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(A) Given the equation $\int \limits_0^{t^2} (f(x) + x^2) dx = \frac{4}{3} t^3$. Applying the Leibniz integral rule by differentiating both sides with

Vedclass · Accepted Answer

$\pi \left(1 - \frac{\pi^3}{16}\right)$

Vedclass · Answer

(A) Given the equation $\int \limits_0^{t^2} (f(x) + x^2) dx = \frac{4}{3} t^3$. Applying the Leibniz integral rule by differentiating both sides with respect to $t$: $\frac{d}{dt} \left( \int \limits_0^{t^2} (f(x) + x^2) dx \right) = \frac{d}{dt} \left( \frac{4}{3} t^3 \right)$. Using the chain rule: $(f(t^2) + (t^2)^2) \cdot \frac{d}{dt}(t^2) = 4t^2$. $(f(t^2) + t^4) \cdot 2t = 4t^2$. Since $t > 0$,we can divide by $2t$: $f(t^2) + t^4 = 2t$. $f(t^2) = 2t - t^4$. To find $f \left(\frac{\pi^2}{4}\right)$,we set $t^2 = \frac{\pi^2}{4}$,which implies $t = \frac{\pi}{2}$ (since $t > 0$). Substituting $t = \frac{\pi}{2}$ into the expression for $f(t^2)$: $f \left(\frac{\pi^2}{4}\right) = 2 \left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^4$. $f \left(\frac{\pi^2}{4}\right) = \pi - \frac{\pi^4}{16}$. Factoring out $\pi$: $f \left(\frac{\pi^2}{4}\right) = \pi \left(1 - \frac{\pi^3}{16}\right)$.

Let $f$ be a continuous function satisfying $\int \limits_0^{t^2} (f(x) + x^2) dx = \frac{4}{3} t^3, \forall t > 0$. Then $f \left(\frac{\pi^2}{4}\right)$ is equal to:

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