Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 July 2014

Lecture 20

The fundamental theorem of calculus

Write $$\int f\left(x\right)dx=A\left(x\right)+c\text{if}\frac{d}{dx}A\left(x\right)=f\left(x\right)\text{.}$$ The fundamental theorem of calculus says $$\text{Area under}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\text{to}\hspace{0.17em}b=A\left(b\right)-A\left(a\right)$$ ${\int }_a^{b}f\left(x\right)dx$ means $$\begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}(f\left(a\right)\mathrm{\Delta }x+f(a+\mathrm{\Delta }x)\mathrm{\Delta }x+\cdots +f(b-\mathrm{\Delta }x)\mathrm{\Delta }x)}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\left(\text{sum of the areas of the little boxes}\hspace{0.17em}\begin{array}{c} } ⏟ Δx \end{array}f(a+\mathrm{\Delta }x)\right)}\end{array}$$ $$\begin{array}{c} a b \end{array}$$ First box has area $f\left(a\right)\mathrm{\Delta }x\text{.}$ Second box has area $f(a+\mathrm{\Delta }x)\mathrm{\Delta }x,$ $\dots \text{.}$

Why is the fundamental theorem of calculus true? $$\text{Area under}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b=A\left(b\right)-A\left(a\right)$$ where $\int f\left(x\right)dx=A\left(x\right)+c\text{.}$

Idea of proof

To show: There exists $A:[a,b]\to ℝ$ such that

(a)	If $c\in [a,b]$ then $A\prime \left(c\right)=f\left(c\right)$
(b)	$A\left(b\right)-A\left(a\right)=$ Area under $f\left(x\right)$ from $a$ to $b\text{.}$

Let $$A\left(x\right)=\text{area under}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}x$$ $$\begin{array}{c} y=f(x) a x b x y A(x) \end{array}$$ To show:

(a)	If $c\in [a,b]$ then $A\prime \left(c\right)=f\left(c\right)$
(b)	$A\left(b\right)-A\left(a\right)=$ Area under $f\left(x\right)$ from $a$ to $b\text{.}$

(a) Assume $c\in [a,b]\text{.}$
To show: $A\prime \left(c\right)=f\left(c\right)\text{.}$ $$\begin{array}{ccc}{\displaystyle A\prime \left(c\right)}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{A(c+\mathrm{\Delta }x)-A\left(c\right)}{\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\left(\frac{\text{area under}\hspace{0.17em}f\left(x\right)}{\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}c+\mathrm{\Delta }x}\right)-\left(\frac{\text{area under}\hspace{0.17em}f\left(x\right)}{\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}c}\right)}{\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\left(\frac{\text{area of little box}}{\text{with height}\hspace{0.17em}f\left(c\right)}\right)}{\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\left(f\left(c\right)\mathrm{\Delta }x\right)}{\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}f\left(c\right)=f\left(c\right)\text{.}}\end{array}$$

(b) To show: $A\left(b\right)-A\left(a\right)=$ area under $f\left(x\right)$ from $a$ to $b\text{.}$ $$\begin{array}{ccc}{\displaystyle A\left(b\right)-A\left(a\right)}& =& {\displaystyle \left(\frac{\text{area under}\hspace{0.17em}f\left(x\right)}{\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b}\right)-\left(\frac{\text{area under}\hspace{0.17em}f\left(x\right)}{\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}a}\right)}\\ & =& {\displaystyle \left(\frac{\text{area under}\hspace{0.17em}f\left(x\right)}{\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b}\right)}\end{array}$$

${\int }_0^2{e}^{x}dx$ $$\begin{array}{c} 1 2 x y y=ex \end{array}$$ $${\int }_0^2{e}^{x}dx=\underset{\mathrm{\Delta }x\to 0}{\text{lim}}(e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+e^{2\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{(2-\mathrm{\Delta }x)}\mathrm{\Delta }x)$$ If $\mathrm{\Delta }x=\frac{1}{3}$ then $$\begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x}}\\ & =& {\displaystyle e^0\frac{1}{3}{e}^{\frac{1}{3}}\frac{1}{3}+e^{\frac{2}{3}}\frac{1}{3}+e^{\frac{3}{3}}\frac{1}{3}+e^{\frac{4}{3}}\frac{1}{3}+e^{\frac{5}{3}}\frac{1}{3}}\\ & =& {\displaystyle \frac{1}{3}(e^0+e^{\frac{1}{3}}+{\left(e^{\frac{1}{3}}\right)}^2+{\left(e^{\frac{1}{3}}\right)}^3+{\left(e^{\frac{1}{3}}\right)}^4+{\left(e^{\frac{1}{3}}\right)}^5)}\\ & =& {\displaystyle \frac{1}{3}\left(\frac{e^{\frac{6}{3}}-1}{e^{\frac{1}{3}}-1}\right)=(e^2-1)\left(\frac{\frac{1}{3}}{e^{\frac{1}{3}}-1}\right)\text{.}}\end{array}$$ If $\mathrm{\Delta }x=\frac{1}{5}$ then $$\begin{array}{ccc}{\displaystyle e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x}& =& {\displaystyle e^0\frac{1}{5}+e^{\frac{1}{5}}·\frac{1}{5}+\cdots +e^{\frac{9}{5}}·\frac{1}{5}}\\ & =& {\displaystyle \frac{1}{5}(e^0+e^{\frac{1}{5}}+{\left(e^{\frac{1}{5}}\right)}^2+{\left(e^{\frac{1}{5}}\right)}^3+\cdots +{\left(e^{\frac{1}{5}}\right)}^9)}\\ & =& {\displaystyle \frac{1}{5}\left(\frac{e^{105}-1}{e^{\frac{1}{5}}-1}\right)=(e^2-1)\frac{\frac{1}{5}}{e^{\frac{1}{5}}-1}\text{.}}\end{array}$$ So $${\int }_0^2{e}^{x}dx=\underset{\mathrm{\Delta }x\to 0}{\text{lim}}(e^0\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x)=\underset{N\to \infty }{\text{lim}}(e^2-1)\left(\frac{1}{N}{e}^{\frac{1}{N}}-1\right)\text{.}$$ Recall: $$\underset{x\to 0}{\text{lim}}\frac{e^x-1}{x}=1\text{.}$$ So $\underset{N\to \infty }{\text{lim}}\frac{e^{\frac{1}{N}}-1}{\frac{1}{N}}=1\text{.}$ So $\underset{N\to \infty }{\text{lim}}\frac{\frac{1}{N}}{e^{\frac{1}{N}}-1}=1\text{.}$ So $${\int }_0^2{e}^{x}dx=\underset{N\to \infty }{\text{lim}}(e^2-1)\left(\frac{\frac{1}{N}}{e^{\frac{1}{N}}-1}\right)=(e^2-1)·1=e^2-1\text{.}$$ Note: $$\int e^{x}dx=e^x+c$$ and $$e^x+e{|}_{x=0}^{x=2}=(e^2+c)-(e^0+c)=e^2-1\text{.}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100423Lect20.pdf and was given on 23 April 2010.

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