CHAPTER 07.07: IMPROPER INTEGRATION: Singular Integrand: Example
In this segment, we're going to take an example for improper integrals. So we're going to take an example of improper integral, and we're going to talk about an integrand where we have singular integrands. Of course, we are only talking about integrals which can be . . . which have a finite value, so the ones which have . . . which can be solved and found within a finite values. So an example of that is given as follows, 5 x, divided by square root of 9 minus x squared, dx, from 0 to 3. Such integrals are important because these are the kind of integrals which you will get for, let's suppose if you have a crack in a particular body, and you are trying to find out what the crack opening displacement is, you get integrands which look of this particular form. So it's important for us to be able to numerically and analytically be able to find out what these values are. So in this case, the example is asking you that, hey, you have this integral, find the integral approximately using two-point Gauss quadrature rule. Now, one of the things which you are seeing here is that this particular integrand here, which is 5 x, divided by square root of 9 minus x squared, it is infinite at x equal to 3, at the upper limit of integration, it is . . . it is infinite, because you get 9 minus 9, you get 0. And the good thing about using, let's suppose Gauss quadrature rule is that we never have to use the endpoints. So if you are familiar with the Gauss quadrature rule, or you already looked at this particular in the previous segment, which you should have, is that the endpoints are never chosen as the quadrature points where these values of the integrands have to be calculated, so we don't have to worry about it becoming infinite at x equal to . . . x equal to 3. So what we're going to do is we're going to take the two-point Gauss quadrature rule and use it to find this integral, and then we will see, go ahead and see that how good it . . . good it is. So let's go ahead and use this two-point Gauss quadrature rule. So we already know that the Gauss quadrature rule, the two-point Gauss quadrature rule for a function which is of the form -1 to 1, is approximately given by c1 times f of x1, plus c2 times f of x2. So that's what the Gauss quadrature rule, two-point Gauss quadrature rule is given by And we have c1 is 1, c2 is also 1, x1 is -1 divided by square root of 3, and we'll take it as -0.5773 in this case, four significant digits, x2 is 1 divided by square root of 3, which is 0.5773,. So that's what the two-point Gauss quadrature rule is, and we want to apply it to the integral which we have. So the first thing which we have to do is that we have to take this integral, 0 to 3, 5 x, divided by 9 minus x squared, square root, dx, and converted your integral going from -1 to +1 so that we can apply the two-point Gauss quadrature rule, so we've got to find out what this should be. So as we have already talked about in Gaussian quadrature rule is that the integral going from a to b, f of x dx can be converted into an integral going from -1 to +1 by this particular formula, b minus a, divided by 2, times -1 to +1, f of b minus a, divided by 2, times x, plus b plus a, divided by 2, dx. So that's how we can convert any integral, definite integral going from a to b to an integral going from -1 to +1. So in this case, let's go ahead and see how we can go about doing this, of course, both the limits of integration have to be finite to be able to use this particular formula. So in this case, we have f of x is given as 5 x, divided by square root of 9 minus x squared, and a is 0, and b is 3. So based on the values of f of x, a, and b, we can rewrite our integral, which is going from 0 to 3, f of x dx, to be b minus a, divided by 2, the integral going from -1 to +1, the value of the function being calculated at b minus a, divided by 2, times x, plus b plus a, divided by 2, dx of course. So that gives us 1.5 integral from -1 to +1, f of 1.5 x, plus 1.5, dx. So that's how we will be able to convert our integral going from 0 to 3 to an integral going from -1 to +1. Now what we're going to do is we're going to apply our Gaussian quadrature formula here. So 1.5 is outside, so I'm going to keep it outside, I'm going to apply the Gaussian quadrature rule formula for this integral here, which will be c1, which is 1, times the value of the function at x1, so instead of x, I substitute x1, plus c2, which is also 1, times the value of the function at x2, which is 1.5 x2, plus 1.5. So that's all I have to do in order to apply the two-point Gaussian quadrature formula here. So what that tells me is that this is equal to 1.5, times the value of the function at 1.5, x1 is -0.5773, so I'm going to substitute that value, plus the value of the function at x2, which is 0.5773, plus 1.5. So this one gives me that, hey, in order to be able to use the Gaussian quadrature formula, I need to calculate the value of the function at 0.6340, and I need to calculate the value of the function at 2.366. Again, keep in mind that both of these values, the arguments of both of these values do need to be between the original lower limit of integration, upper limit of integration, which was 0 and 3. It's only a check, it doesn't mean that they are right. So once we have done that, now we calculate the value of the function at those points, which will be . . . the original function was what? 5 x . . . 5 times x, divided by square root of 9 minus x squared, plus 5 times this x, which is 2.366, divided by square root of 9 minus 2.366 squared, and this turns out to be 1.5 times, the value of the function at 0.6340 turns out to be 1.021, the value of the function at 2.366, turns out to be 6.415, and what we get is 11.24. So that's what we get as the value of the . . . value of that integral, from 0 to 3, to be 11.24. So I'm going to write it down here. That means that 0 to 3, 5 x, divided by square root of 9 minus x squared is approximately turning out to be 11.24, but you can do exact calculations also, 0, 3, 5 x, divided by square root of 9 minus x squared, sorry there should be a dx here, and dx, is exactly equal to 15. So I'm going to leave this as your homework. So we can see there's a vast difference between what we are getting here and what we're getting here by the two-point Gauss quadrature rule formula. We did not face the problem of division by 0, because Gaussian quadrature rule does not use the endpoints as its quadrature points, or its accesses, so, but still we're getting a large amount of error. The relative true error in this case is the exact value minus the approximate value, divided by the exact value, times 100, which is equal to 25.07 percent. So how do we go about reducing the error? We will show that in the next example. But this is the end of this segment. |