 Today, we are going to solve some problems based on numerical quadrature and after that we will consider numerical integration. So, some of simple problems, first problem is we are given a function f x which is piecewise linear. So, f x is defined as equal to x for 0 less than or equal to x less than or equal to half and 1 minus x if 1 by 2 less than or equal to x less than or equal to 1. We want to find an approximation to integral 0 to 1 f x dx by using trapezoidal rule, then composite rule based on partition 0 less than half less than 1. That means, we will be applying trapezoidal rule to interval 0 to half and trapezoidal rule to interval half to 1 and adding it up, then Simpson rule, so basic Simpson's rule and corrected trapezoidal rule. In all these rules, what will come into picture will be value of function f at two end points 0 and 1 and at the midpoint half. In addition for corrected trapezoidal rule, we will need f dash at 0 and f dash at 1. Our function f is such that it vanishes at the two end points, so f of 0 is 0, f of 1 is 0 and f at half is going to be equal to half. f dash at 0 is going to be equal to 1, f dash of 1 is going to be equal to minus 1 because on the interval half to 1, our function is defined as 1 minus x. So, when you take its derivative, so at 1 we are taking the left handed derivative. So, now we are going to substitute in these values in various formulae to obtain approximation to integral 0 to 1 f x dx. So, let us look at first the exact value. So, integral 0 to half x dx plus half to 1 1 minus x dx because that is how our function f is defined. When you integrate, you get the value to be equal to 1 by 4. The first rule is trapezoidal rule. Our formulae is b minus a f a plus f b by 2. In our case a is 0, b is 1, f at 0 is 0, f at 1 is 0. So, the approximation using trapezoidal rule which we get is equal to 0. Now, we want to apply composite trapezoidal rule. We are going to apply it on the partition 0, half and 1. That means on the interval 0 to half, we apply trapezoidal rule. On the interval 0 to half, our function is f x is equal to x. That means it is a linear function and the error is 0 for linear functions for trapezoidal rule. So, using our trapezoidal rule on the interval 0 to half, we are going to get exact value of the integral. Similarly, on the interval half to 1, our function is 1 minus x. So, it is linear. We approximate our integral over half to 1 by trapezoidal rule. So, there again the error is going to be 0. So, composite trapezoidal rule is going to give us exact value. So, it is the using composite trapezoidal rule, you get 1 by 4 into half, 1 by 4 because b minus a. On the interval 0 to half, it is half. So, 1 by 2 and then these two. So, it is 1 by 4. Then we will be looking at value of f of 0 plus value at f of half. At 0, the value is 0. At half, the value is half. So, that is why 1 by 2 plus 1 by 2 to 1. So, again the length of the interval is half divided by 2. That gives us 1 by 4. Now, we will be considering f at half plus f at 1. So, that is going to be equal to 1 by 2 and f at 1 is 0. So, it is 1 by 4. So, for the composite trapezoidal rule, you get the value to be equal to 1 by 4. Next, for Simpson's rule, this is the formula f a is 0, f b is 0, f at a plus b by 2. At the midpoint, it is value is half. So, when you simplify, you get it to be equal to 1 by 3. And in the corrected trapezoidal rule, you have trapezoidal rule plus this term. Trapezoidal rule gives us 0. So, you have only this term and the value which you obtain by corrected trapezoidal rule is 1 by 12 f dash 1 minus f dash 0. So, it is going to be equal to minus 1 by 6. Thus, the best result which we obtain in this example is when you are using composite trapezoidal rule. And that is because our function is piecewise linear. And in the composite trapezoidal rule, we have used our partition such that on each sub interval, our function is linear. If instead of applying composite trapezoidal rule to interval 0 to half and half to 1, if I would have used say interval 0 to 1 by 3, then 1 by 3 to 2 by 3 and 2 by 3 to 1, then our error would not have been equal to 0. Because on the interval 0 to 1 by 3, our function will be linear. On 2 by 3 to 1 also it will be linear, but on 1 by 3 to 2 by 3 it will be piecewise linear. And if you compare the result which you obtained using trapezoidal and Simpson, then Simpson's approximation is a better approximation than the trapezoidal. And the corrected trapezoidal rule that is going to be the worst approximation. The value of the integral is positive and the value which we obtain is negative. So, it is a simple example to compare various our numerical quadrature rules. Now, the next example, we want to construct a rule of the type a 0 f of minus half plus a 1 f of 0 plus a 2 f of half which is exact for polynomials of degree less than or equal to 2. So, that means we want to determine a 0 a 1 a 2 in such a manner