If int_0^{t^2} xf(x)dx = frac{2}{5}t^5, t 0 | vedclass

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(A) Given the equation $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5$. Applying the Leibniz rule for differentiation under the integral sign with respect to

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$\frac{2}{5}$

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(A) Given the equation $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5$. Applying the Leibniz rule for differentiation under the integral sign with respect to $t$: $\frac{d}{dt} \left( \int_0^{t^2} xf(x)dx \right) = \frac{d}{dt} \left( \frac{2}{5}t^5 \right)$. Using the chain rule,we get: $(t^2)f(t^2) \cdot \frac{d}{dt}(t^2) = \frac{2}{5} \cdot 5t^4$. $(t^2)f(t^2) \cdot (2t) = 2t^4$. $2t^3 f(t^2) = 2t^4$. For $t > 0$,we can divide by $2t^3$: $f(t^2) = t$. We need to find $f\left( \frac{4}{25} \right)$. Let $t^2 = \frac{4}{25}$,which implies $t = \frac{2}{5}$ (since $t > 0$). Substituting $t = \frac{2}{5}$ into $f(t^2) = t$,we get: $f\left( \frac{4}{25} \right) = \frac{2}{5}$.

If $\int_0^{t^2} xf(x)dx = \frac{2}{5}t^5, t > 0,$ then $f\left( \frac{4}{25} \right) = $

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