No 1 under this root BC only

A short piece of any curve has horizontal extent $dx$ and vertical extent $dy$, so its length is $ds = \sqrt{(dx)^{2} + (dy)^{2}}$. Unit 8 factored $dx$ out of that root. Here we factor $dt$ instead:

$$ds = \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}\;dt, \qquad L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}\;dt.$$

The $1$ has not been dropped. In $\sqrt{1 + (f'(x))^{2}}$ the $1$ is $\left(\frac{dx}{dx}\right)^{2}$, which is $1$ because $x$ was both the variable and the parameter. Once $t$ takes over, that slot holds $\left(\frac{dx}{dt}\right)^{2}$ like any other, and there is nothing left to be $1$.

The check that confirms it: set $x = t$. Then $\frac{dx}{dt} = 1$, and the parametric integrand becomes $\sqrt{1 + \left(\frac{dy}{dt}\right)^{2}}$, which is exactly Unit 8's. The two formulas are one formula, and the $1$ appears only in the special case.

Two structural consequences follow immediately. The limits are $t$-values, because $dt$ is what was factored out; using $x$-values integrates over the wrong interval. And the integrand is a speed, so it is never negative and the integral never needs an absolute value.

Take $x = 3t^{2}$ and $y = 2t^{3}$ on $[0, 1]$. Then $\frac{dx}{dt} = 6t$ and $\frac{dy}{dt} = 6t^{2}$, so

$$L = \int_0^1 \sqrt{36t^{2} + 36t^{4}}\;dt = \int_0^1 6t\sqrt{1 + t^{2}}\;dt = 2\left(1 + t^{2}\right)^{3/2}\bigg|_0^1 = 2\left(2\sqrt{2} - 1\right) \approx 3.657.$$

Two things about that middle step are worth noticing. Factoring $36t^{2}$ out of the radicand works because $t \geq 0$ on this interval, so $\sqrt{36t^{2}} = 6t$ rather than $6|t|$; on an interval containing negative $t$ that absolute value matters. And the $\sqrt{1 + t^{2}}$ that appears is a coincidence of this curve, not the missing $1$ from Unit 8 reappearing.

Three malformed versions, all CA3, against that $3.657$:

1. Root dropped: $\int_0^1 \left(36t^{2} + 36t^{4}\right)dt = 19.2$. Not a length.
2. Rates added instead of combined: $\int_0^1 \left(6t + 6t^{2}\right)dt = 5$. This is the sum of the legs, which always exceeds the hypotenuse.
3. The $1$ put back: $\int_0^1 \sqrt{1 + 36t^{4}}\;dt \approx 2.42$, which treats $y$ as a function of $x$ with $t$ standing in for $x$. Two different errors at once.

The chord. An arc is at least as long as the straight segment joining its endpoints. Here the curve runs from $(0, 0)$ to $(3, 2)$, so $L \geq \sqrt{9 + 4} = \sqrt{13} \approx 3.606$. The computed $3.657$ clears it by a little, which is right for a curve this gently bent. A candidate answer below the chord is disqualified without any further thought.

The bounding box. An arc that is monotone in both coordinates is at most the sum of the two extents, here $3 + 2 = 5$. So $3.606 \leq L \leq 5$, and both the $19.2$ and the $2.42$ above fall outside that window. Notice that the "rates added" error produces exactly the upper bound, which is the geometric statement that a staircase is longer than a diagonal.

Together those two bounds catch every malformed integrand in the list, and neither requires evaluating anything harder than a square root.

This integral measures how far the particle goes, and that is not always the same as how long the curve is.

Take $x = \cos t$, $y = \sin t$ on $[0, 4\pi]$. The integrand is $\sqrt{\sin^{2}t + \cos^{2}t} = 1$, so the integral is $4\pi$. But the curve is a unit circle, whose length is $2\pi$. The particle goes round twice, and the integral counts both laps because it is adding up distance travelled.

So the question has to be read carefully:

1. "How far does the particle travel on $[t_1, t_2]$?" Integrate over exactly that interval. Retracing counts.
2. "What is the length of the curve?" First find an interval on which the curve is traced once, then integrate over that.

This is 8.2's displacement-against-distance distinction arriving in two dimensions, and it will return in 9.6 with the third member of the family, the displacement vector. It also foreshadows 9.8: a polar curve integrated over too large an interval double-counts its area for exactly this reason.