## 7.1. Riemann Integral

### Examples 7.1.9(b):

*f(x) = x*Riemann integrable on the interval

^{2}*[0,1]*? If so, find the value of the Riemann integral. Do the same for the interval

*[-1, 1]*.

*f(x) = x*is integrable we need to take partitions with more and more points, compute the upper and lower sum, and hope that the numeric answers will get closer and closer to one common value. If that's the case, our guess is that this limit is the integral of

^{2}*f*over the indicated interval.

But of course that's no proof. As it turns out, to prove that this simple function is integrable will be difficult, because we do not have a simple condition at our disposal that could tell us quickly whether this, or any other function, is integrable. That's the bad news; the good news will be that we should be able to generalize the proof for this particular example to a wider set of functions.

Anyhow, here we go. First we should note that in the definition
of upper and lower integral it is not necessary to take the
*sup* and *inf* over *all* partitions.
After all, if *P* is a partition and *P'*
is a refinement of *P* then
*L(f, P') L(f, P)*
and
*U(f, P') U(f, P)*.
Therefore partitions with large norm don't contribute to the
*sup* or *inf* and it is enough to compute
the upper and lower integral by considering partitions with
a small norm only.

Next, take any * > 0* and
a partition *P* with
*|P| < / 2*. Then

where| U(f, P) - L(f, P)| |c_{j}- d_{j}| (x_{j}- x_{j-1})

*c*is the

_{j}*sup*of

*f*over

*[x*and

_{j-1}, x_{j}]*d*is the

_{j}*inf*over that interval.

Since *f* is increasing over *[0, 1]* we know
that the *sup* is achieved on the right side of each subinterval,
the *inf* on the left side. Therefore:

To estimate this sum, we'll apply the Mean Value Theorem for| U(f, P) - L(f, P)|

|c_{j}- d_{j}| (x_{j}- x_{j-1}) = |f(x_{j}) - f(x_{j-1})| (x_{j}- x_{j-1})

*f(x) = x*:

^{2}for|f(x) - f(y)| |f'(c)| |x - y|

*c*between

*x*and

*y*. Since

*|f'(c)| 2*for

*c*in the interval

*[0, 1]*we know that

But the partition|f(x) - f(y)| 2 |x - y|

*P*was chosen with

*|P| < / 2*so that

But then|f(x_{j}) - f(x_{j-1})| 2 |x_{j}- x_{j-1}| 2 / 2 =

because the last sum is telescoping.| U(f, P) - L(f, P)| |f(x_{j}) - f(x_{j})| (x_{j}- x_{j-1})

(x_{j}- x_{j-1}) = (x_{n}- x_{0})

= (1 - 0) =

Since the partition *P* was arbitrary but with small norm - which
we remarked is sufficient for the upper and lower integral - we know that
the upper and lower integral must exist and be equal to one common limit
*L*.

Therefore we know that *f* is integrable and it remains to
find the value *L* of the integral. But that's easy to compute
because now that we know that the function is integrable we can take a
suitable partition to find the value of the integral. So, take the
following partition:

forx_{j}= j/n

*j = 0, 1, 2, ..., n*. Then the upper sum computes to

We know that the upper integral exists and is equal toU(f, P) = c_{j}(x_{j}- x_{j-1}) = f(x_{j}) 1/n

= (j/n)^{2}1/n = 1/n^{3}j^{2}

= 1/n^{3}1/6 n (n+1) (2n+1) = 1/6 (n+1) (2n+1) / n^{2}

*L*. Therefore, the limit as

*n*goes to infinity of the above expression must also converge to

*L*. But then

*L = 1/3*, or in other words:

wherex^{2}dx = 1/3

*a = 0*and

*b = 1*.

As for the interval *[-1, 1]*, we can first play our applet game
to get an idea about the answer, then set out to prove everything just as
we did above.

It seems that the applet says that the integral now should evaluate to
something close to *2/3*. Of course we need some formal proof,
but since that would be similar to the above one we'll leave it as an
exercise.

The proof is somewhat misleading, because it seems to be based on the fact
that *f(x) = x ^{2}* is differentiable. As a generalization
we might conclude that

*differentiable*functions are integrable. That's correct, but another, more general concept can be substituted for differentiability (what might it be?).