chapter_1.3.lyx

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\begin_body

\begin_layout Section
Integrals
\end_layout

\begin_layout Standard
\begin_inset CommandInset label
LatexCommand label
name "sec:Integrals"

\end_inset


\end_layout

\begin_layout Subsection
Introduction and Definition of Integral
\end_layout

\begin_layout Standard
\begin_inset Note Comment
status open

\begin_layout Itemize
I am sure that people have an intuition about what an integral is, but just
 have trouble doing the actual calculations.
 So, we should focus on just explaining calculation rules and calculating
 some examples in the lectures, and leave out some theory.
 But of course, we should leave it in the script so they can read up on
 it.
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Sometimes we face the situation that we are only given the rate of change
 
\begin_inset Formula $\dot{x}(t)$
\end_inset

 of some quantity of interest.
 How can we convert this rate into an absolute value? One simple idea would
 be to start with a fixed value 
\begin_inset Formula $x_{0}:=x(t_{0})$
\end_inset

 and add up the changes 
\begin_inset Formula $\dot{x}(t)\Delta t$
\end_inset

 for small intervals 
\begin_inset Formula $\Delta t$
\end_inset

 until we arrive at the point 
\begin_inset Formula $t_{n}$
\end_inset

 where we want to know the value of 
\begin_inset Formula $x$
\end_inset

:
\begin_inset Formula 
\begin{eqnarray*}
x(t_{n}) & = & \sum_{i=0}^{n-1}\dot{x}(t_{i})\Delta t_{i}\\
 & \mbox{with } & \Delta t_{i}=t_{i+1}-t_{i}.
\end{eqnarray*}

\end_inset

 However, in most cases we will make an error, since the rate of change
 within an interval 
\begin_inset Formula $\Delta t_{i}$
\end_inset

 might vary.
 By only looking at the rate at the interval borders 
\begin_inset Formula $t_{i}$
\end_inset

, we completely ignore that.
 We could make the intervals shorter to keep the error small, but in general
 we will always get an answer 
\begin_inset Formula $x(t_{n})$
\end_inset

 that is slightly wrong.
 How can we avoid those errors? 
\end_layout

\begin_layout Standard
In the previous section, we saw that taking the derivative of a function
 is calculating the rate of change of this function at every point.
 In this situation we were given the function and wanted to know the rate
 of change.
 Now, the situation is reverse.
 Therefore, what we would really like to do is to reverse the process of
 differentiation, i.e.
 to get from the rate of change to the actual function value.
 This is the subject of this chapter.
 It will turn out that integration is not as easy as differentiation.
 However, it has a nice geometrical interpretation that we already read
 of the introductory example above.
 Figure 
\begin_inset CommandInset ref
LatexCommand ref
reference "fig:approxIntegral"

\end_inset

 shows a picture of our approximate integration from above.
 When looking at the sum 
\begin_inset Formula $\sum_{i=0}^{n-1}\dot{x}(t_{i})\Delta t_{i}$
\end_inset

 we can interpret 
\begin_inset Formula $\Delta t_{i}$
\end_inset

 as the width of a rectangle with height 
\begin_inset Formula $\dot{x}(t_{i})$
\end_inset

.
 Summing up their products then means summing up the areas of those rectangles.
 We saw in earlier sections, that taking the derivative is the same as computing
 the rate of change of a function for infinitesimal small steps along the
 
\begin_inset Formula $x$
\end_inset

-axis.
 This infinitesimal small step was incorporated via taking the limit of
 the differential quotient, when letting the step size go to zero.
 Integration involves a similar operation that we just mention here.
 When integrating, you really apply the approximate summation from above,
 but with infinitesimal small 
\begin_inset Formula $\Delta t_{i}$
\end_inset

.
 This means that an integral is the limit of the sum from above when letting
 
\begin_inset Formula $\Delta t_{i}$
\end_inset

 go to zero.
 This limit of the sum is then denoted by a new symbol ''
\begin_inset Formula $\int$
\end_inset

'', a stylized ''S'' for ''
\bar under
s
\bar default
um''.
 The 
\begin_inset Formula $\Delta t$
\end_inset

 is replaced by the symbol 
\begin_inset Formula $dt$
\end_inset

, which indicates that 
\begin_inset Formula $\Delta t$
\end_inset

 is infinitesimally small.
 
\end_layout

\begin_layout Standard
\begin_inset Float figure
placement h
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figs/approxgaussintegral.eps
	lyxscale 50
	width 6cm
	keepAspectRatio

\end_inset


\begin_inset Graphics
	filename figs/gaussintegral.eps
	lyxscale 50
	width 6cm
	keepAspectRatio

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption

\begin_layout Plain Layout
\begin_inset CommandInset label
LatexCommand label
name "fig:approxIntegral"

\end_inset

Left: Geometrical interpretation of the approximate integration: Areas of
 width 
\begin_inset Formula $\Delta t_{i}$
\end_inset

 and height 
\begin_inset Formula $\dot{x}(t_{i})$
\end_inset

 are summed up.
 Right: Geometrical interpretation of integration: The integral of the displayed
 function in the interval 
\begin_inset Formula $[-1,1]$
\end_inset

 is the area under the curve of that function.
 