that there is no error when you use this formula for f to be a polynomial of degree less than or equal to 2. One way of doing this problem is we have got our points minus half, 0 and half. So, you fit a parabola that means a polynomial of degree less than or equal to 2, then you integrate and then you are going to get values of a 0 a 1 a 2. Now, our interpolating polynomial p 2 itself will be equal to function f if f is a polynomial of degree less than or equal to 2 and that is why our numerical integration also will be exact if f is a polynomial of degree less than or equal to 2. We will do it slightly differently, we want that there should not be any error for polynomials of degree less than or equal to 2. So, if we guarantee that there is no error for three functions f x is equal to 1, f x is equal to x and f x is equal to x square, then there will not be any error for a general quadratic polynomial of the form a 0 plus a 1 x plus a 2 x square. Now, when we take these three functions f x is equal to 1, f x is equal to x and f x is equal to x square, we will get three equations in three unknowns a 0 a 1 a 2 and then we determine a 0 a 1 a 2. So, f x is equal to 1, then the left hand side is going to be equal to 2, f x is equal to integral minus 1 to 1 1 d x. So, that will be 2 is equal to f x is a constant function. So, f of minus half is 1, f of 0 is 1, f of half is 1 and hence we get a 0 plus a 1 plus a 2. So, this is our first equation. Next, look at f x is equal to x, when you integrate you are going to get x square by 2, evaluation between 1 and minus 1 that will give us to be 0 and this will be equal to now f x is equal to x. So, f of minus half will be minus half, f of 0 will be 0, f of half will be plus half. So, you get minus a 0 by 2 plus a 2 by 2. So, this equation gives us a 2 has to be equal to a 0. So, now, this is for function f x is equal to x, if f x is equal to x square, then the integration is going to be equal to 2 by 3 and here f of minus half will be minus half. So, f of minus half will be 1 by 4, f of 0 will be 0, f of half also will be equal to 1 by 4 and hence we will get a 0 plus a 2 divided by 4. From this equation, we have got a 2 is equal to a 0. So, using that fact, we get a 0 is equal to a 2 is equal to 4 by 3, then we have not used this equation. So, use this equation to deduce that a 1 is equal to minus 2 by 3 and thus the coefficients a 0, a 1, a 2, they are uniquely determined by the condition that the formula should be exact for polynomials of degree less than or equal to 2. Now, in the next example, we are given a rule that means, we are given the weights, we are given the interpolation points and we are asked to determine the degree of precision of this formula. That means, we want to find the highest degree of the polynomial for which the rule is exact that means, there is no error. Now, when we want to determine degree of precision, what we have to do is we have to proceed in an orderly manner that means check whether there is no error for f x is equal to 1, then look at f x is equal to x. If there is no error for f x is equal to x, then go to f x is equal to x square, but if there is error for f x is equal to x, then that means, the degree of precision is going to be only constant polynomials. It can happen that there is no error for the constant functions, there is no error for function f x is equal to x square, but in that case if there is a error for f x is equal to x, then we say that our degree of precision is going to be only 0. That means, it is going to be exact for only constant polynomial. So, here in this example or in this problem, we are given the interpolation points, we are given the nodes, we are given the weights and we want to determine the degree of precision. So, it is integral x 0 to x 3 f x d x is approximately equal to 3 h by 8 f x 0 plus 3 f x 1 plus 3 f x 2 plus f x 3, where x 0 is any real number and x k are given by x 0 plus k h, k is equal to 1, 2, 3 that means, x 0 x 1 x 2 x 3, they are going to be they form a equidistant partition, which means x 1 minus x 0 is h, x 2 minus x 1 is h, x 3 minus x 2 is equal to h. So, we want to know the degree of precision of this rule that means, the highest degree polynomial for which there is no error. What we are going to do is, we are going to assume that without loss of generality, let x 0 be equal to 0. So, we are just changing the position of our point to start with x 0 is any real number, but we do not lose any generality of the problem, if I can if I assume x 0 to be equal to 0. So, if I assume x 0 is equal to 0, my x 1 is going to be equal to h, x 2 is equal to 2 h, x 3 is equal to 3 h, this is just for the sake of convenience. And then, we will write the formula. So, now, it will become integral 0 to 3 h f x d x is approximately equal to 3 h by 8 and then f at 0 plus 3 times f at h plus 3 times f at 2 h plus f of 3. And then, we will try to find out the formula. So, calculate