MATH 221

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

Last updated: 6 September 2014

Lecture 25

$${\int }_a^{b}f\left(x\right)dx$$ really means $$\begin{array}{c}{\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}(f\left(a\right)\mathrm{\Delta }x+f(a+\mathrm{\Delta }x)\mathrm{\Delta }x+\cdots +f(b-\mathrm{\Delta }x)\mathrm{\Delta }x)}\\ =\underset{\mathrm{\Delta }x\to 0}{\text{lim}}\left(\text{add up the areas of little boxes}\hspace{0.17em}\begin{array}{c} } f(a+kΔx) ⏟Δx \end{array}\right)\end{array}$$ $$\begin{array}{c} a b x y ⏟ Δx \end{array}$$ The first box $\begin{array}{c} } f(a) ⏟Δx \end{array}$ has area $f\left(a\right)\mathrm{\Delta }x\text{.}$

The second box $\begin{array}{c} } f(a+Δx) ⏟Δx \end{array}$ has area $f(a+\mathrm{\Delta }x)\mathrm{\Delta }x\text{.}$

So think of ${\int }_a^{b}f\left(x\right)dx$ as adding up areas from $a$ to $b$ of infinitesimally small boxes $\begin{array}{c} } f(x) ⏟dx \end{array}$ with area $f\left(x\right)dx\text{.}$

${\int }_0^2{e}^{x}dx$ $$\begin{array}{c} 1 2 x y \end{array}$$ Suppose $\mathrm{\Delta }x=\frac{1}{3}$ $$\begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+e^{2\mathrm{\Delta }x}\mathrm{\Delta }x+e^{3\mathrm{\Delta }x}\mathrm{\Delta }x+e^{4\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x}}\\ & =& {\displaystyle e^0\frac{1}{3}+e^{\frac{1}{3}}\frac{1}{3}+e^{\frac{2}{3}}\frac{1}{3}+e^{\frac{3}{3}}\frac{1}{3}+e^{\frac{4}{3}}\frac{1}{3}+e^{\frac{5}{3}}\frac{1}{3}}\\ & =& {\displaystyle \frac{1}{3}(1+e^{\frac{1}{3}}+{\left(e^{\frac{1}{3}}\right)}^2+{\left(e^{\frac{1}{3}}\right)}^3+{\left(e^{\frac{1}{3}}\right)}^4+{\left(e^{\frac{1}{3}}\right)}^5)}\\ & =& {\displaystyle \frac{1}{3}\left(\frac{{\left(e^{\frac{1}{3}}\right)}^6-1}{e^{\frac{1}{3}}-1}\right)}\\ & =& {\displaystyle \frac{1}{3}\left(\frac{e^{\frac{6}{3}}-1}{e^{\frac{1}{3}}-1}\right)}\\ & =& {\displaystyle (e^2-1)\left(\frac{\frac{1}{3}}{e^{\frac{1}{3}}-1}\right)\text{.}}\end{array}$$ Suppose $\mathrm{\Delta }x=\frac{1}{5}$ $$\begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+e^{2\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x}}\\ & =& {\displaystyle e^0\frac{1}{5}+e^{\frac{1}{5}}\frac{1}{5}+e^{\frac{2}{5}}\frac{1}{5}+e^{\frac{3}{5}}\frac{1}{5}+\cdots +e^{\frac{9}{5}}\frac{1}{5}}\\ & =& {\displaystyle \frac{1}{5}(e^0+e^{\frac{1}{5}}+e^{\frac{2}{5}}+e^{\frac{3}{5}}+e^{\frac{4}{5}}+\cdots +e^{\frac{9}{5}})}\\ & =& {\displaystyle \frac{1}{5}(e^0+e^{\frac{1}{5}}+{\left(e^{\frac{1}{5}}\right)}^2+{\left(e^{\frac{1}{5}}\right)}^3+\cdots +{\left(e^{\frac{1}{5}}\right)}^9)}\\ & =& {\displaystyle \frac{1}{5}\left(\frac{{\left(e^{\frac{1}{5}}\right)}^{10}-1}{e^{\frac{1}{5}}-1}\right)}\\ & =& {\displaystyle (e^2-1)\left(\frac{\frac{1}{5}}{e^{\frac{1}{5}}-1}\right)\text{.}}\end{array}$$ Suppose $\mathrm{\Delta }x=\frac{1}{N}$ $$\begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+e^{2\mathrm{\Delta }x}\mathrm{\Delta }x+e^{3\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x}}\\ & =& {\displaystyle e^0\frac{1}{N}+e^{\frac{1}{N}}\frac{1}{N}+e^{\frac{2}{N}}\frac{1}{N}+e^{\frac{3}{N}}\frac{1}{N}+e^{\frac{4}{N}}\frac{1}{N}+\cdots +e^{2-\frac{1}{N}}\frac{1}{N}}\\ & =& {\displaystyle (e^2-1)\left(\frac{\frac{1}{N}}{e^{\frac{1}{N}}-1}\right)\text{.}}\end{array}$$ So $$\underset{\mathrm{\Delta }x\to 0}{\text{lim}}(e^0\mathrm{\Delta }x+e^{\mathrm{\Delta }x}\mathrm{\Delta }x+\cdots +e^{2-\mathrm{\Delta }x}\mathrm{\Delta }x)=\underset{\mathrm{\Delta }x\to 0}{\text{lim}}(e^2-1)\frac{\mathrm{\Delta }x}{(e^{\mathrm{\Delta }x}-1)}=(e^2-1)·1=e^2-1\text{.}$$ Note: $$\begin{array}{ccc}{\displaystyle {\int }_0^2{e}^{x}dx}& =& {\displaystyle e^x+c{|}_{x=0}^{x=2}}\\ & =& {\displaystyle (e^2+c)-(e^0+c)}\\ & =& {\displaystyle e^2+c-e^0-c}\\ & =& {\displaystyle e^2-e^0}\\ & =& {\displaystyle e^2-1\text{.}}\end{array}$$