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
Now we are ready to give the geometrical interpretation of the integral
 
\begin_inset Formula $\int_{t_{0}}^{t_{n}}\dot{x}(t)dt$
\end_inset

: By making the width of the rectangles infinitesimally small, the area
 of all rectangles together is the area under the curve 
\begin_inset Formula $\dot{x}(t)$
\end_inset

.
\end_layout

\begin_layout Subsubsection*
Definition:
\end_layout

\begin_layout Standard
The integral of the function 
\begin_inset Formula $f(x)$
\end_inset

 from 
\begin_inset Formula $a$
\end_inset

 to 
\begin_inset Formula $b$
\end_inset

 is denoted by
\begin_inset Formula $\int_{a}^{b}f(x)dx$
\end_inset

.
 
\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\Diamond$
\end_inset


\end_layout

\begin_layout Standard
We can read of another important property of the integral from the geometrical
 interpretation.
 When computing our approximate integral, we did not make sure that the
 function values of 
\begin_inset Formula $\dot{x}(t)$
\end_inset

 where positive.
 Since 
\begin_inset Formula $\Delta t$
\end_inset

 was always negative, it could in principle happen, that the area 
\begin_inset Formula $\dot{x}(t_{i})\Delta t_{i}$
\end_inset

 has a negative value.
 In this case the area would really be subtracted from the whole area, instead
 of being added.
 Therefore, our approximate integral was really the signed area under the
 curve.
 This carries over to the real integral.
 
\emph on
The integral is the signed area under the graph of 
\begin_inset Formula $f(x)$
\end_inset

.
 All parts of 
\begin_inset Formula $f(x)$
\end_inset

 that lie under the 
\begin_inset Formula $x$
\end_inset

-axis will be subtracted from the whole area.

\emph default
 Therefore, if we really want to compute the area between the graph and
 the 
\begin_inset Formula $x$
\end_inset

-axis, we have to find out where 
\begin_inset Formula $f(x)$
\end_inset

 crosses the 
\begin_inset Formula $x$
\end_inset

-axis and must compute the integrals between those points separately, putting
 a minus in front of the integral where 
\begin_inset Formula $f(x)$
\end_inset

 is under the 
\begin_inset Formula $x$
\end_inset

-axis.
 
\end_layout

\begin_layout Subsubsection*
Example
\end_layout

\begin_layout Standard
If a car is moving at constant speed 
\begin_inset Formula $v,$
\end_inset

 the distance it travels in time 
\begin_inset Formula $t_{0}$
\end_inset

 is given by 
\begin_inset Formula $vt_{0}$
\end_inset

.
 If we draw a plot of 
\begin_inset Formula $v$
\end_inset

 against 
\begin_inset Formula $t_{0}$
\end_inset

, we can see that the distance traveled between times 
\begin_inset Formula $0$
\end_inset

 and 
\begin_inset Formula $t_{0}$
\end_inset

 is given by the following shaded region.
 This is also true if the speed 
\begin_inset Formula $v$
\end_inset

 is not constant, but a function of time 
\begin_inset Formula $v(t)$
\end_inset

.
\end_layout

\begin_layout Standard
Therefore, the distance traveled between times 
\begin_inset Formula $0$
\end_inset

 and 
\begin_inset Formula $t_{0}$
\end_inset

 can be computed by solving the integral 
\begin_inset Formula $\int_{0}^{t_{0}}v(t)dt$
\end_inset

.
 In the case of constant speed, the velocity 
\begin_inset Formula $v$
\end_inset

 does not depend on 
\begin_inset Formula $t$
\end_inset

.
 In this case we can already compute the integral geometrically.
 As already said, it is the area under the constant curve between 
\begin_inset Formula $0$
\end_inset

 and 
\begin_inset Formula $t_{0}$
\end_inset

.
 This is simply 
\begin_inset Formula $vt_{0}$
\end_inset

.
 So far, we do not know how to compute the integral for the case where 
\begin_inset Formula $v$
\end_inset

 depends on 
\begin_inset Formula $t$
\end_inset

.
 We first have to introduce a few rules how to solve integrals.
 This is what we are going to do in the following paragraphs.
\end_layout

\begin_layout Standard
\begin_inset Float figure
placement h
wide false
sideways false
status open

