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 = x_{i} (with endpoints that have the same x-values as before), where x_{i}
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: