Integration

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

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 1 March 2010

Integration

${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ really means

$${\displaystyle \underset{\Delta x\to 0}{\lim }\left(f\left(a\right)\Delta x+f\left(a+\Delta x\right)+\dots +f\left(b-\Delta x\right)\Delta x\right)}$$

$${\displaystyle =\underset{\Delta x\to 0}{\lim }\left(\text{ add up the areas of the little boxes of width}\Delta x\text{ and height}f\left(a+k\Delta x\right)\right)}$$

Example of little box

Example of how multiple little boxes are used to approximate the integral

The leftmost box has area $f\left(a\right)\Delta x.$

The second box has area $f\left(a+\Delta x\right)\Delta x\dots$

So think of ${\displaystyle \underset{b}{\overset{a}{\int }}f\left(x\right)\mathrm{dx}}$ as adding up areas from $a$ to $b$ of infinitesimally small boxes with area $f\left(x\right)\Delta x$

Example

${\displaystyle \underset{0}{\overset{2}{\int }}{e}^{x}dx}$

Suppose $\Delta x=\frac{1}{4}$

$$e^0\Delta x+e^{\Delta x}\Delta x+e^{2\Delta x}\Delta x+e^{3\Delta x}\Delta x+\dots +e^{2-\Delta x}\Delta x$$$$=e^0\frac{1}{4}+e^{\frac{1}{4}}\frac{1}{4}+e^{\frac{2}{4}}\frac{1}{4}+\dots +e^{\frac{7}{4}}\frac{1}{4}$$$$=\frac{1}{4}\left(1+e^{\frac{1}{4}}+{\left(e^{\frac{1}{4}}\right)}^2+{\left(e^{\frac{1}{4}}\right)}^3+\dots +{\left(e^{\frac{1}{4}}\right)}^7\right)$$$$=\frac{1}{4}\left(\frac{{\left(e^{\frac{1}{4}}\right)}^8-1}{e^{\frac{1}{4}}-1}\right)=\frac{1}{4}\left(\frac{e^{\frac{8}{4}}-1}{e^{\frac{1}{4}}-1}\right)=\left(e^2-1\right)\left(\frac{\frac{1}{4}}{e^{\frac{1}{4}}-1}\right)$$

Suppose $\Delta x=\frac{1}{5}$

$$e^0\Delta x+e^{\Delta x}\Delta x+e^{2\Delta x}\Delta x+e^{3\Delta x}\Delta x+\dots +e^{2-\Delta x}\Delta x$$$$=e^0\frac{1}{5}+e^{\frac{1}{5}}\frac{1}{5}+e^{\frac{2}{5}}\frac{1}{5}+\dots +e^{\frac{9}{5}}\frac{1}{5}$$$$=\frac{1}{5}\left(1+e^{\frac{1}{5}}+{\left(e^{\frac{1}{5}}\right)}^2+{\left(e^{\frac{1}{5}}\right)}^3+\dots +{\left(e^{\frac{1}{5}}\right)}^9\right)$$$$=\frac{1}{5}\left(\frac{{\left(e^{\frac{1}{5}}\right)}^{10}-1}{e^{\frac{1}{5}}-1}\right)=\frac{1}{5}\left(\frac{e^{\frac{10}{5}}-1}{e^{\frac{1}{5}}-1}\right)=\left(e^2-1\right)\left(\frac{\frac{1}{5}}{e^{\frac{1}{5}}-1}\right)$$

Suppose $\Delta x=\frac{1}{N}$

$$e^0\Delta x+e^{\Delta x}\Delta x+e^{2\Delta x}\Delta x+e^{3\Delta x}\Delta x+\dots +e^{2-\Delta x}\Delta x$$$$=e^0\frac{1}{N}+e^{\frac{1}{N}}\frac{1}{N}+e^{\frac{2}{N}}\frac{1}{N}+\dots +e^{2-\frac{1}{N}}\frac{1}{N}$$$$=\left(e^2-1\right)\left(\frac{\frac{1}{N}}{e^{\frac{1}{N}}-1}\right)$$

So ${\displaystyle \underset{\Delta x\to 0}{\lim }\left(e^0\Delta x+e^{\Delta x}\Delta x+e^{2\Delta x}\Delta x+e^{3\Delta x}\Delta x+\dots +e^{2-\Delta x}\Delta x\right)}$

${\displaystyle =\underset{\Delta x\to 0}{\lim }\left(e^2-1\right)\frac{\Delta x}{e^{\Delta x}-1}=\left(e^2-1\right)\times 1=e^2-1.}$

Note: $${\displaystyle \underset{0}{\overset{2}{\int }}{e}^{x}dx=e^x+c{|}_{x=0}^{x=2}=\left(e^2+c\right)-\left(e^0+c\right)=e^2+c-e^0-c=e^2-1.}$$

Example

${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{x^2}}\mathrm{dx}$

By adding up little boxes: $$\underset{-1}{\overset{1}{\int }}\frac{1}{x^2}dx=\underset{\Delta x\to 0}{\lim }\left(\frac{1}{{\left(-1\right)}^2}\Delta x+\frac{1}{{\left(-1+\Delta x\right)}^2}\Delta x+\frac{1}{{\left(-1+2\Delta x\right)}^2}\Delta x+\dots +\frac{1}{{\left(-1-\Delta x\right)}^2}\Delta x\right)$$

$$=\underset{\Delta x\to 0}{\lim }\left(\frac{1}{{\left(-1\right)}^2}\Delta x+\frac{1}{{\left(-1+\Delta x\right)}^2}\Delta x+\dots +\frac{1}{{\left(0\right)}^2}\Delta x+\dots +\frac{1}{{\left(-1-\Delta x\right)}^2}\Delta x\right)$$

OOPS! We can't divide by 0.

Thus ${\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{x^2}dx}$ is UNDEFINED.

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

Fundamental Theorem of Calculus

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

1. defined everywhere between $a$ and $b,$
2. continuous everywhere between $a$ and $b,$
3. differentiable everywhere between $a$ and $b.$

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

Why does this work?

Let $A\left(x\right)=$ area under $g\left(x\right)$ from $a$ to $x.$

Then $$\begin{array}{rcl}\frac{dA}{dx}& =& \underset{\Delta x\to 0}{\lim }\frac{A\left(x+\Delta x\right)-A\left(x\right)}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{\left(\text{area of the last little box}\right)}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }\frac{g\left(x\right)\Delta x}{\Delta x}\\ & =& \underset{\Delta x\to 0}{\lim }g\left(x\right)\\ & =& g\left(x\right).\end{array}$$

So ${\displaystyle \int g\left(x\right)dx=A\left(x\right)+c.}$

So $$A\left(b\right)-A\left(a\right)=\left(\text{area under }g\left(x\right)\text{ from }a\text{ to }b\right)-\left(\text{area under }g\left(x\right)\text{ from }a\text{ to }a\right)=\left(\text{area under }g\left(x\right)\text{ from }a\text{ to }b\right).$$

References

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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