1. **Problem Statement:** Find the length of the arc of the curve defined by the function $$y = \int_{-2}^x \sqrt{3t^4 - 1} \, dt, \quad -2 \leq x \leq -1.$$ 2. **Formula for Arc Length:** The length $L$ of a curve $y=f(x)$ from $x=a$ to $x=b$ is given by $$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx.$$ 3. **Find the derivative $\frac{dy}{dx}$:** By the Fundamental Theorem of Calculus, $$\frac{dy}{dx} = \sqrt{3x^4 - 1}.$$ 4. **Substitute into the arc length formula:** $$L = \int_{-2}^{-1} \sqrt{1 + \left(\sqrt{3x^4 - 1}\right)^2} \, dx = \int_{-2}^{-1} \sqrt{1 + 3x^4 - 1} \, dx = \int_{-2}^{-1} \sqrt{3x^4} \, dx.$$ 5. **Simplify the integrand:** $$\sqrt{3x^4} = \sqrt{3} \cdot |x^2| = \sqrt{3} x^2$$ (since $x^2 \geq 0$ for all real $x$). 6. **Evaluate the integral:** $$L = \sqrt{3} \int_{-2}^{-1} x^2 \, dx = \sqrt{3} \left[ \frac{x^3}{3} \right]_{-2}^{-1} = \sqrt{3} \left( \frac{(-1)^3}{3} - \frac{(-2)^3}{3} \right) = \sqrt{3} \left( \frac{-1}{3} - \frac{-8}{3} \right) = \sqrt{3} \cdot \frac{7}{3} = \frac{7\sqrt{3}}{3}.$$ **Final answer:** $$L = \frac{7\sqrt{3}}{3}.$$