Given that frac{d}{d x}left[int_0^{phi(x)} f( | vedclass

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(A) Given the integral equation: $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$.Applying the Leibniz rule for differentiation under the integral sign on

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$\frac{2 \pi}{3}$

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(A) Given the integral equation: $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$.Applying the Leibniz rule for differentiation under the integral sign on both sides with respect to $x$:$\frac{d}{d x}\left[\int_0^{x^3} f(t) d tight] = \frac{d}{d x}\left[x^2 \sin(2 \pi x)ight]$$f(x^3) \cdot \frac{d}{dx}(x^3) = 2x \sin(2 \pi x) + x^2 \cdot \cos(2 \pi x) \cdot 2 \pi$$f(x^3) \cdot 3x^2 = 2x \sin(2 \pi x) + 2 \pi x^2 \cos(2 \pi x)$Dividing both sides by $3x^2$ (for $x 
eq 0$):$f(x^3) = \frac{2x \sin(2 \pi x) + 2 \pi x^2 \cos(2 \pi x)}{3x^2}$$f(x^3) = \frac{2}{3x} \sin(2 \pi x) + \frac{2 \pi}{3} \cos(2 \pi x)$To find $f(8)$,we set $x^3 = 8$,which implies $x = 2$:$f(8) = \frac{2}{3(2)} \sin(4 \pi) + \frac{2 \pi}{3} \cos(4 \pi)$Since $\sin(4 \pi) = 0$ and $\cos(4 \pi) = 1$:$f(8) = \frac{1}{3}(0) + \frac{2 \pi}{3}(1) = \frac{2 \pi}{3}$.

Given that $\frac{d}{d x}\left[\int_0^{\phi(x)} f(t) d t\right]=f(\phi(x)) \cdot \phi^{\prime}(x)$. If $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$,then the value of $f(8)$ is

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