I was looking through some AP Calculus BC problems provided by The College Board, and I came across this one:
At time $t \geq 0$, a particle moving in the xy-plane has velocity vector given by $v(t) = \langle4e^{-t}, \sin(1+\sqrt{t})\rangle$. What is the total distance the particle travels between $t = 1$ and $t = 3$?
I used an integral of the magnitude of the velocity vector like so:
$$\int_{1}^{3}||v(t)||\,\mathrm dt= \int_{1}^{3} \sqrt{\left(-4e^{-t}\right)^{2}+\left(\dfrac{\cos(1+\sqrt{t})}{2\sqrt{t}}\right)^{2}}\,\mathrm dt \approx 1.403$$
But the only choices are $1.861$, $1.983$, $2.236$, and $4.851$, and the correct answer is supposed to be $1.861$. I've checked to make sure I've differentiated correctly and that I've calculated correctly. Is there something wrong with how I calculated the total distance?
Note that
$$\int_{1}^{3}||v(t)||dt= \int_{1}^{3} \sqrt{4e^{-2t}+\sin^2(1+\sqrt{t})}dt$$