The 1, the square, and the root BC only

Take a piece of the curve so short that it is essentially straight. Its horizontal extent is $dx$ and its vertical extent is $dy$, so by the Pythagorean theorem its length is

$$ds = \sqrt{(dx)^{2} + (dy)^{2}}.$$

Factor $dx$ out of the root. Since $\frac{dy}{dx} = f'(x)$,

$$ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\;dx = \sqrt{1 + \big(f'(x)\big)^{2}}\;dx,$$

and adding up the pieces gives

$$L = \int_a^b \sqrt{1 + \big(f'(x)\big)^{2}}\;dx.$$

Every part of that expression now has a source. The $1$ is the $(dx)^{2}$ that was factored out. The square is on $\frac{dy}{dx}$ because the theorem squares both legs. The root is the theorem itself. A formula recalled as a shape loses parts; a formula rebuilt from this triangle does not.

Each part can go missing on its own, and each produces a well-formed integral that evaluates to something plausible. Against $y = x^{3/2}$ on $[0, 5]$, whose true arc length is $\frac{335}{27} \approx 12.407$:

1. The $1$ dropped: $\int_0^5 \sqrt{\frac{9x}{4}}\,dx = 5^{3/2} \approx 11.180$. This measures only the vertical climb.
2. The root dropped: $\int_0^5 \left(1 + \frac{9x}{4}\right)dx = 33.125$. Nearly three times too large, and it is no longer a length.
3. The square dropped: $\int_0^5 \sqrt{1 + \frac{3}{2}\sqrt{x}}\,dx$, which is not even the same kind of expression. Squaring is what makes the two legs combine.

A fourth version substitutes $f$ where $f'$ belongs, giving $\int \sqrt{1 + \big(f(x)\big)^{2}}\,dx$. That measures nothing geometric at all: the length of a curve depends on how steeply it rises, not on how high it is, and a horizontal line high above the axis has the same length as one on it.

First, the straight line. For $y = 3x$ on $[0, 4]$ the formula gives

$$L = \int_0^4 \sqrt{1 + 9}\;dx = 4\sqrt{10} \approx 12.649,$$

and the distance formula from $(0, 0)$ to $(4, 12)$ gives $\sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}$. They agree, as they must. Any remembered formula that fails this test on a line has a part missing, and the test takes ten seconds.

Second, the bounds. Arc length is at least the horizontal extent $b - a$, since the shortest path between the two endpoints is a straight line and the curve is at least as long as its own horizontal shadow. It is also at least the straight-line distance between the endpoints. For the example above, $12.407$ exceeds both $5$ and the chord length $\sqrt{150} \approx 12.247$, which is exactly the ordering to expect.

Those two inequalities catch the dropped $1$ immediately, since $11.180$ is below the chord length and so cannot be an arc length. Nothing about the integrand needs to be re-examined to know it is wrong.

When a curve is more naturally described as $x = g(y)$, the same triangle gives

$$L = \int_c^d \sqrt{1 + \big(g'(y)\big)^{2}}\;dy,$$

with $y$-limits. The choice is 8.5's: use the variable in which the curve is a function. And the check from 8.5 applies without change, since a $dy$ integral whose limits are $x$-values is describing the wrong stretch of curve.

For a parametric path $\big(x(t),\, y(t)\big)$, factoring $dt$ out of $ds$ instead gives

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

which is the distance travelled by a particle moving along that path. There is no $1$ here, and its absence is not an omission: the $1$ in the earlier version was $\left(\frac{dx}{dx}\right)^{2}$, and once $t$ is the parameter neither derivative is automatically $1$.

One consequence carries straight over from 8.2. Distance travelled is not displacement. A path that doubles back covers ground it has already covered, and this integral counts all of it, because the integrand is a speed and a speed is never negative. The net change in position is $\big(x(t_2) - x(t_1),\, y(t_2) - y(t_1)\big)$, and its size is generally smaller.