Theorem: Let X have a pdf $f_X(x)$ and let Y=g(X), where g is monotonic (either strictly increasing or strictly decreasing) so it has an inverse function X=h(Y). Assume that h has derivative h'(y). Then $f_X(h(y))|h'(y)|$
Proof: Here is the proof assuming that g is monotonically increasing.
$F_Y(y)=P(Y \leq y)=P[g(X) \leq y]=P[X \leq h(y)]=F_X[h(y)]$
Now differentiate the cdf, letting x=h(y).
$\frac{d}{dy}F_X[h(y)]=h'(y)f_X[h(y)]$
My problem:
I want to apply the equality $P[g(X) \leq y]=P[X \leq h(y)]$ in the proof to an example to show its validity. Here is what I have tried:
If we look at the standard normal distribution with $\mu=0$ and $\sigma=1$. Then we apply integration limits 0 to 10 for $x$. We have that $g(X)=x^3$
$\int_{0}^{10} \frac{1}{\sqrt{2 \pi}} e^-{\frac{x^2}{2}} dx=0.5$
and for the inverse $\sqrt[3]{x}$ 0 to $\sqrt[3]{10}=1.78$
$\int_{0}^{1.78} \frac{1}{\sqrt{2 \pi}} e^-{\frac{\sqrt[3]{x}^2}{2}} dx=0.466$
But $0.5 \neq 0.466$
What is wrong. Can someone show how I can use the equality $P[g(X) \leq y]=P[X \leq h(y)]$ on this example. And point out what is wrong?