\begin_layout Plain Layout
\align center
\begin_inset Graphics
	filename figs/speedconst.eps
	lyxscale 50
	width 6cm
	keepAspectRatio

\end_inset


\begin_inset Graphics
	filename figs/speedvary.eps
	lyxscale 50
	width 6cm
	keepAspectRatio

\end_inset


\end_layout

\begin_layout Plain Layout
\begin_inset Caption

\begin_layout Plain Layout
\begin_inset CommandInset label
LatexCommand label
name "fig:approxIntegral"

\end_inset

Distance traveled at time in the time interval 
\begin_inset Formula $[0,3]$
\end_inset

 with constant speed (left) and with varying speed (right).
\end_layout

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Standard
Most rules for solving simple integrals are actually the inverted rules
 for differentiation.
 Before reversing this rules we must make sure that integrating is indeed
 the inverse of differentiation.
\end_layout

\begin_layout Subsubsection*
Proposition (Integrating is the inverse of differentiating)
\end_layout

\begin_layout Standard
Suppose that the function 
\begin_inset Formula $F(x)$
\end_inset

 is defined by 
\begin_inset Formula $F(x)=\int_{0}^{x}f(s)ds$
\end_inset

, i.e.
 
\begin_inset Formula $F(x)$
\end_inset

 is the area under the curve from 
\begin_inset Formula $0$
\end_inset

 to 
\begin_inset Formula $x$
\end_inset

.
 Then, 
\begin_inset Formula $F'(x)=f(x),$
\end_inset

 provided that 
\begin_inset Formula $f(x)$
\end_inset

 is continuous at 
\begin_inset Formula $x.$
\end_inset

 
\end_layout

\begin_layout Subsubsection*

\series medium
\emph on
\bar under
Proof:
\series default
\emph default
\bar default
 
\end_layout

\begin_layout Standard
\begin_inset Note Note
status open

\begin_layout Subsubsection*

\series medium
- draw picture as well
\end_layout

\begin_layout Subsubsection*

\series medium
- if pressed for time, leave this out
\end_layout

\end_inset


\end_layout

\begin_layout Standard
For simplicity, suppose that 
\begin_inset Formula $f(s)$
\end_inset

 is increasing near 
\begin_inset Formula $x.$
\end_inset

 Then
\end_layout

\begin_layout Standard
\begin_inset Formula 
\begin{eqnarray*}
f(x)\Delta x & \leq\int_{x}^{x+\Delta x}f(s)ds & \leq f(x+\Delta x)\Delta x\\
f(x) & \leq\frac{\int_{x}^{x+\Delta x}f(s)ds}{\Delta x} & \leq f(x+\Delta x).
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Standard
Now, we can observe that 
\begin_inset Formula 
\[
F(x+\Delta x)-F(x)=\int_{x}^{x+\Delta x}f(s)ds,
\]

\end_inset

so
\begin_inset Formula 
\[
f(x)\leq\frac{F(x+\Delta x)-F(x)}{\Delta x}\leq f(x+\Delta x).
\]

\end_inset


\end_layout

\begin_layout Standard
If we let 
\begin_inset Formula $\Delta x\rightarrow$
\end_inset

0, 
\begin_inset Formula $f(x+\Delta x)\rightarrow f(x)$
\end_inset

 by continuity of 
\begin_inset Formula $f(x)$
\end_inset

, and 
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\noun off
\color none

\begin_inset Formula $\frac{F(x+\Delta x)-F(x)}{\Delta x}\rightarrow F'(x)$
\end_inset

 by definition of derivative.
 In summary,
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
f(x)\leq F'(x)\leq f(x),
\]

\end_inset

 which implies that 
\begin_inset Formula $F'(x)=f(x).$
\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\Box$
\end_inset


\end_layout

\begin_layout Standard
The proposition shows, that integrating is really inverse differentiation.
 That is why the integral is also sometimes called 
\emph on

\begin_inset Index idx
status collapsed

\begin_layout Plain Layout
anti-derivative
\end_layout

\end_inset

anti-derivative.

\emph default
 So far we have only discussed so called 
\emph on

\begin_inset Index idx
status collapsed

\begin_layout Plain Layout
definite integral
\end_layout

\end_inset

definite integrals,
\emph default
 i.e.
 integrals that are computed between two points on the 
\begin_inset Formula $x$
\end_inset

-axis.
 Often, we can also inverse the process of differentiation first and plug
 in the borders of the integral later.
 The notation for this kind of integral is the same except that the boundary
 points are left out at the integral sign.
 Now, we can collect all our kinds of integrals in a mathematical definition.
\end_layout

\begin_layout Subsubsection*
Definition (Anti-Derivative)
\end_layout

\begin_layout Standard
Any function 
\begin_inset Formula $F(x)$
\end_inset

 with the property that 
\begin_inset Formula $F'(x)=f(x)$
\end_inset

 is called an 
\emph on
anti-derivative
\emph default
 or 
\emph on
indefinite integral
\emph default
 of 
\begin_inset Formula $f(x)$
\end_inset

