If for a continuous function f(x), int_{-pi}^ | vedclass

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(A) Given the equation: $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$Splitting the integral,we get: $\int_{-\pi}^{t} f(x) dx + \int_{-\pi}^{t} x dx =

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$\pi$

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(A) Given the equation: $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$Splitting the integral,we get: $\int_{-\pi}^{t} f(x) dx + \int_{-\pi}^{t} x dx = \pi^2 - t^2$Evaluating the second integral: $\int_{-\pi}^{t} x dx = \left[ \frac{x^2}{2} \right]_{-\pi}^{t} = \frac{t^2}{2} - \frac{(-\pi)^2}{2} = \frac{t^2}{2} - \frac{\pi^2}{2}$Substituting this back: $\int_{-\pi}^{t} f(x) dx + \frac{t^2}{2} - \frac{\pi^2}{2} = \pi^2 - t^2$Rearranging the terms: $\int_{-\pi}^{t} f(x) dx = \pi^2 - t^2 - \frac{t^2}{2} + \frac{\pi^2}{2} = \frac{3}{2}\pi^2 - \frac{3}{2}t^2 = \frac{3}{2}(\pi^2 - t^2)$Differentiating both sides with respect to $t$ using the Leibniz rule: $\frac{d}{dt} \left[ \int_{-\pi}^{t} f(x) dx \right] = \frac{d}{dt} \left[ \frac{3}{2}(\pi^2 - t^2) \right]$By the Fundamental Theorem of Calculus: $f(t) = \frac{3}{2}(0 - 2t) = -3t$Therefore,$f\left(-\frac{\pi}{3}\right) = -3 \left(-\frac{\pi}{3}\right) = \pi$

If for a continuous function $f(x),$ $\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$ for all $t \ge -\pi,$ then $f\left(-\frac{\pi}{3}\right)$ is equal to

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