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

. The first integrated

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

. 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

and

as a scatter plot. Noting that

y1=x.^2

and

y2=2*x

and the plots are correct.

Instead of manually typing in

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=[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

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

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

that is 0.5 less than 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

is a constant, the difference in the

values

is zero across the board as expected.

Instead of manually typing in:

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

. 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

is a straight line and is independent of

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

is equal to

2*x4

i.e.

y4=2*x4

Integration vs Differentiation

The integration of

y1=2*x1

with respect to

gives

y2=x2^3

and the differentiation of

y3=x3^3

with respect to

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

x1step=0.0001
[m1,n1]=size(x1);
x1=[1:x1step: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';

y2=[x1step]*cumsum(y1);
y2=y2(2:end);
x2=[x1step]*[[2:m1]+[x1step/2]]';
figure(2);
scatter(x,y,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','2=int(y1,x1)','location','northwest');

x3step=0.0001
x3=[1:x3step:10]';
[m3,n3]=size(x3);
y3=x3.^2;
figure(3);
scatter(x3,y3,100,'fill','r');
grid('minor');
xlabel('x');
ylabel('y');
ax=gca
ax.YAxisLocation='right';

y4=[diff(y3)]./x3step
x4=x3step*[[2:m3]-[x3step/2]]';
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=x3^2','y4=diff(y3,x3)','location','NorthWest');