.
 A indefinite integral is denoted by 
\begin_inset Formula $F(x)=\int f(s)ds$
\end_inset

.
 Once we have the indefinite integral, calculating any definite integral
 can be computed via: 
\begin_inset Formula 
\[
\int_{a}^{b}f(s)ds=F(b)-F(a).
\]

\end_inset


\end_layout

\begin_layout Standard
If 
\begin_inset Formula $F(x)$
\end_inset

 is an anti-derivative, so is 
\begin_inset Formula $F(x)+C$
\end_inset

 for any constant 
\begin_inset Formula $C.$
\end_inset

 This constant does not matter, because if evaluate the integral, the constants
 cancel: 
\begin_inset Formula $F(b)+C-(F(a)+C)=F(b)-F(a)+C-C=F(b)-F(a).$
\end_inset


\end_layout

\begin_layout Standard
In many cases, the shorthand 
\begin_inset Formula $[F(x)]_{a}^{b}=F(b)-F(a)$
\end_inset

 is used.
 
\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\Diamond$
\end_inset


\end_layout

\begin_layout Standard
The relationship between integrating and differentiating allows us to calculate
 integrals.
 Furthermore, once we have calculated an indefinite integral, we can easily
 check whether it is correct by just differentiating it again.
 If we get the original function we started with, everything is correct,
 otherwise there is a mistake somewhere.
 One could even just guess the integral, and then verify whether the guess
 is correct.
 So, even if integrating can be hard sometimes, checking whether the result
 is correct is almost always easy.
\end_layout

\begin_layout Subsubsection*
Example
\end_layout

\begin_layout Standard
Even without knowing any rule how to integrate so far, we know the anti-derivati
ve of 
\begin_inset Formula $f(x)=e^{x}$
\end_inset

 is again 
\begin_inset Formula $F(x)=e^{x}+C$
\end_inset

, since 
\begin_inset Formula $\left(e^{x}\right)'=e^{x}$
\end_inset

.
 The area under the graph of 
\begin_inset Formula $f(x)$
\end_inset

 between 
\begin_inset Formula $0$
\end_inset

 and some other point is given by the integral 
\begin_inset Formula 
\begin{eqnarray*}
\int_{0}^{T}f(x)dx & = & F(T)-F(0)\\
 & = & e^{T}-e^{0}\\
 & = & e^{T}-1.
\end{eqnarray*}

\end_inset

 For 
\begin_inset Formula $T\rightarrow\infty$
\end_inset

 the integral diverges, which means that the area under 
\begin_inset Formula $f(x)=e^{x}$
\end_inset

 is not finite.
\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Subsubsection*
Exercise
\end_layout

\begin_layout Standard
\begin_inset Marginal
status collapsed

\begin_layout Plain Layout
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
fbox{
\backslash
bf{
\backslash
Large 
\backslash
sf E}}
\end_layout

\end_inset


\end_layout

\end_inset

Calculate the derivative of 
\begin_inset Formula $F(x)=\frac{1}{n+1}x^{n+1}$
\end_inset

 and thus verify that 
\begin_inset Formula $F(x)$
\end_inset

 is the anti-derivative of 
\begin_inset Formula $f(x)=x^{n}.$
\end_inset

 For which values of 
\begin_inset Formula $n$
\end_inset

 is this not valid?
\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Standard
If you have solved the exercise, you have verified that the indefinite integral
 of 
\begin_inset Formula $f(x)=x^{n}$
\end_inset

 is 
\begin_inset Formula $F(x)$
\end_inset

 for all 
\begin_inset Formula $n\not=-1$
\end_inset

.
 This is the only rule we will introduce here.
 More advanced rules will be introduced in the next section.
 Like for differentiation, there are a few anti-derivatives that we will
 not prove here and which you should just remember, because they occur very
 often when calculating integrals.
\end_layout

\begin_layout Standard
\align center
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\series bold
Anti-Derivatives of important functions
\end_layout

\begin_layout Itemize
\begin_inset Formula $f(x)=a\Rightarrow F(x)=ax+C$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $f(x)=x^{a}\Rightarrow F(x)=\frac{1}{a+1}x^{a+1}+C$
\end_inset

 for 
\begin_inset Formula $a\not=-1$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $f(x)=\sin(x)\Rightarrow F(x)=-\cos(x)+C$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $f(x)=\cos(x)\Rightarrow F(x)=\sin(x)+C$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $f(x)=e^{x}\Rightarrow F(x)=e^{x}+C$
\end_inset


\end_layout

\begin_layout Itemize
\begin_inset Formula $f(x)=\frac{1}{x}\Rightarrow F(x)=\ln(x)+C$
\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection*
Example (Approximating Integrals with Matlab)
\end_layout

\begin_layout Standard
Using a similar idea as for taking numerical derivatives we can also numerically
 compute approximate values of integrals in matlab.
 This idea is to use the definition of an integral as the sum of the areas
 of rectangles with height 
\begin_inset Formula $f(x)$
\end_inset

 and width 
\begin_inset Formula $\Delta x$
\end_inset

 in the limit of infinitesimally small 
\begin_inset Formula $\Delta x$
\end_inset

.
 In order to give you a function, that you can reuse, we will use a bit
 more advanced matlab programming.
 Assume you are given a matlab function 
\begin_inset Formula $f$
\end_inset

 which returns the function values at the points that you call it on.
 You can realize that in matlab via a so called 
\emph on
function handle.
 
\emph default
A function handle is a variable, which is really a function.
 That means you can simply call the variable as if it would be a function.
 The important point is, that you can write code that uses a function for
 which you do not know what it will be exactly.
 For now, we simply a assign a function to a function handle 
\family typewriter
f
\family default
.
 This seems completely pointless now, but will be become clear later.
 You can assign functions to variable by using the ``
\family typewriter
@
\family default
''-symbol.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

 >> f = @exp % assign f to be the exponential function
\end_layout

\begin_layout Plain Layout

f =
\end_layout

\begin_layout Plain Layout

    @exp 
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Now we go on with approximating our integral.
 First, we choose a certain small step width 
\begin_inset Formula $\Delta x$
\end_inset

, integral border between which we want to compute the integral, and base
 points at which we will evaluate the function to get the height of the
 rectangles.
 
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

>> dx = .01; % set stepwidth
\end_layout

\begin_layout Plain Layout

>> a = 1; b = 2; % set integral borders
\end_layout

\begin_layout Plain Layout

>> x = [a:dx:b]; % get base points
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Next, we evaluate the function 
\family typewriter
f
\family default
 (which is 
\begin_inset Formula $\exp$
\end_inset

 at the moment) at the base points.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

>> fval = f(x); % evaluate function handle at x
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Now, we use the fact that 
\begin_inset Formula 
\begin{eqnarray*}
\int_{a}^{b}f(x)dx & \approx & \sum_{i=1}^{n}f(x_{i})\Delta x\\
 & = & \Delta x\sum_{i=1}^{n}f(x_{i}),
\end{eqnarray*}

\end_inset

where is the number of our base points 
\begin_inset Formula $x_{i}$
\end_inset

 and 
\begin_inset Formula $f(x_{i})$
\end_inset

 are the values contained in the array 
\family typewriter
fval
\family default
.
 Translating that formula into matlab yields:
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

>> I = sum(fval)*dx; % compute approximate integral
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Now, we can use the fact that we did not directly specify what function
 
\family typewriter
f
\family default
 exactly is to write a matlab function which we can reuse.
 For that purpose, we create a file with the name of the function by:
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

>> edit integrate.m
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
In that file we specify the function.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

function I = integrate(f,a,b,dx)
\end_layout

\begin_layout Plain Layout

    x = [a:dx:b]; % get base points
\end_layout

\begin_layout Plain Layout

    fval = f(x); % evaluate function handle at x     
\end_layout

\begin_layout Plain Layout

    I = sum(fval)*dx; % compute approximate integral
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
The code above defines a function named 
\family typewriter
integrate
\family default
 that takes four input arguments: the function handle 
\family typewriter
f
\family default
, the integral boundaries 
\family typewriter
a
\family default
 and 
\family typewriter
b
\family default
 and the step width 
\family typewriter
dx
\family default
.
 It returns the approximate value of the integral 
\begin_inset Formula $\int_{a}^{b}f(x)dx$
\end_inset

 as we compute it above.
 However, we can call it on different functions and different integral boundarie
s now, without having to redefine 
\family typewriter
a
\family default
 and 
\family typewriter
b
\family default
 all the time.
 Our example from above can be reproduced via:
\begin_inset Newline newline
\end_inset


\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

>> I = integrate(@exp,1,2,.01)
\end_layout

\begin_layout Plain Layout

I =
\end_layout

\begin_layout Plain Layout

    4.7213 
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
Other uses of the function could look like
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{verbatim}
\end_layout

\begin_layout Plain Layout

>> I = integrate(@cos,0,2,.01)
\end_layout

\begin_layout Plain Layout

I =
\end_layout

\begin_layout Plain Layout

    0.9122 
\end_layout

\begin_layout Plain Layout

>> I = integrate(@log,1,3,.001)
\end_layout

\begin_layout Plain Layout

I =
\end_layout

\begin_layout Plain Layout

    1.2964 
\end_layout

\begin_layout Plain Layout


\backslash
end{verbatim}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Subsection
Evaluating and Transforming Integrals
\end_layout

