Chapter 35: Definite Integrals

Definite integrals formalize the "area" idea.

Note - The Definite Integral of f(x) from x = a to x = b

﻿$\int_{a}^{b}f(x)dx$﻿

b - Upper Bound

a - Lower Bound

x - Integrand

d - Variable of Integration

The Integral sign ﻿$\int$﻿ is also related to the letter S.

﻿$\int_{a}^{b}f(x)dx = \lim_{n\rightarrow \infty }$﻿ ﻿$\sum_{k=1}^{n}f(a+k\frac{b-a}{n})(\frac{b-a}{n})$﻿ Riemann sum with n equally sized intervals

Example

﻿$\int_{a}^{b}\pi dx =\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\pi \cdot (\frac{1-0}{n})$﻿

﻿$=\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\pi (\frac{1}{n})=\lim_{n\rightarrow \infty }\pi \cdot (\frac{1}{n})\cdot n$﻿

﻿$= \pi$﻿

Thus, the total area is ﻿$\pi$﻿

Fact - ﻿$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$﻿

Example - ﻿$\int_{0}^{1}1-x^{2}dx=\frac{2}{3}$﻿

﻿$\int_{0}^{1}1-x^{2}dx=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}f(0+k\frac{1-0}{N})\cdot (\frac{1-0}{N})$﻿

﻿$=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}f(\frac{k}{N})(\frac{1}{N})$﻿

﻿$=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}\left [ 1-(\frac{k}{N})^{2} \right ](\frac{1}{N})$﻿

﻿$=\lim_{N\rightarrow \infty }\frac{1}{N}\sum_{k=1}^{N}\left [ 1-(\frac{k}{N})^{2} \right ]=\lim_{N\rightarrow \infty }\frac{1}{N}(\sum_{k=1}^{N}1-\sum_{k=1}^{N}(\frac{k}{N})^{2})$﻿

﻿$=\lim_{N\rightarrow \infty }\frac{1}{N}(N-\frac{1}{N^{2}}\sum_{k=1}^{N}k^{2})$﻿

﻿$=\lim_{N\rightarrow \infty }\frac{1}{N}(N-\frac{1}{N^{2}}\frac{N(N+1)(2N+1)}{6})$﻿

﻿$=\lim_{N\rightarrow \infty }1-\frac{1}{N^{3}}(\frac{2N^{3}+3N^{2}+N}{6})$﻿

﻿$=1-\frac{2}{6}=1-\frac{1}{3}=\frac{2}{3}$﻿

Discuss - ﻿$\int_{0}^{1}10xdx=5$﻿ using ﻿$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$﻿

Rules for Definite Integrals

﻿$\int_{a}^{b}f(x)dx= -\int_{b}^{a}f(x)dx$﻿

﻿$\int_{a}^{a}f(x)dx= 0$﻿ (No area)

﻿$\int_{a}^{b}kf(x)dx= k\int_{a}^{b}f(x)dx$﻿

﻿$\int_{a}^{b}f(x)+g(x)dx= \int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$﻿

﻿$\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx= \int_{a}^{c}f(x)dx$﻿

If ﻿$f(x)$﻿ has a min and max on ﻿$[a,b]$﻿

﻿$(min f)(a-b)\leq \int_{a}^{b}f(x)dx\leq (max f)(a-b)$﻿

If ﻿$f(x)\leq g(x)on [a,b]$﻿ then

﻿$\int_{a}^{b}f(x)dx\leq \int_{a}^{b}g(x)dx$﻿

Geometry and Area

Example - Use geometry to calculate ﻿$\int_{a}^{b}f(x)dx$﻿

where ﻿$f(x)=\left\{\begin{matrix} 1 & 0\leq x \leq 1 \\ x & 1\leq x\leq 2 \end{matrix}\right.$﻿

﻿$\int_{0}^{2}f(x)dx=1\cdot 1+1\cdot 1+\frac{1}{2}\cdot 1\cdot 1$﻿

﻿$=2+\frac{1}{2}=\frac{5}{2}$﻿

Example - Calculate ﻿$\int_{0}^{4}\sqrt{4^{2}-x^{2}dx}$﻿

﻿$\int_{0}^{4}\sqrt{4^{2}-x^{2}dx}=\frac{1}{4}(\pi \cdot4^{2} )$﻿

﻿$=4 \pi$﻿

Summary

﻿$\int_{a}^{b}f(x)dx=$﻿ Area under graph

﻿$=\lim_{N\rightarrow \infty }\sum_{k=1}^{N}f(a+k\frac{b-a}{N})(\frac{b-a}{N})$﻿