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

We are given that for a continuous function $$f(x)$$, the equation $$\int_{-\pi}^{t} (f(x) + x) dx = \pi^2 - t^2$$ holds for all $$t \geq -\pi$$. We need to find the value of $$f\left(-\frac{\pi}{3}\right)$$.

Since the equation is true for all $$t \geq -\pi$$, we can differentiate both sides with respect to $$t$$ to find an expression for $$f(t)$$. According to the Fundamental Theorem of Calculus, if $$F(t) = \int_{a}^{t} g(x) dx$$, then $$F'(t) = g(t)$$. Here, the left side is $$\int_{-\pi}^{t} (f(x) + x) dx$$, so we let $$g(x) = f(x) + x$$.

Differentiating both sides of the given equation with respect to $$t$$:

Left side derivative: $$\frac{d}{dt} \left[ \int_{-\pi}^{t} (f(x) + x) dx \right] = f(t) + t$$.

Right side derivative: $$\frac{d}{dt} \left[ \pi^2 - t^2 \right] = -2t$$.

So, we have:

$$f(t) + t = -2t$$

Now, solve for $$f(t)$$:

Subtract $$t$$ from both sides:

$$f(t) = -2t - t$$

$$f(t) = -3t$$

Therefore, the function is $$f(t) = -3t$$.

Now, substitute $$t = -\frac{\pi}{3}$$ to find $$f\left(-\frac{\pi}{3}\right)$$:

$$f\left(-\frac{\pi}{3}\right) = -3 \times \left(-\frac{\pi}{3}\right)$$

Simplify:

$$f\left(-\frac{\pi}{3}\right) = -3 \times -\frac{\pi}{3} = 3 \times \frac{\pi}{3} = \pi$$

To verify, substitute $$f(x) = -3x$$ into the original integral:

$$\int_{-\pi}^{t} (f(x) + x) dx = \int_{-\pi}^{t} (-3x + x) dx = \int_{-\pi}^{t} (-2x) dx$$

Integrate $$-2x$$:

$$\int -2x dx = -2 \times \frac{x^2}{2} = -x^2$$

Evaluate from $$-\pi$$ to $$t$$:

$$\left[ -x^2 \right]_{-\pi}^{t} = -t^2 - \left( -(-\pi)^2 \right) = -t^2 - (-\pi^2) = -t^2 + \pi^2 = \pi^2 - t^2$$

This matches the right side of the given equation, confirming that $$f(x) = -3x$$ is correct.

Hence, $$f\left(-\frac{\pi}{3}\right) = \pi$$, which corresponds to option A.

So, the answer is Option A.