for f x is equal to 1, integral 0 to 3 h f x d x and the formula using formula, we are going to get a value. So, check whether these two are equal, if they are equal then we will go to f x is equal to x and we will continue till we get a function or we get a power of x for which the two values are not equal that will determine the degree of precision of our rule. So, this is by assuming x 0 is equal to 0, if I assume x 0 is equal to 0, integral x 0 to x 3 f x d x becomes integral 0 to 3 h f x d x, which is approximately equal to 3 h by 8 f of 0 plus 3 f h plus 3 f of 2 h plus f of 3 when you put f x is equal to 1, this is our left hand side L h s. So, L h s is 3 h, R h s will be 3 h by 8, f x is constant 1. So, it will be 1 plus 3 plus 3 plus 1. So, that is 8. So, you get 3 a. So, there is no error for f x is equal to 1. Next, for f x is equal to x, the integral is going to be x square by 2 evaluated between 0 and 3 h. So, that gives us 9 h square by 2, R h s will be 3 h by 8 f of 0. Now, it is going to be 0, then it will be 3 h plus 6 h plus 3 h. So, that again is going to be 9 h square by 2. Consider f x is equal to x square, the exact integral is going to be 27 h cube by 3. That means, it is going to be 9 h cube. The right hand side will be 3 h by 8, no contribution from this term. Here, it will be 3 h square, this will be 4 h square multiplied by 3. So, that will be 12 h square plus here it is going to be 3 h square. So, that is 9 h square. So, when you simplify, you get it to be 9 h cube. So, again no error and f x is equal to x cube. So, the exact integral will be 81 h raise to 4 by 4 and the right hand side also is the same. Next, f x is equal to x raise to 4. So, this is the first time when the left hand side is not equal to right hand side and hence the degree of precision is going to be equal to 3. So, here this is the first time when the left hand side is not equal to right hand side. So, this is an example of a Newton Quartz formula. We have considered Newton Quartz formula for like trapezoidal rule, when we had considered two equidistant points in the interval a to b. Then, we had Simpson's rule. So, we had the two end points and the mid point and then we fitted a polynomial of degree less than or equal to 2. Now, if you consider interval a b and look at four equidistant points including the two end points. So, you have four points, you fit a cubic polynomial and then you integrate. So, because you are fitting a cubic polynomial, if the function itself is a cubic polynomial, then there is no error and hence no error in the integration formula and if you now here what was given to us, it was given to us the formula and then we decided what is the degree of precision. So, this is Simpson's 38th rule, a special case of a Newton Quartz formula. If we were given Simpson's rule, the direct formula and asked to determine the degree of precision, then we would have again obtained no error for 1 x x square x cube and as we have noticed before that this is something unexpected. You are fitting a polynomial of degree 2. So, there should not be any error for quadratic polynomial, but we get a result that no error for cubic polynomial as well. So, this is where as here for the we are fitting cubic polynomial and no error for cubic polynomial. Now, we are going to look at composite trapezoidal rule. So, we will take a special example where we will apply composite trapezoidal rule to find an approximate value of our integral a to b f x d a. We have got a formula for the error in the composite trapezoidal rule. It involves second derivative of the function. Now, the function which we are going to integrate is f x is equal to 1 by x on the interval 1 to 7. So, we choose our interval 1 to 7 because our function f x is equal to 1 by x. It has got a singularity at 0. So, whatever interval you are choosing for integrating that should not include 0. For this function f x is equal to 1 by x, we can calculate its second derivative and then in the error formula we have got second derivative and then we have got power of h and some constant. So, we will try to determine the length of the sub interval so that we achieve the desired accuracy. Now, once we determine h, h is going to be equal to b minus a by n. So, it will also tell us equivalently the number of sub intervals. Next, we will consider the same example for composite Simpson's rule and then find the step size h and the number of sub intervals to achieve the same degree of precision. So, our function is d x by x integral 1 to 7 and we want to determine the step size h such that the error is less than 4 into 10 raised to minus 8 in the case of composite trapezoidal rule. In the case of composite trapezoidal rule error is given by minus f double dash c divided by 12 h square into b minus a where h is b minus a divided by n. Point c is going to be some point in the interval a to b. In this example, b minus a is going to be 7 minus 1. So, it is 6 f x is 1 by x f dash x is minus 1 by x square and f double dash x is 2 by x cube. Point c, we know that it exists, but we do not know what that point c is. So, one dominates f double dash c by norm f double dash infinity. 1 upon x cube is going to be a decreasing function on the interval 1 to 7. So, the maximum will be attained at the left hand point and that gives us norm f double dash infinity to be less than or equal to 2. So, our error, take the modulus. So, modulus of the error will be less than or equal to 2 by 12 b minus a is going to be 6. So, you are going to have 6 into 2 12. So, that will get cancelled. So, you will get h square should be less than 4 into 10 raise to minus 8 and h is going to be equal to nothing but 6 by n. So, modulus of error is less than or equal to h square, which is less than, which we want to be less than 4 into 10 raise to minus 8. So, that is what we want to get. So, h is equal to minus 8 and hence h should be less than 2 times 2 into 10 raise to minus 4, h is equal to 6 by n. So, that gives you 6 by n should be less than 2 into 10 raise to minus 4 or n should be bigger than 3 into 10 raise to 4. So, that is 30,000. So, if you choose n to be bigger than 30,000, then the error will be less than 4 into 10 raise to minus 8 in the composite trapezoidal rule. In the case of composite Simpson's rule, you have fourth derivative of the function h by 2 raise to 4 b minus a by 180. So, calculate the fourth derivative, that is 24 by x raise to 5. Again, the maximum will be attained at the left end point. So, norm f 4 infinity will be 24, h by 2 raise to 4. So, I write it as h raise to 4 and 2 raise to 4 is 16, 180 from here and b minus a is 16. So, that 6 is here. So, this is going to be equal to h raise to 4 divided by 20. Suppose we want this to be less than again the same number 4 into 10 raise to minus 8, then modulus of the error will be less than 4 into 10 raise to minus 8 provided h raise to 4 is less than 80 into 10 raise to minus 8, which gives us h to be less than 80 raise to 1 by 4 10 raise to minus 2. So, we have got 3 raise to 4 is 81. So, that is why 80 raise to 1 by 4 will be less than 3 into 10 raise to minus 2 and that will mean that h is 6 by n. So, our n should be bigger than 2 into 10 raise to 2, that means 200. So, here we have like for that composite trapezoidal rule, you will be evaluating your function if there are n intervals, then you will be evaluating it at n plus 1 points. In case of composite Simpson, we evaluate the function at the n plus 1 partition points and in addition at the midpoint, that means 2 n plus 1. Now, in order to achieve the desired accuracy in case of trapezoidal rule, we need number of subintervals to be 30,000, that means we will be evaluating our function 30,000 plus 1 times. For the Simpson's rule, we have got 200. So, we will be evaluating it 200 into 2 plus 1. So, that means it is going to be 401. So, it is much less than in case of the composite trapezoidal rule. This is about the function evaluation and then you evaluate the function, you multiply by weight and then you are going to add all these numbers. So, this illustrates that the higher the order of H, the formula it should be preferred that we had composite trapezoidal composite Simpson. If your number of intervals is the same, then in the case of composite Simpson, we had double the computation, but if you fix a desired accuracy, that will be achieved with much less effort in case of composite Simpson's rule as compared to composite trapezoidal rule provided your function f is sufficiently differentiable. You need your function to be 4 times differentiable. If your function is only twice differentiable, then we will not get H raise to 4 term in the composite Simpson's rule, then the order of convergence gets reduced to H square. So, for smooth function, the higher the power of H will be available and that formula will be more efficient as compared to formula with lower power of H. So, these were some of the simple examples to illustrate our theory. Now, we are going to consider what is known as Romberg integration. If we remember the corrected composite trapezoidal rule, so corrected trapezoidal rule was obtained by considering cubic hermite polynomial. That means, we looked at a cubic polynomial which interpolates our function f at point a and point b and also the derivatives at point a and point b. So, the formula which one gets, it involves f a f b f dash a f dash b. Now, the term which contains the derivative, it is of the form constant times f dash a minus f dash b. So, if we consider composite corrected trapezoidal rule, that means our interval a b, it is divided into n equal parts on each sub interval you apply corrected trapezoidal rule. Because of this f dash a minus f dash b, when we apply it to sub intervals and add it up, all the derivative terms gets cancelled and then what remains is only two end derivatives f dash