So I was given the following prompt:
The instantaneous rate of change of the vector-valued function $g(t)$ is given by $r(t)=\langle\sqrt{t^2+1}, \sin(t^2)\rangle$. If $g(7)=\langle\sqrt{2},\pi\rangle$, find $g(0)$.
I guess I'm a bit confused about the integration that would be involved with this problem. I understand that I'm going to have to integrate both given functions, which I would then plug $0$ in to, but I'm a bit confused about what this integration would look like. Some clarification would be appreciated!
Edit: I understand what the integral formula would look like here, it's the actual integration that's confusing to me, I'm not coming up with a normal integral when I integrate.
Best Answer
You have
$$g(7) - g(0) = \int_0^7 r(t) dt.$$ hence
$$g(0) = g(7) - \int_0^7 r(t) dt = \left(\sqrt 2 - \int_0^7 \sqrt{t^2+1} dt,\pi - \int_0^7\sin t^2 dt\right)$$