\begin_layout Subsubsection
Integrals and Sums
\end_layout

\begin_layout Standard
Above, we have seen that the integral of 
\begin_inset Formula $f$
\end_inset

 is the limit of a sum of infinitesimal narrow rectangles with heights given
 by the function values 
\begin_inset Formula $f(x)$
\end_inset

.
 If 
\begin_inset Formula $f$
\end_inset

 is the sum of two functions 
\begin_inset Formula $f(x)=g_{1}(x)+g_{2}(x)$
\end_inset

, we can split this sum into two.
 Therefore, the integral of a sum of two functions is the sum of their integrals
:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{a}^{b}(f(x)+g(x))dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx.
\]

\end_inset

This rule is of course also valid for a sum of an arbitrary number of functions.
\end_layout

\begin_layout Standard
By the same reasoning, constant factors can be pulled out of the integral,
 because constant factors can be pulled out of sums:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{a}^{b}c\cdot f(x)dx=c\cdot\int_{a}^{b}f(x)dx.
\]

\end_inset


\end_layout

\begin_layout Subsubsection
Integrals of Polynomials
\end_layout

\begin_layout Standard
Now we are equipped with enough rules to be able to calculate the anti-derivativ
e of polynomials.
 We show how to do this with a small example.
\end_layout

\begin_layout Subsubsection*
Example
\end_layout

\begin_layout Standard
We calculate the anti-derivative of 
\begin_inset Formula $f(x)=2x^{2}+x+1$
\end_inset

: 
\begin_inset Formula 
\begin{eqnarray*}
F(x) & = & \int f(x)dx\\
 & = & \int2x^{2}+x+1dx\\
 & = & \int2x^{2}dx+\int xdx+\int1dx\\
 & = & \frac{2}{3}x^{3}+\frac{1}{2}x^{2}+x.
\end{eqnarray*}

\end_inset

We can use this anti-derivative to compute the integral 
\begin_inset Formula $\int_{1}^{2}f(x)dx$
\end_inset

:
\begin_inset Formula 
\begin{eqnarray*}
\int_{1}^{2}f(x)dx & = & F(2)-F(1)\\
 & = & \frac{16}{3}+1+2-\frac{2}{3}-\frac{1}{2}-1\\
 & = & \frac{37}{6}.
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Subsubsection*
Example
\end_layout

\begin_layout Standard
Before coming to more advanced rules for calculating integrals, we show
 a more sophisticated example: What is the integral of 
\begin_inset Formula $f(x)=|x|$
\end_inset

 from 
\begin_inset Formula $-10$
\end_inset

 to 
\begin_inset Formula $10$
\end_inset

? At first glance, it is not clear how we should integrate 
\begin_inset Formula $|x|$
\end_inset

.
 Therefore, we apply a simple trick.
 Since an integral is simply an infinite sum of infinitesimal small rectangle,
 it does not matter if we split an integral into two ''sums''.
 Therefore, we can split the integral 
\begin_inset Formula $\int_{-10}^{10}|x|dx$
\end_inset

 into 
\begin_inset Formula $\int_{-10}^{10}|x|dx=\int_{-10}^{0}|x|dx+\int_{0}^{10}|x|dx$
\end_inset

.
 Between those boundaries, we can use the piecewise definition of 
\begin_inset Formula $|x|$
\end_inset

:
\begin_inset Formula 
\begin{eqnarray*}
\int_{-10}^{10}|x|dx & = & \int_{-10}^{0}-x\, dx+\int_{0}^{10}x\, dx\\
 & = & [-\frac{1}{2}x^{2}]_{-10}^{0}+[\frac{1}{2}x^{2}]_{0}^{10}\\
 & = & \frac{100}{2}+\frac{100}{2}=100.
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Subsubsection
Integrals and Change of Variables
\end_layout

\begin_layout Standard
Sometimes, it is easy to integrate 
\begin_inset Formula $f(s),$
\end_inset

 but we actually want to integrate 
\begin_inset Formula $g(x)=f(x+s)$
\end_inset

 .
 The new function 
\begin_inset Formula $g(x)$
\end_inset

 can be seen as a shifted version of 
\begin_inset Formula $f(x)$
\end_inset

, so we can integrate by shifting the limits of integration as well.
 This leads the rule:
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{a}^{b}f(x+s)dx=\int_{a+s}^{b+s}f(x)dx.
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Comment
status open

\begin_layout Plain Layout
Draw picture on the board, especially point out how the shift works, i.e.
 that a shift by t is equivalent to using x MINUS t instead of t.
\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection*
Example
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{0}^{10}(x-5)^{2}dx=\int_{-5}^{5}x^{2}dx=\frac{2}{3}5^{3}
\]