a and f dash b. So, corrected composite trapezoidal rule is trapezoidal rule plus a term which contains the derivatives at the two end points. So, it is of the form integral a to b f x d x is equal to the trapezoidal rule plus h square by 12 f dash a minus f dash b plus term of the order of h raise to 4. Here, the error is of the order of h raise to 4, but one needs to know what is f dash a and what is f dash b. So, you have order of convergence h square in the trapezoidal rule. In the correction trapezoidal rule, you have got order of convergence h raise to 4 with the rider that you should know what is f dash a and what is f dash b. Now, one wants to know whether one can do something and get a formula, which will have order of convergence or which will have error to be of the order of h raise to 4, but what should involve only function values because the derivative values they are not available. So, we look at this corrected trapezoidal rule more carefully. You have integral a to b f x d x is equal to t n plus c 1 h square plus term of the order of h raise to 4. What is c 1? c 1 is f dash a minus f dash b divided by 12. That means, our c 1 is independent of the partition. That means, it does not matter the what partition I am looking at. What is c 1 is? It is only f dash a minus f dash b upon 12. If I change my partition, the term will remain the same, c 1 will remain the same. So, why do not I do like this? That I look at a partition with n intervals. So, I will have such integral a to b f x d x to be equal to trapezoidal rule plus a term plus a term of the order of h raise to 4. Now, instead of n intervals, I will consider a partition with n by 2 intervals. That means, earlier our sub intervals they had length to be equal to h. Now, I will look at length to be 2 h. So, then I will have integral a to b f x d x is equal to t to n. So, that means, that is the trapezoidal rule based on partition with n by 2 interval plus I will have c 1 the same constant and instead of h square, I will have 2 h square because now the length of the partition is 2 h plus 2 h. So, this is the term of the order of h raise to 4. So, I have got t n plus c 1 h square t n by 2 plus c 1 2 h square. So, now, I will take a combination of t n and t n by 2, so as to get rid of the term which contains h square. So, let me explain. So, we have integral a to b f x d x is t n plus c 1 h square plus term of the order of h raise to 4. Integral a to b f x d x is t n by 2 and denotes the number of intervals in the sub partition plus c 1 2 h square plus term of the order of h raise to 4. Now, what one can do is multiply the first equation by 4 and subtract the two equations. So, left hand side will be 4 times integral a to b f x d x minus integral a to b f x d x. So, that is going to be 3 times integral a to b f x d x. Here, it will be 4 t n minus t n by 2 the numerator here. Look at this term. It is going to be c 1 times 4 h square minus c 1 times 4 h square. So, this will get cancelled and then term of the order of h raise to 4. Big o of h raise to 4 that means, it is less than or equal to constant times h raise to 4. So, this is and then I divide by 3 throughout. So, I get integral a to b f x d x is equal to 4 t n minus t n by 2 divided by 3 plus term of the order of h raise to 4. So, I have got t n 1. So, integral a to b f x d x is equal to t n 1 plus a term of the order of h raise to 4. Now, what we are going to do is we are considering this partition say t 0 then t 1 t 2 t n minus 2 t n minus 1 t n and this is going to be of length h. So, t n is equal to t n minus 1 t is going to be h by 2 f a plus f b plus h times summation f t i, i goes from 1 to n minus 1. When we look at t n by 2 then it is going to be h times f of f f plus f b because wherever there is h now it is going to be 2 h plus 2 h times summation i goes from 2 to n minus 2 i even f of t i. So, here t n by 2 and t n they have some points they are in common like for the t n by 2 what comes into picture is a b and all even order t i's. When you want to calculate t n then to these points we add points t 1 t 3 up to t n minus 1. So, that means, whatever was the work done for t n by 2, what we are going to do is we are going to do calculation of the function values at t i's for i even that work one uses and then one looks at the combination 4 t n minus t n by 2 divided by 3 whatever formula you get that is going to give you the error to be of the order of h raise to 4. So, this is the first step of Romberg integration. So, if your function f is going to be 4 times differentiable then we obtain a formula which involves only function values no derivative values which gives us the error to be less than or equal to constant times h raise to 4. Now, whether suppose my function f is 6 times differentiable is it possible to obtain error obtain a formula in which case error is less than or equal to constant times h raise to 6. So, such a thing is possible. So, we