Find the length of the curve $$ y=\int_{-3}^{x} \sqrt{7 t^{4}-1} \mathrm{dt},-3 \leq x \leq-1 $$. The length of the curve is $$ \square $$. (Type an exact answer, using radicals as needed.)

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To find the length of the curve $$ y = \int_{-3}^{x} \sqrt{7 t^{4} - 1} \, dt $$ for $$ -3 \leq x \leq -1 $$, follow these steps:

1. **Find the derivative $$ \frac{dy}{dx} $$:**
   $$
   \frac{dy}{dx} = \sqrt{7x^4 - 1}
   $$

2. **Use the formula for the arc length:**
   $$
   L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
   $$
   Substituting $$ \frac{dy}{dx} $$ into the formula:
   $$
   L = \int_{-3}^{-1} \sqrt{1 + 7x^4 - 1} \, dx = \int_{-3}^{-1} \sqrt{7x^4} \, dx
   $$

3. **Simplify the integrand:**
   $$
   \sqrt{7x^4} = |x^2| \sqrt{7} = x^2 \sqrt{7} \quad (\text{since } x^2 \geq 0)
   $$
   So the integral becomes:
   $$
   L = \sqrt{7} \int_{-3}^{-1} x^2 \, dx
   $$

4. **Evaluate the integral:**
   $$
   \int_{-3}^{-1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-3}^{-1} = \left( \frac{(-1)^3}{3} \right) - \left( \frac{(-3)^3}{3} \right) = \left( -\frac{1}{3} \right) - \left( -9 \right) = \frac{26}{3}
   $$

5. **Multiply by $$ \sqrt{7} $$:**
   $$
   L = \sqrt{7} \times \frac{26}{3} = \frac{26\sqrt{7}}{3}
   $$

**Final Answer:**
$$
\boxed{\dfrac{26\,\sqrt{7}}{3}}
$$
The length of the curve is $$ \frac{26\sqrt{7}}{3} $$.

Find the length of the curve .
The length of the curve is .
(Type an exact answer, using radicals as needed.)

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