I have the curve defined by $$x=t^2-t \;\;\;\; y=2t-1\;\;\;\; 1 \le t \le 2$$ Now I need to find the length, so I take the derivatives of both of those: $${dx\over dt} = 2t-1 \;\;\;\; {dy\over dt} = 2$$ Easy enough, and now I put that into the length equation: $$L = \int_1^2{\sqrt{(2t-1)^2 +(2)^2}dt}$$ That is where I'm starting to get stuck, but I'll show you what I have so far: $$u-sub: \;\;\; u=2t-1 \;\;\; {du\over 2}=dt$$ $$L = {1\over2}\int_1^2{\sqrt{(u)^2 +(2)^2}\;dt}$$
Now I'll do a trig sub: $$u=2tan\theta \;\;\; {du\over 2}=sec^2\theta \;d\theta$$ $$L = {1\over4}\int_1^2{\sqrt{4tan^2\theta +4}\;sec^2\theta \;d\theta}$$ Get rid of that gross constant: $$L = {1\over2}\int_1^2{\sqrt{tan^2\theta +1}\;sec^2\theta \;d\theta}$$
Use a fancy-dancy trig identity: $$L = {1\over2}\int_1^2{\sqrt{sec^2\theta}\;sec^2\theta \;d\theta}$$
equals:
$$L = {1\over2}\int_1^2{sec^3\theta \;d\theta}$$
Sorry if that is messy (can't figure out how to align), but if I use $\int sec^3 \theta d\theta = {1\over2}sec\theta tan\theta + ln(sec\theta + tan\theta)$ and do the rest after that, am I on the correct path? The reason I ask is because I end up getting $0.9747$ for the length but according to an online calculator, if I evaluated everything correctly, I should get $2.85869785445055$ If anyone would like to see the rest of what I did I can show it.
Thank you!
When you do a substitution you have to substitute also the integration bounds. If $u=2t-1$, then $u(1)=1$ and $u(2)=3$, so your knew integral in $u$ goes from $1$ to $3$. Then $u=2\tan \theta$ and so when $u=1$ you have $1=2\tan \theta$ and so $\theta= \arctan\left(\tfrac12\right)$ and then you do the same for $u=3$ so you have an integral not from $1$ to $2$ but between the bounds you found this way.