0
$\begingroup$

I'm trying to compute $\int_0^4 x^2 dx$ using 2-point Gaussian quadrature. This should give an exact result since for 2 points, we can exactly handle polynomials of degree 2*2-1=3.

By the usual technique, we must first transform this integral to the range $[-1,1]$. This is done using the formula (see change of interval section on this Wikipedia page: https://en.wikipedia.org/wiki/Gaussian_quadrature) by

$\int_0^4 f(x) dx = \frac{4-0}{2} \int_{-1}^1 f(\frac{4-0}{2}x + \frac{4+0}{2}) dx = 2 \int_{-1}^1 f(2x+2) dx$,

and for my function $f(x)=x^2$, this gives

$\int_0^4 x^2 dx = \int_{-1}^1 (2x+2)^2 dx$.

Now 2-point Gaussian quadrature tells me that $\int_{-1}^1 f(x) dx = f(\frac{1}{\sqrt{3}}) + f(-\frac{1}{\sqrt{3}})$ (these weights and points are listed in the first table on the Wikipedia page I linked to above). Now, if I actually evaluate this, I get

$\int_0^4 x^2 dx = \int_{-1}^1 (2x+2)^2 dx = (\frac{2}{\sqrt{3}} + 2)^2 + (-\frac{2}{\sqrt{3}} + 2)^2 = \frac{32}{\sqrt{3}}$.

However, the original integral gives $\int_0^4 x^2 = \frac{x^3}{3}|_0^4 = \frac{256}{3}$ and so my Gaussian quadrature approach is clearly miles off of the actual answer! Given that $x^2$ has degree $\leq 3$, a 2-point Gaussian quadrature should give the exact answer and so my question is basically what have I done wrong?!?!?

Thanks.

$\endgroup$
  • $\begingroup$ Just out of curiousity.. can I ask why you're using an approximation method on an elementary function? Practice, perhaps? I only ask because when utilizing Gaussian quadrature approximation, you risk the chance of generating an error.. one that is proportional to the function's $2n^{th}$ derivative, where $n$ is the number of points. Just something to keep in mind if you attempt higher-ordered functions. $\endgroup$ – Charles Jul 31 '17 at 15:08
2
$\begingroup$

First off, you forgot the factor of $2$ in the transformation, i.e.

$$\int_{0}^{4} x^2 dx = 2\int_{-1}^{1}(2x+2)^2 dx$$

Secondly your calculation is wrong, you should get

$$\left(\frac{2}{\sqrt{3}} + 2\right)^2 + \left(-\frac{2}{\sqrt{3}} + 2\right)^2 = \frac{32}{3}$$

Finally, $4^3 = 64$. If you correct all your mistakes you'll get the correct answer of $64/3$ using both methods.

$\endgroup$
  • $\begingroup$ Thank you! Sigh....it appears I'm having one of those days..... $\endgroup$ – user11128 Jul 31 '17 at 15:48
  • $\begingroup$ @user11128 unnecessary mistakes like these happen to me all the time. The important part is to understand the concepts imho. $\endgroup$ – user159517 Jul 31 '17 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.