Prove that the arc length of the curve y = f(x) between

Prove that the arc length of the curve $y = f(x)$ between points with abscissas $x = a$ and $x = b$ equals the arc length of the curve $y = f(x - h)$ between points with abscissas $x = a + h$ and $x = b + h$. Give a geometric interpretation.

We compute both arc lengths as integrals and show they coincide via a shift of variable.

Step 1: Arc length of $y_1=f(x)$

\[L_1=\int_a^b\sqrt{1+[f'(x)]^2}\,dx\]

Step 2: Arc length of $y_2=f(x-h)$

Since $\dfrac{dy_2}{dx}=f'(x-h)$:

\[L_2=\int_{a+h}^{b+h}\sqrt{1+[f'(x-h)]^2}\,dx\]

Step 3: Substitute $u=x-h$

Then $dx=du$, and the limits $x=a+h,\,b+h$ become $u=a,\,b$:

\[L_2=\int_a^b\sqrt{1+[f'(u)]^2}\,du = L_1\]

since the integration variable is a dummy variable. \[\boxed{L_1=L_2}\]

Step 4: Geometric interpretation

The curve $y_2=f(x-h)$ is $y_1=f(x)$ shifted horizontally by $h$ (right if $h>0$, left if $h<0$). Horizontal translation is an isometry: it preserves distances and slopes at corresponding points — at $x=c$ on $y_1$ the slope is $f'(c)$, and at the corresponding point $x=c+h$ on $y_2$ the slope is again $f'\big((c+h)-h\big)=f'(c)$. Since arc length depends only on this intrinsic slope profile and not on horizontal position, sliding the whole curve sideways cannot change its length — like sliding a length of wire across a table.

Step 5: Combined with vertical translation

Together with the analogous vertical-translation result, arc length is invariant under any combination of horizontal and vertical shifts:

\[\boxed{L[f(x)\text{ on }[a,b]] = L[f(x-h)+c\text{ on }[a+h,b+h]]}\]