have got a asymptotic series expansion integral a to b f x d x is equal to t n plus c 1 h square plus c 2 h raise to 4 plus c k h raise to 2 k plus term of the order of h raise to 2 k plus 2. So, we want the function to be sufficiently differentiable to k plus 2 times differentiable. These c i's the coefficients they are going to be independent of h and hence independent of h n and that means it is independent of the partition. In fact, the c i's are some constants alpha i's multiplied by the k minus first derivative at a minus k minus first derivative at point b. Now, this asymptotic expansion it is known as Euler Maclaurin series expansion and the coefficients alpha i's. So, for that what comes into picture are Bernoulli polynomial, but I do not want to get into details of those things. It is just I want to tell you that if the function is sufficiently differentiable then we have got a asymptotic series expansion. Now, using composite trapezoidal rule we got rid of the term h square and then obtained a result which has got error to be less than or equal to constant times h raise to 4. So, this asymptotic series expansion it tells us that it is possible now to get rid of the term h raise to 4 and obtain a error to be less than or equal to h raise to 6 and so on. So, this is known as Romburg integration. So, look at this expansion T n plus c 1 h square plus c 2 h raise to 4 plus term of the order of h raise to 6 when instead of considering n intervals I look at n by 2 intervals I have T n by 2 plus c 1. Now, our intervals are of length 2 h. So, it is 2 h square plus c 2 2 h raise to 4 plus c 2 h raise to 4 plus term of the order of h raise to 6. Now, what we are doing is we are considering 4 times this minus this and divided by 3. So, that was our T n 1. So, when you do that you are going to get integral a to b f x d x is equal to T n 1 plus this term gets cancelled because you are multiplying here by 4 and subtracting. So, nothing here this term is going to have c 2 h raise to 4 common 4 minus this term. So, 4 minus 16 divided by 3 plus term of the order of h raise to 6 and thus you have T n 1 plus c 2 dash h raise to 4 plus term of the order of h raise to 6 where now c 2 dash is minus 9 c 2. So, you have integral a to b f x d x is equal to T n 1 plus c 2 dash h raise to 4 plus term of the order of h raise to 6 c 2 dash independent of h. So, integral a to b f x d x is equal to T n 1 plus c 2 dash will be equal to T n by 2 1 consider partition with n by 2 intervals c 2 dash will not change what will change will be instead of h you will have 2 h raise to 4 plus term of the order of h raise to 6. Now, we want to get rid of these two terms. So, we will multiply this equation by 16 subtract these two equations and divide by 15. So, you will have integral a to b f x d x is equal to 16 T n 1 minus T n by 2 1 divided by 15 this term will go away and you are left with term of the order of h raise to 6. So, this is going to be your T n 2 plus term of the order of h raise to 6 T n 1 what it involves is 1 plus c 2 dash T n T n by 2 these are the composite trapezoidal rules T n by 2 1 will involve T n by 2 and T n by 4. So, T n 2 it is going to be based on trapezoidal rule based on n intervals trapezoidal rule based on n by 2 intervals and trapezoidal rule based on n by 4 intervals and their combinations that is going to give us a formula which has error to be less than or equal to h raise to 6. So, integral a to b f x d x is 16 T n 1 minus T n by 2 1 divided by 15 plus term of the order of h raise to 6. So, diagrammatically here you have T n here you have T n by 2 T n by 4 these are all composite trapezoidal rules then consider 4 times this minus 1 times this divided by 3 that will give you T n 1 next consider 4 times this minus 1 times T n by 4 T n by 4 divided by 3 that will give you T n by 2 1 after obtaining this consider 16 times this minus this term divided by 15 that is going to give you T n 2. More general case will be T n by 2 T n 0 is equal to T n T n is composite trapezoidal rule define T n m by this formula that T n m is equal to 4 raise to m T n m minus 1 minus T n by 2 m minus 1 upon 4 raise to m minus 1 m is equal to 1 2 up to k. So, the final thing T n k is going to give you the error to be less than or equal to constant times h raise to 2 k plus 2 or integral a to b f x d x is equal to T n m plus term of the order of h raise to 2 m plus 2. If m is equal to 0 that means the composite trapezoidal rule we know that it is of the order of h square then m is equal to 1 order of h raise to 4 and so on. So, thus if our function f is sufficiently differentiable then based on composite trapezoidal rule associated with different partitions we can construct approximation to integral a to b f x d x with the error of the order of h raise to 2 k plus 2 where k is equal to 0 1 2 and so on. So, this is the result of the next time we are going to consider numerical differentiation. Thank you.