# Find error term of a quadrature

The three-point quadrature rule with error term is given by $$\int_{-1}^1f(x)dx=\frac59f\left(\frac{-\sqrt{15}}5\right)+\frac89f(0)+\frac59f\left(\frac{\sqrt{15}}5\right)+kf^{(6)}(c).$$ Find $k$.

After using Lagrange's interpolation (interpolating $f$ at $\frac{-\sqrt{15}}5,0,\frac{\sqrt{15}}5$), I found that the error term should be of the form $$\int_{-1}^1\frac{f'''(c_x)}6x(x^2-\frac53)dx.$$ However, I can't use the mean value theorem, because $x(x^2-\frac53)$ changes sign in $[-1,1]$. So how to continue from here?

I think this is somewhat related to the Simpson's error terms. But the textbook I'm using (Sauer) omitted the proof.

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If you can take it as given that the error term has this form, then you can calculate $k$ by substituting a function with constant sixth derivative, $f(x)=x^6$. Then
$$\frac27=\int_{-1}^1x^6\,\mathrm dx=2\cdot\frac59\left(\frac{\sqrt{15}}5\right)^6+k\cdot6!\;,$$
$$k=\frac1{6!}\left(\frac27-\frac{10}9\left(\frac{\sqrt{15}}5\right)^6\right)=\frac1{15750}\;.$$
But I think $c$ depends on $x$. – asepe Apr 15 '12 at 17:57
@asepe: I don't understand what that means. The only equation in which you use a variable called $c$ is the first one, and in that equation there's no free variable $x$; the variable $x$ occurs in that equation only as an integration variable. Thus it's not clear to me what it would mean for $c$ to depend on $x$. – joriki Apr 15 '12 at 18:19