Integration

## Integration

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

$lim Δx→0 f a Δx+f a+Δx +…+f b-Δx Δx$

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

## Example

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

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

$e 0 Δx+ e Δx Δx+ e 2Δx Δx+ e 3Δx Δx+…+ e 2-Δx Δx$$= e 0 1 4 + e 1 4 1 4 + e 2 4 1 4 +…+ e 7 4 1 4$$= 1 4 1+ e 1 4 + e 1 4 2 + e 1 4 3 +…+ e 1 4 7$$= 1 4 e 1 4 8 -1 e 1 4 -1 = 1 4 e 8 4 -1 e 1 4 -1 = e 2 -1 14 e 1 4 -1$

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

$e 0 Δx+ e Δx Δx+ e 2Δx Δx+ e 3Δx Δx+…+ e 2-Δx Δx$$= e 0 1 5 + e 1 5 1 5 + e 2 5 1 5 +…+ e 9 5 1 5$$= 1 5 1+ e 1 5 + e 1 5 2 + e 1 5 3 +…+ e 1 5 9$$= 1 5 e 1 5 10 -1 e 1 5 -1 = 1 5 e 10 5 -1 e 1 5 -1 = e 2 -1 15 e 1 5 -1$

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

$e 0 Δx+ e Δx Δx+ e 2Δx Δx+ e 3Δx Δx+…+ e 2-Δx Δx$$= e 0 1 N + e 1 N 1 N + e 2 N 1 N +…+ e 2- 1 N 1 N$$= e 2 -1 1N e 1 N -1$

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

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

Note: $∫ 0 2 e x dx= e x +c | x=0 x=2 = e 2 +c - e 0 +c = e 2 +c- e 0 -c= e 2 -1.$

## Example

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

By adding up little boxes: $∫ -1 1 1 x 2 dx= lim Δx→0 1 -1 2 Δx+ 1 -1+Δx 2 Δx+ 1 -1+2Δx 2 Δx+…+ 1 -1-Δx 2 Δx$

$= lim Δx→0 1 -1 2 Δx+ 1 -1+Δx 2 Δx+…+ 1 0 2 Δx+…+ 1 -1-Δx 2 Δx$

OOPS! We can't divide by 0.

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

Note: $∫ -1 1 1 x 2 dx = ∫ -1 1 x -2 dx= x -1 -1 +c | x=1 x=-1 = 1 -1 -1 +c + -1 -1 -1 +c =-1+c-1-c=-2.$ So this is a case when $∫ a b df dx dx≠f b -f a .$ 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 $∫ a b df dx dx=f b -f a$ (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 where $\int g\left(x\right)dx=A\left(x\right)+c.$

Why does this work?

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

Then $dA dx = lim Δx→0 A x+Δx -A x Δx = lim Δx→0 area of the last little box Δx = lim Δx→0 g x Δx Δx = lim Δx→0 g x = g x .$

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

So