Assignment 13
In class (and in the book, in the chapter Axioms) we began a calculation of \(\int_{[0,1]}x\), assuming that \(\int\) is some integral satisfying the axioms and that \(x\) is integrable. We carried out the process by dividing \(I\) into \(N\) equal subintervals \(I_1,I_2\ldots, I_N\) each of length \(1/N\), and then computing a lower estimate on each and an upper estimate on each. Combining these with the axioms of integration, we found
\[\sum_{i=1}^N\frac{i-1}{N^2}\leq\int_{[0,1]}x\leq\sum_{i=1}^N\frac{i}{N^2}\]
Exercise 1 (Integrating \(x\))
Call the lower estimate \(L_N\) and the upper estimate \(U_N\). Prove that as \(N\to\infty\), so long as one of these sums converges so does the other, and their values are equal. Thus, the constant sequence \(\int_{[0,1]}x\) is squeezed between \(L_N\) and \(U_N\), so in the limit must also take their common value!
Next, prove that \(U_N\to \frac{1}{2}\) as \(N\to\infty\) *Hint: use previous homework, where we did summation by parts to find a formula for \(\sum_{i=1}^N i=1+2+3\cdots+N\).
Solution. Writing them out, \[L_N = \frac{0}{N^2}+\frac{1}{N^2}+\frac{2}{N^2}+\cdots+\frac{N-1}{N^2}\] \[U_N =\frac{1}{N^2}+\frac{2}{N^2}+\cdots+\frac{N-1}{N^2}+\frac{N}{N^2}\]
Thus, the two sums are just ‘shifted copies’ of one another and thier differencde telescopes: \(U_N-L_N=\frac{N}{N^2}-\frac{0}{N^2}=\frac{1}{N}\), so
\[U_N=L_N+\frac{1}{N}\]
Thus, assuming either sum converges they both do, and by the limit theorems \[\lim U_N= \lim L_N+\lim\frac{1}{N}=\lim L_N\]
So we just need to find the value of one of the sums. Taking \(U_N\), we notice we can factor out the \(N^2\) in the denominator as a constant from the sum, leaving us with a sum of the first \(N\) consecutive integers. But we previously figured out this sum! \[U_N=\frac{1}{N^2}(1+2+3+\cdots+N)=\frac{1}{N^2}\frac{N(N+1)}{2}\] Simplifying and taking a limit as \(N\to\infty\), \[\lim U_N = \lim \frac{N+1}{2N}=\frac{1}{2}\lim\left(1+\frac{1}{N}\right)=\frac{1}{2}\left(1+\lim\frac{1}{N}\right)=\frac{1}{2}\]
For all \(N\) we knew \(L_N\leq \int_{[0,1]}x\leq U_N\), but now we’ve shown \(\lim L_N=\lim U_N=\frac{1}{2}\)! So, the integral is trapped between these by the squeeze theorem, and \[\int_{[0,1]}x=\frac{1}{2}\]
Throughout this entire calculation we’ve only used the axioms, and the assumption that \(x\) is integrable. Thus, this exercise proves a pretty strong result: no matter how you try to precisely define integrals, there is an unambiguous choice for the value of \(\int_{[0,1]}x\). If its defined at all, it must equal exactly \(1/2\).
Exercise 2 (Integrating \(x\), Part II) Extend the above result to show that if \(x\) is integrable on \([a,b]\), then \[\int_{[a,b]}x=\frac{b^2}{2}-\frac{a^2}{2}\] (Its OK if you assume in your proof that \(0< a < b\) to cut down on worrying about negative numbers, the proof of the general case is not much more work)
Hint: first, generalize the work we did together in the book above, from the interval \([0,1]\) to a general interval \([0,c]\) and prove that \(\int_{[0,c]}x=\frac{c^2}{2}\). Then use the fact that you know this for all \(c>0\) and the subdivision axiom to get what you want.
