I used the formula r(t) = r0(1-t) + r1(t) and got t, 2t, 3t for my r(t) vector, and i plugged in the values of x y and z, then plugged them into |r'(t)| and now its one big messy integral and I'm not sure if it was set up correctly, and if it is I'm not sure how to solve it. I have a feeling it shouldn't be this complicated. Even an online integral solver wouldn't solve it. I followed the example in the book and it looked like I was doing it the right way..
$$\int_0^1te^{2t\times 3t}\sqrt{1+4+9}dt$$ $$=\sqrt{14}\int_0^1te^{6t^2}dt$$ $$=\frac{\sqrt{14}}{12}\int_{t=0}^{t=1}e^{6t^2}d(6t^2)$$ $$=\frac{\sqrt{14}}{12}e^{6t^2}\bigg|_0^1$$ $$=\frac{\sqrt{14}}{12}\left(e^6-1\right)$$