Quick description
This method - a combination of the divide and conquer strategy and the base times height bound - is almost embarrassingly obvious: the Riemann integral is defined to be the infimum over all sums of the form
over all dissections
Therefore, any such sum gives us an upper bound for the integral. Often one can get useful bounds by taking very simple sums with
equal to 1 or 2. (When
we are using the base times height technique.)
Prerequisites
First-year calculus
Example 1
The integral test for convergence is a classic example of this method: if is a monotone non-increasing function, then
converges if and only if the integral
is finite. Indeed from taking upper and lower Riemann sums one obtains the bounds

Note that this method can fail completely if is non-monotone, and in particular if it oscillates at a wavelength significantly smaller than
.
Example 2
Suppose you are asked to prove that the integral tends to
as
tends to infinity. One can use relatively sophisticated tools such as the monotone convergence theorem to prove this (since the pointwise limit of the functions
is the function that is
when
and
when
). However, one can also prove it simply and directly as follows. We use the fact that the functions
decrease as
increases. Therefore, for any
we can split the interval
into the three parts
and
and thereby obtain the upper bound
which is at most
At this point, all we have to do is choose
to make this quantity small. We could argue abstractly that for every
the quantity tends to
as
which proves the result we wanted, or we could be slightly more ambitious and try to find
that minimizes, or at least comes close to minimizing, the expression in question.
Let us briefly see what happens if we go for the second approach. As is often the case, if we want to have a reasonable idea of what the minimum is, we should try to choose so that the two parts of the expression are equal. How do we get
to equal
? If
then
so we need to take a larger
Bearing in mind that
is close to
a pretty good choice is
in which case
With a little bit of work, one can prove that this choice of
leads to an upper bound for the integral of at most
which tends to
Example 3
Suppose one wants to show that the contour integral goes to zero in magnitude as
, where
is the semicircular contour
. (This is a special case of Jordan's lemma, but pretend for now that one is not aware of this lemma.) A direct application of the base times height bound, using the fact that
is bounded in magnitude by
, gives an upper bound of the base
times the height
(for
); but unfortunately
does not go to zero as
.
But one can do better by noting that the bound is usually quite inefficient. Indeed, if
, then
. Thus as soon as
is even moderately large, e.g.
, then
is actually quite small, e.g.
.
This suggests performing the following Riemann sum: in the region (say), use the upper bound
; in the remaining region
, use the cruder bound
. The length of the curve in the first region can be bounded crudely by the length
of the whole curve; meanwhile, the length of the curve in the second region can be bounded by
for some absolute constant
. This leads to a net upper bound of

which now does successfully go to zero as .