Solution. The solution above immediately generalizes from the interval \([0,1]\) to \([0,c]\). Still taking our upper and lower sums \(U_N,L_N\) to be evenly partitioning the interval into \(N\) pieces, each piece \(I_i\) is now of width \(c/N\), with endpoints \(I_i=[c(i-1)/N, ci/N]\). Since \(x\) is increasing over the interval \([0,c]\) it takes its max value at the right endpoint and its min value at the left endpoint, so \[L_N=\sum_{i=1}^N f(c(i-1)/N)\frac{c}{N}=\sum_{i=1}^N\frac{c(i-1)}{N}\frac{1}{N}=c^2\sum_{i=1}^N\frac{i-1}{N^2}\] \[U_N=\sum_{i=1}^N f(ci/N)\frac{c}{N}=\sum_{i=1}^N\frac{ci}{N}\frac{1}{N}=c^2\sum_{i=1}^N\frac{i}{N^2}\]
But these sums are exactly a constant \(c\) times the two sums we did in the previous problem! There we showed the limits were both equal to 1/2, so by the limit theorems we see \[\lim U_N=\lim L_N = c^2\lim\sum_{i=1}^N\frac{i}{N^2}=c^2\frac{1}{2}=\frac{c^2}{2}\]
Now to deal with an arbitrary interval \([a,b]\). If \(a,b>0\) then we can work by subdivision to see \[\int_{[0,b]}x=\int_{[0,a]}x+\int_{[a,b]}x\] And solving for the integral we want
\[\int_{[a,b]}x=\int_{[0,b]}x-\int_{[0,a]}x\]
But we know both of these! \[\int_{[0,b]}x=\frac{b^2}{2}\hspace{1cm}\int_{[0,a]}x=\frac{a^2}{2}\]
All together then, \[\int_{[a,b]}x=\frac{b^2}{2}-\frac{a^2}{2}\]
Exercise 3 (Integrating \(x^2\)) Prove that if \(x^2\) is integrable on \([0,c]\) that its value must be \[\int_{[0,c]}x^2=\frac{c^3}{3}\]
Use this to deduce that for any interval \([a,b]\), (feel free to just do the case \(a,b\geq 0\)) \[\int_{[a,b]}x^2 = \frac{b^3}{3}-\frac{a^3}{3}\]
Hint: follow the similar process to what we did in class/the book/the previous exercise, for \(x\): using the axioms to bound by sums, and then using the summation by parts formula from earlier in the course to calculate the limit
Solution. The same ideas carry through without change here: starting on an arbitrary interval \([0,c]\) with \(c>0\), we define the upper and lower sums by subdividing into \(N\) constant width partitions \(I_i=[c(i-1)/N, ci/N]\), and bounding \(f(x)=x^2\) by its maximal and minimal values on each partition, respectively.
Since \(x^2\) is increasing, it takes its minimal value on \(I_i\) at the left endpoint and its maximal value on the right endpoint: so for all \(x\in I_i\), \[\left(\frac{c(i-1)}{N}\right)^2\leq x^2\leq \left(\frac{ci}{N}\right)^2\]
Following exactly as we did in the book, by the axiom that integrals respect inequalities and the axiom that we know how to integrate constants, \[\frac{c^2(i-1)^2}{N^2}\frac{c}{N}=\int_{I_i}\left(\frac{c(i-1)}{N}\right)^2\leq\int_{I_i}x^2\leq\int_{I_i}\left(\frac{ci}{N}\right)^2=\frac{c^2i^2}{N^2}\frac{c}{N}\]
Using the subdivision axiom we can then sum all of these up to get bounds on the integral \(\int_{[0,c]}x^2\):
\[\sum_{i=1}^N\frac{c^3}{N^3}(i-1)^2\leq\int_{[0,c]}x^2\leq \sum_{i=1}^N\frac{c^3}{N^3}i^2\]
Call the lower estimate \(L_N\) and the upper estimate \(U_N\). Notice these are the same sum, but shifted by one term, so their difference is again easy to compute: \[U_N-L_N=\frac{c^3}{N^3}\left(N-0\right)=\frac{c^3}{N^2}\]
So, as \(N\to\infty\) this difference goes to zero, and as long as one of the sums converges so does the other, and their limiting value is equal. This means we need only calculate one of the sums: sticking with \(U_N\) we factor out all of the constants to reveal a sum of the squares of the first \(N\) positive integers. But we know this sum! (From our earlier section on partial summation)
\[U_N =\frac{c^3}{N^3}\sum_{i=1}^N i^2 = \frac{c^3}{N^3}\frac{N(N+1)(2N+1)}{6}\]
We can evaluate this straighforwardly with the limit laws, knowing that limits of the form \(\frac{aN+b}{N}\) exist:
\[\lim U_N =\frac{c^3}{6}\left(\lim \frac{N}{N}\right)\left(\lim\frac{N+1}{N}\right)\left(\lim\frac{2N+1}{N}\right)\] \[=\frac{c^3}{6}(1)(1)\left(\frac{1}{2}\right)=\frac{c^3}{3}\]
Extending to an arbitrary interval \([a,b]\subset (0,\infty)\) is then exactly as in our previous problem: by subdivision \[\int_{[0,b]}x=\int_{[0,a]}x+\int_{[a,b]}x\] And solving for the integral we want
\[\int_{[a,b]}x=\int_{[0,b]}x-\int_{[0,a]}x=\frac{b^3}{3}-\frac{a^3}{3}\]