Yeah, so welcome to our today's tutorial.

Today we will consider exercise four. I don't know what's wrong here.

Hope it remains as it is. So exercise four. Here we shall a little bit practice the integration

with different quadrature rules and to this end we have given three different functions.

They are all polynomials and we should do a numerical integration using either Newton

codes quadrature or Gauss Legendre quadrature. So let me just copy this.

So here we go. So in the very first part here we have to integrate the function f1 using Newton

Gauss quadrature, Newton codes sorry, quadrature with two quadrature points which is the trapezoidal

rule and let me furthermore open the lecture notes because there we have this nice overview

of the weights and gratitude points. So this is here. So this we can keep in mind here.

Okay. So the first function here. This is a linear function so you can also say a polynomial

first order and first let us compute the analytical solution because this is also

asked for. So the analytical solution here. It's just a standard integration. The limits

are given here. It's the interval from minus one to one. So what do we get after integration here?

Xi squared plus I'm not happy with just Xi squared. If you take the derivative

of Xi squared what do you get? Yes that's it. That's what I wanted to know. Slight different

result could follow here. Okay so what do we get? One divided by two? No.

So it's actually

so now it should yes it's two.

When we discuss the accuracy of Newton-Coltz gratitude we distinguish between even and odd

polynomials. Do you see what could be the reason for that? Here by the analytical treatment. What

happens to this term where we have after integration? Exactly it cancels out because if you take

one and minus one in general to an even power then it just cancels out in this interval from

minus one to one. So and the even exponent here actually results from an odd exponent before

integration. So actually the odd exponent does not play a role. Okay so this is the analytical

solution and now we can integrate with Newton-Coltz and we choose two integration points.

Would you please look up which degree of polynomial is possible to be integrated

exactly by the choice here. Just have a look to your notes. The morning we discussed Gauss

quadrature. This was this formula two times the number of quadrature points minus one. This holds

for Gauss. N minus one. Why N minus one? Yeah maybe let's have a look here. This was in notes of last

time. So

just as a side remark here.

So if we have an even polynomial then yeah it's two minus one. If we have an odd polynomial it's

two but two is not odd. So in fact we can integrate by these two

quadrature points a linear function exactly. So and let us try whether this works. It should work of course.

So the approximation is that we

evaluate the first node and then we evaluate the function at the second node, sorry second quadrature point.

So which is in our case

to be evaluated at

here at minus one and at one and the associated weights are one in both cases. So we have to write here

the function one by two evaluated at minus one multiplied by one plus

so I put it here in a very formal way of course one can abbreviate that and what is the result here?

What do we get? Two. So as expected. Okay

so NEWTON CODE with two integration points

what do we get? 2. so as expected. okay so Newton codes with two integration

points

so based on this finding we can have a look at the next function which is here

a cubic function yeah and for sure with the two so here we are 4.2 with the two

integration points we will for sure not be able to integrate it exactly. how many

quadrature points do we need to integrate a cubic polynomial exactly?