Revision history [back]

Anonymous, the first thing I noticed was your bounds of integration. I posted a question about this last night as I was struggling with how we know. Wes had a nice answer and Justin clarified a bit in class for me. It is all about the time interval, so typically when integrating along a line segment, you are going from t=0 to t=1, thus $\int_0^1$.

Then when you use u-sub, don't forget to change your bounds. Let's start here: $$u=2t+2t^2$$ $$\frac{du}{2(1+2t)}=dt$$

When t=1, u=0. When t=1, u=4. Giving you: $$3\int_0^4 (\cos(u)(1+2t)\frac{1}{2(1+2t)})du$$ $$\frac{3}{2}\int_0^4 (\cos(u))du$$ $$\frac{3}{2}\sin(u)\biggr|_0^4$$ $$\frac{3}{2}\sin(4)-\frac{3}{2}\sin(0)$$ $$\frac{3}{2}\sin(4)-0$$ $$\frac{3\sin(4)}{2}$$

Which is the answer! Hope this helps, everything else was spot on!