# Tutorial Video

# Integration Revision

Let’s plot a curve as a a series of bars and a line.

`x1=[1:10]';`

y1=2*x1;

figure(1);

bar(x1,y1);

hold('on');

plot(x1,y1,'LineWidth',3);

hold('off');

grid('minor');

xlabel('x');

ylabel('y');

ax=gca;

ax.YAxisLocation='right';

We seen before that integration is the sum of the area under the curve (the sum of the bars). We can plot the area under the curve with respect to `x1`

. The first integrated `y2`

value is going to be the sum of all the bars at `x1(1)`

i.e. `x1(1)`

and this is going to correspond to a new value of `x2=1.5`

because we are shifted over by `0.5`

, the mid point of the step size in `x1`

. The next value is going to be the sum of the bars at `x1=2`

i.e. `x1(1)+x(2)`

and correspond to a new value of `x2=2.5`

and so on and so forth.

`y2=[sum(y1(1));...`

sum(y1(1:2));...

sum(y1(1:3));...

sum(y1(1:4));...

sum(y1(1:5));...

sum(y1(1:6));...

sum(y1(1:7));...

sum(y1(1:8));...

sum(y1(1:9));...

sum(y1(1:10))];

x2=[[1:10]+0.5]';

figure(2);

scatter(x1,y1,100,'fill','r');

hold('on');

scatter(x2,y2,100,'fill','b');

hold('off');

grid('minor');

xlabel('x');

ylabel('y');

ax=gca;

ax.YAxisLocation='right';

legend('y1=2*x1','y2=int(y1,x1)','location','northwest');

We can plot `x1`

, `y1`

and `x2`

, `y2`

as a scatter plot. Noting that `y1=x.^2`

and `y2=2*x`

and the plots are correct.

Instead of manually typing in

`y2=[sum(y1(1));...`

sum(y1(1:2));...

sum(y1(1:3));...

sum(y1(1:4));...

sum(y1(1:5));...

sum(y1(1:6));...

sum(y1(1:7));...

sum(y1(1:8));...

sum(y1(1:9));...

sum(y1(1:10))];

We can use

`y2=cumsum(y1);`

# Differenciation

What we are now going to do is look at the plot `y3=1`

i.e. a constant with respect to `x3`

`x3=[1:10]';`

[m3,n3]=size(x3);

y3=ones(m3,n3);

figure(3);

scatter(x3,y3,100,'fill','r');

hold('on');

scatter(x4,y4,100,'fill','b');

hold('off');

grid('minor');

xlabel('x');

ylabel('y');

ax=gca;

ax.YAxisLocation='right';

legend('y1=1','2=int(y1,x1)','location','northwest');

Now instead of looking at the area under the line, we are going to look at the difference between the nearest `y3`

values i.e. the difference between a bar with the bar on its left. For instance the tenth bar at `x3=10`

and the 9th bar at `x3=9`

is going to correspond to a new value of `x4`

that is 0.5 less than 9 i.e. `x4=8.5`

. The next value is going to be the difference of the points at `x3=9`

and `x3=8`

and correspond to a new value of `x4`

that is 0.5 less thanÂ i.e. `x4=7.5`

and so on and so forth.

We will plot this as a scatter plot on the same chart.

`y4=[[y3(1)];...`

[y3(2)-y3(1)];...

[y3(3)-y3(2)];...

[y3(4)-y3(3)];...

[y3(5)-y3(4)];...

[y3(6)-y3(5)];...

[y3(7)-y3(6)];...

[y3(8)-y3(7)];...

[y3(9)-y3(8)];...

[y3(10)-y3(9)]];

x4=[[1:10]-0.5]';

figure(4);

scatter(x3,y3,100,'fill','r');

hold('on');

scatter(x4,y4,100,'fill','b');

hold('off');

grid('minor');

xlabel('x');

ylabel('y');

ax=gca;

ax.YAxisLocation='right';

legend('y3=1','y4=diff(y3,x3)','location','NorthWest');

In this case since `y3`

is a constant, the difference in the `y3`

values `y4`

is zero across the board as expected.

Instead of manually typing in:

`y4=[[y3(1)];...`

[y3(2)-y3(1)];...

[y3(3)-y3(2)];...

[y3(4)-y3(3)];...

[y3(5)-y3(4)];...

[y3(6)-y3(5)];...

[y3(7)-y3(6)];...

[y3(8)-y3(7)];...

[y3(9)-y3(8)];...

[y3(10)-y3(9)]];

We can use

`y4=[y3(1);diff(y3)];`

x4=[[1:10]-0.5]';

A value for `x3=0`

was not selected and is the reason for `y4(0)=1`

opposed to its actual value of `0`

. As this is an outlier it is better to ignore this first value and instead look at:

`y4=[diff(y3)];`

x4=[[2:10]-0.5]';

Let’s instead change the `y3=1`

(constant) to `y3=x3`

. Now the bar chart from right to left looks like a set of stairs and as we climb down the step size remains constant:

As a result `y4`

is a straight line and is independent of `x4`

.

Okay so let’s now change `y3=x3.^2`

. Here the bar chart from right to left stepping down we see the first step is the larges, the the second step, then the third and it is a much shallower step down on the left hand side:

Actually we can see that every value of `y4`

is equal to `2*x4`

i.e. `y4=2*x4`

# Integration vs Differentiation

The integration of `y1=2*x1`

with respect to `x1`

gives `y2=x2^3`

and the differentiation of `y3=x3^3`

with respect to `x3`

gives `y4=2*x4`

i.e. as you can clearly see they are the inverse process of one another.

We can think of integration as adding the sum of all the bars as we move from left to right – climbing the stairs i.e. starting at the origin and our first plot point being `0+(step size/2`

).

And we can think of differentiation as being the difference in the bar height as we step down from the right to left. The first step from the right being `end-(step size)/2`

.

Of course as we seen with integration the above is only an approximation as we used finite relatively large step sizes.

# Uncertainty in Results

In the above we had the bin width equaling to the finite bin width of 1. For integration and differentiation we use an infinitely small bin width. Let’s constrict our bin width

# Rules of Integration and Differenciation

Taking

For integration we elevate the original power by 1 and then divide by the new power.

For differentiation we multiply by the original power and then devalue the power by 1:

# Symbolic

We can set x to be a symbol and write down our starting equations with respect to x:

`syms('x')`

z1=1

z2=x

z3=x^2

z4=x^3

Now we can differentiate the functions with respect to x:

`dz1=diff(z1,x)`

dz2=diff(z2,x)

dz3=diff(z3,x)

dz4=diff(z4,x)

Now we can integrate the functions with respect to x:

`iz1=int(z1,x)`

iz2=int(z2,x)

iz3=int(z3,x)

iz4=int(z4,x)