## Examples 7.1.11(a): Riemann Lemma

Is the function

When we proved before that the
integral of *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]*(since this is the same example as before, using Riemann's Lemma will hopefully simplify the solution).*f(x) = x*exists and is equal to

^{2}*1/3*we showed that

for(*) U(f, P) - L(f, P)

*every*partition

*P*with small enough norm. Using Riemann's Lemma it is enough to find

*one*partition such that inequality (*) holds. Therefore, take the partition

for the intervalP = {j/n, j = 0, 1, 2, ..., n}

*[0, 1]*. On every subinterval

*[(j-1)/n, j/n]*the maximum of

*f*occurs on the right, the minimum on the left. Therefore:

But since the last expression converges to zero as| U(f, P) - L(f, P) | | f(j/n) - f((j-1)/n)| |1/n|

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

= 1 / n^{3}| j^{2}- j^{2}+ 2 j - 1| 2 / n^{3}j

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

*n*goes to infinity, Riemann's Lemma shows that

*f(x) = x*is indeed integrable over the interval

^{2}*[0, 1]*.

The remainder of this example is just as before (exercise!).

So, Riemann's lemma has indeed simplified our computation, because
we are now able to pick *one partition* that is
best suited for our particular function and/or interval.