- We will show that antiderivatives can be used to calculate areas of plane domains, in particular areas below graphs.
- For this, we will need to use the
*Sigma notation* *Definition.*If \(m\le n\) are integers and \(a_m,\dotsc,a_n\) are real numbers, then \[ \sum_{i=m}^na_i=a_m+a_{m+1}+a_{m+2}+\dotsb+a_{n-1}+a_n. \]- With function notation, this can be written as \[ \sum_{i=m}^nf(i)=f(m)+f(m+1)+f(m+2)+\dotsb+f(n-1)+f(n). \]

- For this, we will need to use the
- Exercises. Appendix E: 2, 4, 8, 12, 16, 18, 20, 22, 24, 26
- Find \(\sum_{i=1}^n1\).
- By definition, the number \(\sum_{i=1}^n1\) is the number \(1\) added together \(n\) times.
- That is, we have \(\sum_{i=1}^n=1+1+\dotsb+1=n\).

*Theorem.*Let \(c\) be a number. Then we have the following formulas.- \(\sum_{i=m}^nca_i=c\sum_{i=m}^na_i\).
- \(\sum_{i=m}^na_i+b_i=\sum_{i=m}^na_i+\sum_{i=m}^nb_i\).

- Find \(\sum_{i=1}^ni=1+2+\dotsb+n\).
- There is an anecdote that when the German Mathematician Karl Friedrich Gauß (1777-1855) was 10 years old, his elementary school teacher made everyone in the class add up the numbers from 1 to 100. 10-year old Gauß came up with the formula which we'll prove, he didn't have to add up the numbers by hand, and thus he was done way earlier than the others.
- The trick is to write up the sum both in ascending and descending order: \[\begin{alignat*}{5} \sum_{i=1}^ni&=1&&+2&&+\dotsb&&+(n-1)&&+n \\ \sum_{i=1}^ni&=n&&+(n-1)&&+\dotsb&&+2&&+1 \end{alignat*}\]
- Adding together the two lines, we get \[ 2\sum_{i=1}^ni=(n+1)+(n+1)+\dotsb+(n+1)+(n+1)=n(n+1), \] which gives \(\sum_{i=1}^ni=\frac{n(n+1)}{2}\).
- Therefore, little Gauß could write \(\sum_{i=1}^{100}i=\frac{100\cdot101}{2}=5050\). Sehr gut!

- Let's now find \(\sum_{i=1}^ni^2=1^2+2^2+\dotsb+n^2\).
- Let \(S\) denote the desired sum.
- First, let's write out the followig
*telescoping sum*, or collapsing sum. \[ \sum_{i=1}^n((i+1)^3-i^3)=(2^3-1^3)+(3^3-2^3)+(4^3-3^3)+\dotsb+((n+1)^3-n^3)=(n+1)^3-1^3=n^3+3n^2+3n. \] - On the other hand, we can use the Theorem to write the following. \[ \sum_{i=1}^n((i+1)^3-i^3)=\sum_{i=1}^n(3i^2+3i+1)=3\sum_{i=1}^ni^2+3\sum_{i=1}^ni+\sum_{i=1}^n1 =3S+\frac32n(n+1)+n=3S+\frac32n^2+\frac52n. \]
- Equating the 2 right sides, we get the following \[ n^3+3n^2+3n=3S+\frac32n^2+\frac52n. \]
- This gives \[ 3S=n^3+\frac32n^2+\frac12n, \] and therefore \[ S=\frac{2n^3+3n^2+n}{6}=\frac{n(n+1)(2n+1)}{6}. \]

- We'll be using the following formulas.
- \(\sum_{i=1}^n1=n\).
- \(\sum_{i=1}^ni=\frac{n(n+1)}{2}\).
- \(\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}\).
- \(\sum_{i=1}^ni^3=\left(\frac{n(n+1)}{2}\right)^2\).

- Exercises. Appendix E: 30, 32, 36, 44, 48