2 Finding length of curve problems

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1. Find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.

y = the integral from -2 to x of the SQUARE ROOT (3t^4-1)dt, -2 < x < -1

note that the "<" is actually less than or equal to, don't know how to post that.

For this one, do I just plug the x in? x_x I'm really clueless on how to start

2. Find the length of the curve.

y = the integral of 0 to x of SQUARE ROOT (cos(2t))dt from x = 0 to x = pi/4The problem with me is I know how to do it in terms of y and x, but I am terrible at parametrics.

If you can, please help me with any of the above =) thanks
 
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Heres the arc length formula:

[tex]\int_a^b \sqrt{1+\frac{dy}{dx}} dx[/tex]. Sub in the requirements, easy enough to get.
 
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O sorry didnt actually read it through well. For the first one,

[tex]\int^{-2}_{x} \sqrt{3t^4 -1} dt[/tex]. If there wasn't an X there, but instead a normal number like you normally see, you would find the integral and then sub in b into it, and - the integral with a subbed in. In this case just sub in X.
 
how do you actually type in the integral sign and stuff?
 
Here's one source:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php
which will help with formatting the integrals

And another source for starters: (crash course in LaTeX at these forums:
https://www.physicsforums.com/misc/howtolatex.pdf
 
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Oh, and to see specifically how Gib Z did it, click to quote him, and take a look at what he has.

However, the formula has a small mistake in it...
(So, I copied and pasted from the quote so I could change it more simply)
[tex]\int_a^b \sqrt{1+(\frac{dy}{dx})^2} dx[/tex].
There's supposed to be a squared in there...
 
Btw, rather than actually having to quote me, just click on my latex pictures, that'll show up what I typed to show that code.