find the length of the curve $y= \int_{-2}^x\sqrt{3t^4-1} \, dt$

Question

I'm stuck for a while not sure how to continue, or if there's a mistake that I did that prevent me to continue.

$$y= \int_{-2}^x\sqrt{3t^4-1} \ dt \ , \ -2≤x≤-1$$

$$ y = F(x) - F(-2) $$ $$ y' = f(x) - f(-2) $$ $$ y' = \sqrt{3x^4-1} - \sqrt{47} $$ $$ (y')^2 = 3x^4-1 -2 \sqrt{(3x^4-1)\ 47} + 47$$ $$ 1+(y')^2 = 3x^4 -2 \sqrt{(3x^4-1)\ 47} + 47$$

Now how can I integrate this: $$ \int_{-2}^{-1} \sqrt{3x^4 -2 \sqrt{(3x^4-1)\ 47} + 47} $$ And I cant get it to $\int_{-2}^{-1} \sqrt{ [\sqrt{3x^4-1} \ - \sqrt{47}]^2}$

Answer 1

Notice by the FTC

$$ y' = \sqrt{3x^4-1}$$

and Thus,

$$ (y')^2 +1 = 3x^4 $$

it follows that

$$ \mathcal{L} = \int\limits_{-2}^{-1} \sqrt{3}x^2 dx = ... $$

Answer 2

Your problem is in the line $$ y' = (F(x) - F(-2))'\ne f(x) - f(-2) $$

the result should be $$y'=f(x)$$ since $(F(-2))'=0.$