Set up, but do not evaluate, an integral that represents the length of the following curve.
$$\left\{\begin{array}{rl} x &= t-t^2\\ y &= \frac{4}{3}t^{3/2}\end{array}\right. \qquad\qquad 1 \leq t \leq 2$$
My solution:
First of all, I find derivative
$$\dfrac{dx}{dt}=1-2t$$
and
$$\dfrac{dy}{dt}=2\sqrt{t}$$
Since parameter $t$ ranges from $t=1$ to $t=2$ we see that
$$(x(1),y(1))=(0,\frac{4}{3})$$
and
$$(x(2),y(2))=(-2, \frac{8\sqrt{2}}{3})$$
It is easy to see in integrating formula we should begin from $x=-2 \ (t=2)$ to $x=0 \ (t=1)$. So length of this parametric curve is $$L=\int \limits_{2}^{1}\sqrt{(1-2t)^2+4t}\ dt$$ But it is obvious that this integral has negative value. What's wrong?
Why should upper limit be 2 and lower limit be 1. I can not grasp this question for quite long period of time. Please help.
ADDITION: Suppose we have an ellipse in parametric equations $x=a\cos \theta, \ y=b\sin \theta, \ \theta \in[0,2\pi]$. And we want to find its area. Is we compute the integral $\int \limits_{0}^{2\pi}{b\sin \theta (a\cos \theta)' d\theta}$ we will get the negative. Here we have ambiguity with arc length.