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Standard
Similarly, we might want to integrate a function 
\begin_inset Formula $g(x)=f(sx)$
\end_inset

 where the 
\begin_inset Formula $x$
\end_inset

-axis has been rescaled.
 In this case, we also have to adjust the limits of integration, but also
 get an extra factor in front of the integral.
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{a}^{b}f(sx)dx={\frac{1}{s}\int}_{as}^{bs}f(x)dx
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Note Comment
status open

\begin_layout Plain Layout
Draw picture, point out where the extra factor comes from.
\end_layout

\end_inset


\end_layout

\begin_layout Paragraph
Example
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{0}^{3}(2x)^{2}dx={\frac{1}{2}\int}_{0}^{6}x^{2}dx=2\cdot\frac{1}{3}\cdot8\cdot27
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{0}^{1}e^{2x}dx={\frac{1}{2}\int}_{0}^{2}e^{x}dx=\frac{1}{2}\left(e^{2}-1\right)
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\int_{0}^{2\pi}\sin\left(\frac{x}{2}\right)={2\int}_{0}^{\pi}\sin(x)dx=2\cdot2.
\]

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\begin_layout Standard
The following box provides a summary of all transformation rules, that we
 have seen so far.
\end_layout

\begin_layout Standard
\align center
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
status open

\begin_layout Plain Layout

\series bold
Transformations rules for integrals
\end_layout

\begin_layout Itemize

\bar under
Summation Rule:
\bar default
 The integral of a sum of is the sum of the integrals: 
\begin_inset Formula 
\begin{eqnarray*}
\int_{a}^{b}\sum_{i=1}^{n}g_{i}(x)dx & = & \sum_{i=1}^{n}\int_{a}^{b}g_{i}(x)dx.
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Itemize
Constant factors can be pulled out of the integral:
\begin_inset Formula 
\begin{eqnarray*}
\int_{a}^{b}c\cdot f(x)dx & = & c\cdot\int_{a}^{b}f(x)dx.
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Itemize
Integrating from 
\begin_inset Formula $a$
\end_inset

 to 
\begin_inset Formula $c$
\end_inset

 is the same as integrating from 
\begin_inset Formula $a$
\end_inset

 to 
\begin_inset Formula $b$
\end_inset

, and then 
\begin_inset Formula $b$
\end_inset

 to 
\begin_inset Formula $c$
\end_inset

, and adding the two integrals (This is useful for evaluating integrals
 over piecewise functions):
\begin_inset Formula 
\begin{eqnarray*}
\int_{a}^{b}\cdot f(x)dx & +\int_{b}^{c}f(x)dx= & \int_{a}^{c}f(x)dx.
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Itemize

\bar under
Shift Rule
\bar default
:
\begin_inset Formula 
\[
\int_{a}^{b}f(x+s)dx=\int_{a+s}^{b+s}f(x)dx.
\]

\end_inset


\end_layout

\begin_layout Itemize

\bar under
Rescaling Rule
\bar default
:
\end_layout

\begin_layout Plain Layout
\begin_inset Formula 
\[
\int_{a}^{b}f(sx)dx=\frac{1}{s}\int_{as}^{bs}f(x)dx
\]

\end_inset


\end_layout

\end_inset


\end_layout

\begin_layout Subsubsection
Integrals and Statistics
\end_layout

\begin_layout Standard
Integrals are extremely important in probability theory and statistics.
 The expected value of a random variable 
\begin_inset Formula $X$
\end_inset

 is simply the integral 
\begin_inset Formula $E(X)=\int_{\Omega}x\cdot p(x)dx$
\end_inset

, where 
\begin_inset Formula $p(x)$
\end_inset

 is the probability density function of 
\begin_inset Formula $x.$
\end_inset

 The subscript 
\begin_inset Formula $\Omega$
\end_inset

 indicates that we choose the boundaries of the integral such that all possible
 values of the random variable are covered.
 The integral is often not too hard to do.
 Unfortunately, for our favorite distribution, the 
\emph on

\begin_inset Index idx
status collapsed

\begin_layout Plain Layout
Normal distribution
\end_layout

\end_inset

Normal distribution
\emph default
 or 
\emph on

\begin_inset Index idx
status collapsed

\begin_layout Plain Layout
Gaussian
\end_layout

\end_inset

Gaussian
\emph default
, calculating the expectation involves some techniques that are beyond the
 scope of this course
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
namely either transforming to polar coordinates or using contour integration
\end_layout

