In addition to two-dimensional areas and three-dimensional volumes, the integral can be used to compute one-dimensional lengths. The idea, once again, is to approximate the length by a sum and to take the limit as the number of summands approaches to infinity.

More precisely, we want to calculate the length of the graph of a function f (x) from x = a to x = b. This length can be expressed as the sum of the lengths of the graph from x = a + (i - 1)Δx to x = a + iΔx, for i = 1,…, n, where Δx = (b - a)/n. We approximate the lengths of these smaller curves by line segments segments with the same endpoints, having lengths of

Making a further approximation, we replace these segments with segments tangent to the graph at x = xi (with endpoints that have the same x-values as before), where xi is some number in the interval [a + (i - 1)Δx, a + iΔx]. The length of one of these new segments is equal to

 = Δx

This is illustrated below.

This approximation is valid as Δx approaches zero, since the original segment was a secant line for the curve whose endpoints approach the associated point of tangency. Consult the geometric definition of the derivative for more detail.

Summing the lengths of these tangent segments gives an approximation to the length of the graph over the whole interval:

 Δx

Taking the limit as n→∞ (where the segments approximating the curve become shorter and shorter), we have the following expression for the exact length of the curve:

 dx