Given the equation $\int_{0}^{36}f\left(\frac{tx}{36}\right)dt=4\alpha f(x)$, we need to find $\beta^\alpha$ for $f(x)$ being a parabola passing through $(2,1)$ and $(-4,\beta)$.
1. **Identify $f(x)$:** Assume $f(x)=ax^2+bx+c$. Since $f(x)$ is a standard parabola (symmetric around $y$-axis), $b=0$. This gives $f(x)=ax^2+c$.
2. **Apply the points:**
(a) At $(2,1)$: $4a+c=1$
(b) At $(-4,\beta)$: $16a+c=\beta$.
Solving these:
$c=1-4a$
$\beta=16a+c=16a+(1-4a)=12a+1$.
3. **Work with the integral:**
Substitute $u=\frac{tx}{36}$, then $t=36u/x$, and $dt=36/x \, du$. The limits for $t$ from 0 to 36 imply $u$ goes from 0 to 1. The integral becomes:
$\int_{0}^{1}f(xu)\frac{36}{x}du=4\alpha f(x)$.
Thus,
$\frac{36}{x}\int_{0}^{1} (ax^2u^2+c)du=4\alpha(ax^2+c)$.
4. **Evaluate the integral:**
$\int_{0}^{1}(ax^2u^2+c)du= ax^2\int_{0}^{1}u^2du + c\int_{0}^{1}du$.
Calculate $\int_{0}^{1}u^2du=\frac{1}{3}$ and $\int_{0}^{1}du=1$.
Thus,
$\frac{36}{x}(ax^2\frac{1}{3}+c)=4\alpha(ax^2+c)$.
5. **Resulting equation:**
$12(ax^2/3+c)=4\alpha(ax^2+c)$.
At $x=0$, $12c=4\alpha c\rightarrow \alpha=3$.
6. **Substitute $\alpha$ in $c$ and $\beta$:**
$\beta=12a+1=4$, as $1-4a=4\rightarrow a=-\frac{3}{4}$.
$\beta=4$, $\alpha=3$ implies $\beta^\alpha=4^3=64$.
7. **Validation:**
The computed value 64 fits the range (64,64).
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