Integral (Reimann)

Partition and Refinement

Definition

Def Partition of Interval Given interval $[a,b]\subset \mathbb{R}$, a set $P$ is a partition of $[a,b]$ if

• $P\subseteq [a,b]$.
• $P$ is finite.
• $a\in P$ and $b\in P$.

By convention, label the elements in $P$ as: $a=x_0<x_1<\cdots <x_n=b$

Definition

Def Refinement A partition $Q$ is called refinement of $P$ if $P\subseteq Q$.

Proposition

Prop If $P$ is a partition of $[a,b]$ then $P$ is a refinement of $\{a,b\}$.

Proposition

Prop Suppose $P$ and $Q$ are partitions of $[a,b]$, then $P\cup Q$ is a refinement of $P$ and $Q$.

Riemann-Darboux Integrability

Definition

Def Lower Sum and Upper Sum Suppose $f:[a,b]\to \mathbb{R}$ is a bounded function, and elements of partition $P$ on $[a,b]$ be $x_0,x_1\dots x_n$. For each $i\in \mathbb{N}$, $1\le i\le n$, we define$m_i=\inf \{f(x):x_{i-1}\le x\le x_i\},M_i=\sup \{f(x):x_{i-1}\le x\le x_i\}$Then the lower sum $L(f,P)$ and upper sum $U(f,P)$ are defined by $L(f,P)={\sum }_{i=1}^n{m}_i(x_i-x_{i-1}),U(f,P)={\sum }_{i=1}^n{M}_i(x_i-x_{i-1})$

Proposition

Prop For all $\epsilon >0$, there is some partition $P$ such that $U(f,P)-L(f,P)<\epsilon$

Riemann-Darboux Integrability

Suppose $f:[a,b]\to \mathbb{R}$ is a bounded function, $f$ is integrable if $L_a^b(f)=U_a^b(f)$where $L_a^b(f)=\sup \{L(f,P):P\text{ is a partition over }[a,b]\}$ and $U_a^b(f)=\inf \{U(f,P):P\text{ is a partition over }[a,b]\}$. And we define the integral of $f$ be ${\int }_a^{b}f(x)=L_a^b(f)=U_a^b(f)$

Lemma

Lemma Suppose $f:[a,b]\to \mathbb{R}$ be bounded function, $P$ and $Q$ are partitions of $[a,b]$. Then,

• $L(f,P)\le U(f,p)$.
• If $Q$ is a refinement of $P$ then $L(f,P)\le L(f,Q)$ and $U(f,p)\ge U(f,Q)$.
• $L(f,P)\le U(f,Q)$.

Proof The first proposition arises from $m_i\le M_i$. And $L(f,P)\le L(f,Q)\le U(f,Q)$ implies the third proposition.

Theorem

Continuous function on $[a,b]$ are integrable on $[a,b]$. (Continuity is sufficient but not necessary.)

Proposition

Suppose $f$ is integrable on $[a,b]$, define $F:[a,b]\to \mathbb{R}$ by $F(x)={\int }_a^{x}f$, then $F$ is continuous.

The first fundamental theorem of calculus

Suppose $f$ is continuous at $c\in (a,b)$. Then $F(x)={\int }_a^{x}f$ is differentiable at $c$ and $F^{\prime }(c)=f(c)$

The second fundamental theorem of calculus

If $f$ is integrable on $[a,b]$ and $f=g^{\prime }$ then ${\int }_a^{b}f=g(b)-g(a).$