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

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 $\int_{-2}^{-1} \sqrt{ [\sqrt{3x^4-1} \ - \sqrt{47}]^2}$

• If $F(x)$ is a function of $x$ so that $F(-2)$ is constant, and if $y=F(x)-F(-2)$, then you cannot have $y’=f(x)-f(-2)$, because the linearity of the derivative goes as $$\bigl[F(x)-F(-2)\bigr]’=F’(x)-\bigl[F(-2)\bigr]’$$ and not as $F’(x)-F’(-2)$. Commented Jun 9, 2018 at 1:06

Notice by the FTC

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

and Thus,

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

it follows that

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

• Thanks, because it's represented by a function I forgot that it's, in fact, a constant. Commented Jun 9, 2018 at 0:04

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

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