8y+15)}{\sqrt{2y}}$$

However, when I tried to integrate this pdf:  $\int_{9/8}^{25/8} {-\frac{3(8y+15)}{\sqrt{2y}}dy} =-94\neq1$

It turns out that the integral does not equal to 1, which suggests that there is something wrong with my answer. Any help would be highly appreciated.
$$\int_{-\sqrt{8y}}^{\sqrt{8y}} f(x)\ dx = \int_{-\sqrt{8y}}^{3} f(x)\ dx +\int_{3}^{\sqrt{8y}} f(x)\ dx$$

The first term is zero because $f(x)=0$ in that range. If $3\le\sqrt{8y}\le5$, then you can go on and integrate the second term.
$$\int_{-\sqrt{8y}}^{\sqrt{8y}} f(x)\ dx = \int_{-\sqrt{8y}}^{3} f(x)\ dx +\int_{3}^{\sqrt{8y}} f(x)\ dx$$

The first term is zero because $f(x)=0$ in that range. If $3\le\sqrt{8y}\le5$, then you can go on and integrate the second term.