## Examples 7.1.6(a):

*f(x) = x*for

^{2}-1*x*in the interval

*[-1, 1]*. Find:

- The left and right sums where the interval
*[-1, 1]*is subdivided into 10 equally spaced subintervals. - The upper and lower sums where the interval
*[-1, 1]*is subdivided into 10 equally spaced subintervals. - The upper and lower sums where the interval
*[-1,1]*is subdivided into*n*equally spaced subintervals.

Note, in particular, that right and left Riemann sums are
the same (an accident ?). The upper and lower sums are similarly
computed, noting that in the interval *[0, 1]* the
largest value of *x ^{2}-1* over any subinterval
inside

*[0, 1]*is always the right endpoint, while the smallest value occurs on the left endpoint. For subintervals inside

*[-1, 0]*it is just the other way around.

Note that this time the values don't agree but we have that
*L(f, P) U(f, p)*,
which is again no accident.

To answer the last question we can not use the above Java applet
because we don't know a numeric value for *n*. Here
is the appropriate manual computation for the, say, the upper sum:
We are looking for *n* equally spaced subintervals
of *[-1, 1]* so that our partition consists of the
points:

wherex_{j}= -1 + j * 2/n

*j = 0, 1, ..., n*and

*| P | = 2/n*.

**Case 1:** *n* is even, i.e. *n = 2 N*
for some integer *N*

The renumbered partition now is:

wherex_{j}= -1 + j * 1/N

*j = 0, 1, ..., N, N+1, ..., 2 N*. In that case the point

*0*is part of the partition (take

*j = N*). The point to notice before we can begin the computation is that the function is decreasing over the interval

*[-1, 0]*and increasing over the interval

*[0, 1]*. That implies that for all partition points less than 0 the maximum of

*f*occurs on the left endpoint, for points bigger than 0 the maximum occurs on the right. Hence:

In appropriate "sigma" notation we therefore have:U(f, P) =

We have used the results on the sum of square integers mentioned in the chapter on Induction to simplify the various sums.U(f, P) =

=

=

=

=

=

=

=

We can verify our formula by looking at the above applet. There the
*10*-upper sum computes to -1.119999. In our formula we
need to let *N = 5* because *10 = n = 2 N*.
Substituting *N = 5* gives the exact value of -1.12
for the *10*-th upper sum.

The case where *n* is odd, as well as the computation of
the lower sum, is left as an exercise. There's nothing new in those
computations (but for *n* odd there's something special
happening at 0), it's just somewhat tedious. You can check whichever
formulas you come up with against the numeric answers of the above
applet.