Gauss Point

A Gauss point is used for approximating integrals. For an example of its use, let’s approximate the integration of the following function:

$y = \frac{1}{2}x^{3} - \frac{1}{10}x^{2} + \frac{1}{2}x + 2$

You might have used the trapezium rule before to approximate integrals, by calculating the two values at either end of the integral and drawing a trapezium underneath them and working out the area of the trapezium:

\begin{align} \int_{-1}^{1}y\ dx & \approx \frac{1}{2}(0.9 + 2.9) \times 2 \\ & = 3.8 \end{align}

Note that, in order to calculate this approximation, we’ve had to evaluate the function in two places, at $$x = -1$$ and at $$x = 1$$. Gauss noted that to get the same accuracy, we actually only need to evaluate the function in one place, at $$x = 0$$:

The reason we only need this one point is because every trapezium that passes through here has the exact same area:

Note for this linear approximation, this only works if we use the point at $$x = 0$$, so $$x = 0$$ is called the Gauss point and its corresponding weight is $$2$$ because we multiply the value at $$x = 0$$ by $$2$$ to approximate the area.

\begin{align} \int_{-1}^{1}y\ dx & \approx 2 \times 2 \\ & = 4 \end{align}

We can use more than one Gauss point to make higher order approximations of integrals, for example we can use a cubic approximation using Gauss points at $$x = -\frac{1}{\sqrt{3}}$$ and $$x = +\frac{1}{\sqrt{3}}$$ both with weights of $$1$$. If we run this through the function in Figure 1, we get the answer $$3.9\dot{3}$$, which (as the function is cubic) is the exact answer.

Gaussian quadrature is preferred over other methods of numerical integration because it involves evaluating the fewest points on the function to get the accuracy you need, which can offer substantial time savings when doing intensive calculations such as Finite Element Analysis. However, remember that it is only an approximation, and if you have a finite element with complex loading applied to it that would ordinarily give rise to big discontinuities or high order variations in stress, then consider that a simple cubic approximation might not appropriately reflect this.