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(C) Given the equation: $\int_0^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$.Let $u = \frac{tx}{36}$, then $du = \frac{x}{36} dt$, so $dt = \fr

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$64$

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(C) Given the equation: $\int_0^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)$.Let $u = \frac{tx}{36}$, then $du = \frac{x}{36} dt$, so $dt = \frac{36}{x} du$.When $t=0, u=0$ and when $t=36, u=x$.The integral becomes: $\int_0^x f(u) \cdot \frac{36}{x} du = 4\alpha f(x)$.$\int_0^x f(u) du = \frac{4\alpha x f(x)}{36} = \frac{\alpha x f(x)}{9}$.Differentiating both sides with respect to $x$ using the Leibniz rule:$f(x) = \frac{\alpha}{9} [f(x) + x f'(x)]$.$f(x) = \frac{\alpha}{9} f(x) + \frac{\alpha x}{9} f'(x)$.$(1 - \frac{\alpha}{9}) f(x) = \frac{\alpha x}{9} f'(x) \Rightarrow (9 - \alpha) f(x) = \alpha x f'(x)$.$\frac{f'(x)}{f(x)} = \frac{9 - \alpha}{\alpha} \cdot \frac{1}{x}$.Integrating both sides: $\ln|f(x)| = (\frac{9}{\alpha} - 1) \ln|x| + C$.$f(x) = c x^{(\frac{9}{\alpha} - 1)}$.Since $f(x)$ is a standard parabola, the exponent must be $2$, so $\frac{9}{\alpha} - 1 = 2 \Rightarrow \frac{9}{\alpha} = 3 \Rightarrow \alpha = 3$.Thus, $f(x) = cx^2$.Since it passes through $(2, 1)$, $1 = c(2)^2 \Rightarrow c = \frac{1}{4}$.So, $f(x) = \frac{x^2}{4}$.For the point $(-4, \beta)$, $\beta = \frac{(-4)^2}{4} = \frac{16}{4} = 4$.Finally, $\beta^{\alpha} = 4^3 = 64$.

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 . . . . . . .

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