Apparently either I've forgotten some basic rule about integrals (it has been a while since I've taken a basic calc class) or something is wrong with this problem in pearson mylab.
This was the problem:
Evaluate $\int_C\frac{x^2}{y^{4/3}}ds$ where C is the curve $x=t^2,y=t^3$ for $-3\le{t}\le-1$
The answer that it finally accepted was $\frac1{27}(85^{3/2}-13^{3/2})$
I've been wracking my brain trying to figure out why it's not $\frac1{27}(13^{3/2}-85^{3/2})$
After all, (unless I suddenly can't work out integrals, which seems unlikely given that I did a bunch of very similar problems before and after this with no problem) the antiderivative should work out to $\frac1{27}(4+9t^2)^{3/2}$ and then you just work out the values at $t=-1$ and $t=-3$ and subtract the first from the second.
Am I missing something obvious or is this problem actually just borked?
Your antiderivative is incorrect because you are missing a minus sign. Once you substitute the parametrization you should have $$\int_{-3}^{-1} \sqrt{4t^2+9t^4} \ dt.$$ In order to pull the $t^2$ out of the square root, we need to make it $|t|$ (recall that the square root function by definition returns a nonnegative number, so $\sqrt{a^2}=|a|$, not just $a$). Since $t$ is negative on the interval $[-3,-1]$, $|t|=-t$ on this interval. Therefore the integral becomes: $$\int_{-3}^{-1}-t\sqrt{4+9t^2} \ dt.$$