Recipe

Find the arc length of a curve

Use when: the exact (or numerically approximated) length of y=f(x) on [a,b] is needed, not just an estimate

Goal: the arc length integral fully set up and evaluated, exactly when the integrand has an elementary antiderivative, or numerically otherwise

Taught in Arc length

1. 1

   Compute f'(x); it's needed squared, so simplify it first if that makes the square root easier later.

   Example

   f(x) = \dfrac{x^3}{6}+\dfrac{1}{2x} on [1,2]: f'(x) = \dfrac{x^2}{2}-\dfrac{1}{2x^2}.

   Why: f'(x) is exactly the "rise" the Pythagorean theorem needs to pair with a "run" of 1 inside the arc length integrand.

2. 2

   Square f'(x), add 1, and simplify algebraically before taking the square root, since the simplified form often reveals a perfect square.

   Example

   1+\left(\frac{x^2}{2}-\frac{1}{2x^2}\right)^2 = 1+\frac{x^4}{4}-\frac12+\frac{1}{4x^4} = \frac{x^4}{4}+\frac12+\frac{1}{4x^4} = \left(\frac{x^2}{2}+\frac{1}{2x^2}\right)^2

   Why: the algebra done here, before ever taking a square root, is what decides whether the arc length integral ends up elementary or needs numerical evaluation.

3. 3

   Take the square root of the simplified expression from Step 2.

   Example

   \sqrt{\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)^2} = \frac{x^2}{2}+\frac{1}{2x^2} \quad (\text{positive on } [1,2])
   • the expression under the root simplifies to a perfect square, or otherwise integrates elementarily — integrate directly in Step 4.
   • no elementary antiderivative exists for the square root of 1+[f'(x)]^2 (the common outcome for most functions) — evaluate the integral numerically instead, using a method from earlier in this course.

   Why: this square root resists elementary antidifferentiation more often than it yields to it, so checking which case applies is a genuine fork, not a formality.

4. 4

   Set up L=\displaystyle\int_a^b\sqrt{1+[f'(x)]^2}\,dx and evaluate it (symbolically if Step 3 gave an elementary form, numerically otherwise).

   Example

   L = \int_1^2\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)dx = \left[\frac{x^3}{6}-\frac{1}{2x}\right]_1^2

   Why: this integral is the sum of infinitely many straight-segment approximations to the curve, each one a Pythagorean-theorem hypotenuse shrinking toward zero width.

5. 5

   Compute the final number and check it's at least as large as b-a, the straight-line distance underneath the curve.

   Example

   \left(\frac{8}{6}-\frac14\right)-\left(\frac16-\frac12\right) = \frac{13}{12}-\left(-\frac13\right) = \frac{13}{12}+\frac{4}{12} = \frac{17}{12}

   \boxed{L=\dfrac{17}{12}}, which is larger than b-a=1, as a curve's length must be.

   Why: a curve can never be shorter than the straight horizontal run beneath it, since \sqrt{1+[f'(x)]^2}\geq1 everywhere.