# Composite Lagrangian Quadrature rule for sin(x)

Suppose we want to estimate $\int_0^{h}f(x) dx$ using Lagrangian interpolating polynomials and with nodes $x_0 = 0,x_1 = \frac{2}{3}h$. Thus we compute the quadrature rule

$$Q(f) = a_0f(0)+a_1f(\frac{2}{3}h).$$

Now using the fact that $$a_k = \int_0^hl_k(x)dx$$

we can find $$Q(f) = \frac{h}{4}(f(0)+3f(\frac{2}{3}h)).$$

Then here comes the part where I get confused. We are then asked to find the composite quadrature rule for $\int_0^1f(x)dx$ and use it to estimate $\int_0^1 \cos(x)dx$ using a step size $h=0.5$.

I am not so sure how to proceed from here. Cheers in advance for any help.

## 1 Answer

Split the interval $[0,1]$ into two $[0,1/2)\cup [1/2,1]$, so that

$$\int_0^1{\rm d}x~f(x) = \int_0^{1/2}{\rm d}x~f(x) + \int_{1/2}^1{\rm d}x~f(x) = \int_0^{h}{\rm d}x~f(x) + \int_{h}^{2h}{\rm d}x~f(x) \tag{1}$$

with $h=1/2$. The second integral in the "problematic" one, to solve it, use the change of variables $y = x - h$, ${\rm d}x = {\rm d}y$ it then becomes

$$\int_0^1{\rm d}x~f(x) = \int_0^{h}{\rm d}x~f(x) + \int_{0}^{h}{\rm d}y~f(y + h) \tag{2}$$

Now both integrals can be approximated with the rule you described. In both cases $f(x) = \cos(x)$, so that $\cos(y+h) = \cos(y + 1/2)$