Let a differentiable function f satisfy the equation $$\int_{0}^{36}f(\frac{tx}{36})dt=4\alpha f(x)$$. If y=f(x) is a standard parabola passing through the points (2, 1) and $$(-4,\beta)$$, then $$\beta^{\alpha}$$ is equal to______.

Given $$\int_0^{36} f\left(\frac{tx}{36}\right)dt = 4\alpha f(x)$$ 

the parabola $$y=f(x)$$ passes through $$(2,1)$$ and $$(-4,\beta)$$, and we seek $$\beta^\alpha$$.

$$f(x)=ax^2$$ and substituting into the integral yields $$\int_0^{36} a\left(\frac{tx}{36}\right)^2dt = 4\alpha\cdot ax^2$$, which becomes $$\frac{ax^2}{36^2}\int_0^{36}t^2dt = 4\alpha\,ax^2$$.

$$\frac{ax^2}{1296}\cdot\frac{36^3}{3} = 4\alpha\,ax^2$$, so $$\frac{36^3}{3\times1296} = 4\alpha \implies \frac{46656}{3888} = 4\alpha \implies 12 = 4\alpha \implies \alpha = 3$$.

The parabola passes through $$(2,1)$$, so $$f(2)=a\cdot4=1\implies a=\tfrac14$$. 

At $$x=-4$$ it follows that $$\beta=f(-4)=\tfrac14\cdot16=4$$.

Therefore, $$\beta^\alpha = 4^3 = 64$$ and the final answer is 64.