Two different integration questions are baffling me right now, and I have no idea how to approach them:

The first deals with finding the length of the curve $$ y = \int_{-2}^x sqrt(3t^4-1)dt$$

The second is (probably) a very simple integration problem that I can't get because my mind is now muddled with more complicated equations. It's part of a larger problem also involving length of a curve, but I can work it out after I figure this out:

$$ \int_t^1 sqrt(1+ (4/9)x^{(-2/3)})dx $$

Your help is much appreciated! I've been trying to solve them for the past hour and haven't gotten anywhere.

  • $\begingroup$ The first one is an "elliptic integral" ... look it up. $\endgroup$ – GEdgar Mar 25 '14 at 0:42
  • $\begingroup$ We haven't covered elliptic integrals in the course I'm taking, so there should be a simpler way to solve it. Do I need to solve the integral and then use the formula for arc length, L = integral sqrt(1+(dy/dx)^2? $\endgroup$ – user3361007 Mar 25 '14 at 0:55
  • $\begingroup$ So, since you cannot evaluate the integral, maybe you can compute $dy/dx$ without doing it! $\endgroup$ – GEdgar Mar 25 '14 at 2:29
  • $\begingroup$ Oh, I think I get it! because the function being integrated is already dy/dx, so I just plug that into my equation! Will the bounds -2 and x affect it at all if it's within the range -2<=x<=-1? $\endgroup$ – user3361007 Mar 25 '14 at 4:55

For your second problem, if $x$ is positive, we have $$\sqrt{1+\frac{4}{9}x^{-2/3}}=\frac{1}{3x^{1/3}}\sqrt{9x^{2/3}+4}.$$ To integrate, let $u=9x^{2/3}+4$. If we are integrating over an interval where $x$ is negative, $\frac{1}{3x^{1/3}}$ should be replaced by $-\frac{1}{3x^{1/3}}$.

  • $\begingroup$ So then I get 1/18du = (1/3)x^(-1/3), and (1/18)integral u^(1/2), and finally: $$ (1/27)(9x^{2/3}+4)^{3/2} $$ Is that right? $\endgroup$ – user3361007 Mar 25 '14 at 1:04
  • $\begingroup$ Yes, apart from the $+C$. $\endgroup$ – André Nicolas Mar 25 '14 at 1:14

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