254 Integrals
norm JP1 and write f(xk) Oxk as an approximation to the amount of work
done in pulling the particle from the left end to the right end of the kth
subinterval. The sum(4.653) Tlf(xk) Oxkshould then be a good approximation to our answer W and hence we
should have(4.66) + W = lim I f (xk) 1 xk = f 'f(x) dx.
Our statements about (4.653) and (4.66) were necessarily vague and
optimistic because the quantity W that we are trying to calculate has not
yet been defined. We must recognize the fact that we cannot prove
correctness of a formula for W when we have no definition or other
information that tells us what W is. In the absence of another definition
or other information, we must adopt the principle that our work with
partitions and Riemann sums provides the motivation for the definition
whereby W is defined by the formula(4.661) W = f ab f(x) dxwhenever f is a function for which the integral exists as a Riemann integral.
The above ideas will now be applied to basic problems. The Newton
(1642-1727) law of universal gravitation says that if two particles of mass
in, and m2 are concentrated at distinct points Pl and P2, then these
particles attract each other with a force whose magnitude is proportional
to the product of the masses and inversely proportional to the square of
the distance between them. Suppose we have a particle of mass m,
concentrated permanently at the
MI f 1 origin, Figure 4.67, and that we
0 a x b x have a "test particle" of unit mass
Figure 4.67 that we wish to move along the
positive x axis. There is then a
constant k, which depends only upon the units used to measure mass,
force, and distance, such that the force on the test particle has magnitude
km,/x2 when the particle is at distance x from the origin. The work Wa,b
required to move the particle from the point a (that is, the point with
coordinate a) to the point b is then to be calculated from the formula(4.671) Wa b =
1 bkml
dx.
a x2From this we find the remarkably simple formula