If $f:[0,\infty)\to\mathbb{R}$ is a decreasing differentiable function, prove that $\lim_{x\to\infty} x^4 f(x)=0$.

I am trying to solve the following problem:

Let $f:[0,\infty)\to\mathbb{R}$ be a decreasing differentiable function such that $f(0)=1$, $\lim_{x\to\infty} f(x)=0$ and $\int_0^{\infty}x^4f'(x)dx < \infty$. Prove that $\lim_{x\to\infty} x^4 f(x)=0$.

I tried to use a technique that I saw in another problem: Because $f$ is decreasing and $f(x)\to 0$ as $x\to\infty$, we know that $f$ is nonnegative. If we have that$\int_0^{\infty}x^3f(x)dx < \infty$ and $x>0$, then\begin{align}0&\leq (5^{1/4}x)^4f(5^{1/4}x) = 5x^4f(5^{1/4}x) = 5f(5^{1/4}x) \int_x^{5^{1/4}x} t^3dt = 5 \int_x^{5^{1/4}x} t^3f(5^{1/4}x)dt \leq \\ &\leq 5 \int_x^{5^{1/4}x} t^3f(t)dt = 5\left( \int_0^{5^{1/4}x} t^3f(t)dt - \int_0^{x} t^3f(t)dt\right)\to 0 \text{ as } x\to\infty.\end{align}Notice that we used the hypothesis that $f$ is decreasing in the inequality. By taking $u=5^{1/4}x$, we finish the proof.

However, I don't know how to prove that $\int_0^{\infty}x^3f(x)dx$ is finite with the given hypothesis.