The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted
, which we can parameterize by the vector-valued function,

for
, which has differential

Then with
, we have



Complete the square in the quadratic term of the integrand:
, then in the integral we substitute
:


Make another substitution of
:

Integrate by parts, taking




So, we have by the fundamental theorem of calculus that



It can be written as 12/100
or reduced to 6/50
or reduced even further to 3/25