${\int }_{-1}^1\frac{1}{x^2}dx$ $$\begin{array}{c} -1 1 x 1 y \end{array}$$ By adding up little boxes: $$\begin{array}{ccc}{\displaystyle {\int }_{-1}^1\frac{1}{x^2}dx}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}(\frac{1}{{\left(-1\right)}^2}\mathrm{\Delta }x+\frac{1}{{(-1+\mathrm{\Delta }x)}^2}\mathrm{\Delta }x+\frac{1}{{(-1+2\mathrm{\Delta }x)}^2}\mathrm{\Delta }x+\cdots +\frac{1}{{(1-\mathrm{\Delta }x)}^2}\mathrm{\Delta }x)}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}(1·\mathrm{\Delta }x+\frac{1}{{(-1+\mathrm{\Delta }x)}^2}\mathrm{\Delta }x+\cdots +\underset{\text{OOPS!!}}{\frac{1}{0^2}}\mathrm{\Delta }x+\cdots +\frac{1}{{(1-\mathrm{\Delta }x)}^2}\mathrm{\Delta }x)}\end{array}$$ So ${\int }_{-1}^1\frac{1}{x^2}dx$ is UNDEFINED.

Note: $$\begin{array}{ccc}{\displaystyle {\int }_{-1}^1\frac{1}{x^2}dx}& =& {\displaystyle {\int }_{-1}^1{x}^{-2}dx}\\ & =& {\displaystyle \frac{x^{-1}}{-1}+c{|}_{x=-1}^{x=1}}\\ & =& {\displaystyle (\frac{1^{-1}}{-1}+c)-(\frac{{\left(-1\right)}^{-1}}{-1}+c)}\\ & =& {\displaystyle -1+c-1-c}\\ & =& {\displaystyle -2\text{.}}\end{array}$$ So this is a case where $${\int }_a^b\frac{df}{dx}dx\ne f\left(b\right)-f\left(a\right)\text{.}$$ i.e. $$\text{adding up areas of little boxes}$$ and $$\text{doing the indefinite integral and plugging in}$$ give different answers.

The fundamental theorem of calculus says $${\int }_a^b\frac{df}{dx}dx=f\left(b\right)-f\left(a\right)$$ (is not a lie) provided $f\left(x\right)$ doesn't do anything bad between $a$ and $b\text{.}$ It should be

(a)	defined everywhere between $a$ and $b,$
(b)	continuous everywhere between $a$ and $b,$
(c)	differentiable everywhere between $a$ and $b\text{.}$

The fundamental theorem of calculus says $$\text{Area under}\hspace{0.17em}g\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b=A\left(b\right)-A\left(a\right)$$ where $\int g\left(x\right)dx=A\left(x\right)+c\text{.}$

Why does this work?

Let $A\left(x\right)=\text{area}$ under $g\left(x\right)$ from $a$ to $x\text{.}$ Then $$\begin{array}{ccc}{\displaystyle \frac{dA}{dx}}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{A(x+\mathrm{\Delta }x)-A\left(x\right)}{\mathrm{\Delta }x}\begin{array}{c} a x x y y=g(x) \end{array}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\left(\text{area of last little box}\right)}{\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{g\left(x\right)\mathrm{\Delta }x}{\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}g\left(x\right)=g\left(x\right)\text{.}\begin{array}{c} a x Δ x + x x y y=g(x) \end{array}}\end{array}$$ So $$\int g\left(x\right)dx=A\left(x\right)+c\text{.}$$ So $$\begin{array}{ccc}{\displaystyle A\left(b\right)-A\left(a\right)}& =& {\displaystyle \left(\text{area under}\hspace{0.17em}g\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b\right)-\left(\text{area under}\hspace{0.17em}g\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}a\right)}\\ & =& {\displaystyle \text{area under}\hspace{0.17em}g\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 25 from a series of handwritten lecture notes for the class MATH 221 given on November 6, 2000.

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