\end_inset

.
\end_layout

\begin_layout Paragraph
Examples
\end_layout

\begin_layout Enumerate

\series bold
The Expectation of a Uniform Distribution:
\series default

\begin_inset Newline newline
\end_inset

The pdf of a Uniform Distribution on 
\begin_inset Formula $[0,1]$
\end_inset

 is 
\begin_inset Formula $f(x)=1.$
\end_inset

 The expectation is 
\begin_inset Formula $\mathbb{E}[X]=\int_{0}^{1}xdx=\frac{1}{2}$
\end_inset


\end_layout

\begin_layout Enumerate

\series bold
The Variance of a Uniform Distribution
\series default

\begin_inset Newline newline
\end_inset

The Variance is 
\begin_inset Formula $\mathbb{V}[X]=\mathbb{E}[X^{2}]-[\mathbb{E}[X]]^{2}.$
\end_inset

 So, for the Uniform distribution, 
\begin_inset Formula $\mathbb{V}[X]=\int_{o}^{1}x^{2}dx-\frac{1}{4}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}.$
\end_inset


\end_layout

\begin_layout Enumerate

\series bold
The expectation function 
\begin_inset Formula $\mathbb{E}$
\end_inset

 is a linear function for any distribution:
\begin_inset Newline newline
\end_inset


\series default
Consider the random variables 
\begin_inset Formula $X$
\end_inset

 and 
\begin_inset Formula $Y$
\end_inset

 and their joint distribution 
\begin_inset Formula $p(x,y)$
\end_inset

.
 The expectation of 
\begin_inset Formula $X$
\end_inset

 is given by 
\begin_inset Formula $\mathbb{E}[X]=\int xp(x)dx$
\end_inset

, the expectation of 
\begin_inset Formula $Y$
\end_inset

 by 
\begin_inset Formula $\mathbb{E}[y]=\int yp(y)dy$
\end_inset

.
 By not specifying any integral boundaries we silently assume that the integrals
 are taken over the whole range of possible values of each random variable.
 The single distributions 
\begin_inset Formula $p(x)$
\end_inset

 and 
\begin_inset Formula $p(y)$
\end_inset

, called 
\emph on

\begin_inset Index idx
status collapsed

\begin_layout Plain Layout
marginal distribution
\end_layout

\end_inset

marginal distributions
\emph default
 or simply 
\emph on

\begin_inset Index idx
status collapsed

\begin_layout Plain Layout
marginals
\end_layout

\end_inset

marginals
\emph default
, can be computed from the joint distribution 
\begin_inset Formula $p(x,y)$
\end_inset

 by 
\begin_inset Formula $p(x)=\int p(x,y)dy$
\end_inset

 and 
\begin_inset Formula $p(y)=\int p(x,y)dx$
\end_inset

.
 This computation is usually referred to as ``integrating out variables''.
 What it means is, that when writing 
\begin_inset Formula $p(x)$
\end_inset

 we do not care about the value of 
\begin_inset Formula $y$
\end_inset

.
 For that reason we have to collect all the probability mass of the 
\begin_inset Formula $y$
\end_inset

s that occur together with that 
\begin_inset Formula $x$
\end_inset

.
 This is exactly what we do by integrating.
 
\begin_inset Newline newline
\end_inset

Even though, we do not know anything about the distribution 
\begin_inset Formula $p(x,y)$
\end_inset

, by simply using properties of integrals, we can show that the expectation
 is a linear function.
 For that purpose we have to show that (i) the expectation of the sum of
 
\begin_inset Formula $X$
\end_inset

 and 
\begin_inset Formula $Y$
\end_inset

 is the sum of the expectations and (ii) the expectation of scaled versions
 of 
\begin_inset Formula $X$
\end_inset

 is the scaled expectation of 
\begin_inset Formula $X$
\end_inset

.
 For the sake of compactness, we merge (i) and (ii) in a single computation.
 We show that 
\begin_inset Formula $\mathbb{E}[aX+bY]=a\mathbb{E}[X]+b\mathbb{E}[Y]$
\end_inset

 for 
\begin_inset Formula $a,b\in\mathbb{R}$
\end_inset

.
\begin_inset Formula 
\begin{eqnarray*}
\mathbb{E}[aX+bY] & = & \int\int(ax+by)p(x,y)\, dxdy\\
 & = & \int\int axp(x,y)\, dxdy+\int\int byp(x,y)\, dxdy\\
 & = & a\int\int xp(x,y)\, dxdy+b\int\int yp(x,y)\, dxdy\\
 & = & a\int x\int p(x,y)\, dydx+b\int y\int p(x,y)\, dxdy\\
 & = & a\int xp(x)\, dx+b\int yp(y)dy\\
 & = & a\mathbb{E}[X]+b\mathbb{E}[Y]
\end{eqnarray*}

\end_inset


\end_layout

\begin_layout Standard
\align right
\begin_inset Formula $\lhd$
\end_inset


\end_layout

\end_body
\end_document