- We'll show an alternative way of proving our formulas, that of
*Mathematical induction*.- It might be a little more incomfortable that little Gauß's proof, but it will give a streamlined approach to proving other formulas.

*Mathematical induction.*- Let \(m\) be an integer, and let \(S_n\) be a statement involving an integer \(n\ge m\).
- Suppose that \(S_m\) is true,
- and suppose that if \(S_n\) is true, then so is \(S_{n+1}\).
- Then \(S_n\) is true for all \(n\ge m\).

- Proof of \(\sum_{i=1}^ni=\frac{n(n+1)}{2}\) using Mathematical Induction.
- We'll have \(m=1\).
- \(S_1:\sum_{i=1}^1i=\frac{1\cdot 2}{2}=1\).
- Suppose that \(S_n:\sum_{i=1}^ni=\frac{n(n+1)}{2}\). Then to prove \(S_{n+1}:\) \[ \sum_{i=1}^{n+1}i=\sum_{i=1}^ni+(n+1)=\frac{n(n+1)}{2}+(n+1)=\frac{n^2+n+2(n+1)}{2}=\frac{(n+1)(n+2)}{2} \] therefore \(S_{n+1}\) is true.
- Therefore, Mathematical induction shows that \(S_n:\sum_{i=1}^ni=\frac{n(n+1)}{2}\) is true for all \(n\ge1\).
- Exercises. Appendix E: 38, 42, 50

- Let's try to solve the
*area problem*.- Suppose that \(f(x)\ge0\) and the domain is \([a,b]\).
- Find the area of the region \(S\) that lies under the curve \(y=f(x)\) from \(a\) to \(b\).
- The solution is clear, if \(S\) is a rectangle, a triangle, or a shape that can be decomposed to those ones, since they have area formulas.
- But what about, \(f(x)=x^2\), over \([0,1]\)?

- What you can do is to
*approximate*the area the following way.- Cut up the interval \([a,b]\) to \(n\)
*subintervals*\[ I_1=[a,a+\Delta x],\,I_2=[a+\Delta x,a+2\Delta x],\dotsc,I_n=[a+(n-1)\Delta x,b] \] of equal length. - We denote the length of one piece by \(\Delta x=\frac{b-a}{n}\).
- Then let's sum up the areas of the rectangles with base the subintervals \(I_i\) and height the value of \(f(x)\) at the rightmost point \(f(a+i\Delta x)\).
- Click here for an illustration.
- As \(n\) increases, the sum seems to approach \(\frac13\).

- Cut up the interval \([a,b]\) to \(n\)

- Let us now prove that the area under \(y=f(x)=x^2\) for \(0\le x\le1\) is \(\frac13\).
- We are approximating this area with the sum of the rectangles with base \(\Delta x=\frac{b-a}{n}=\frac{1}{n}\), and height \(f(a+i\Delta x)=f(\frac{i}{n})=\frac{i^2}{n^2}\).
- That is, the approximated area is \(R_n=\sum_{i=1}^n\frac{1}{n}\frac{i^2}{n^2}\).
- We can calculate this using a formula we've seen: \[ R_n=\sum_{i=1}^n\frac{i^2}{n^3}=\frac{1}{n^3}\sum_{i=1}^ni^2=\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}. \]
- Then we can take the limit of this: \[ \lim_{n\to\infty}R_n=\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}\frac{1(1+n^{-1})(2+n^{-1})}{6}=\frac13. \]
- Exercise. Show that we get the same limit if we make the height of the rectangles the left endpoints \(f(a+(i+1)\Delta x)=f(\frac{i-1}{n})=\frac{(i-1)^2}{n^2}\).
- Exercise. Appendix E: 26.