Where am I going wrong? I want to prove that it's accurate for deg 2 polynomials or less and my answer doesn't suggest that? Any tips on where I'm wrong ? Ive attached an image of my work below!
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Please try to type your formulas, it makes your question better accessible. Also, it may serve as a review of your calculations, thus enabling to find errors before you need to actually ask the question. – Lutz Lehmann Apr 08 '18 at 14:36
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Your principal problem is that you want to find parameters for $$ \int_0^1f(x)dx=w_0f(x_0)+w_1f(1) $$ but your following equations are for $$ \int_0^1f(x)dx=w_0f(0)+w_1f(x_0) $$ which is the mirrored situation of the task. The solutions are related, but not identical.
The equation to compare coefficients to find the conditions should have been $$ A+\frac12B+\frac13C=w_0(A+Bx_0+Cx_0^2)+w_1(A+B+C)\\ \left\{\begin{aligned} 1&=w_0+w_1\\[.5em] \frac12&=w_0x_0+w_1\\[.5em] \frac13&=w_0x_0^2+w_1 \end{aligned}\right. $$ a said before, similar to the ones you found, but not identical.
Lutz Lehmann
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Where is the point i go wrong? Is it in my formula for A, B and C ? @LutzL – IanWalter Apr 08 '18 at 14:53
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No, your computation for the mirrored problem is correct, you would have to mirror the solution back. – Lutz Lehmann Apr 08 '18 at 16:19
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Oh I see! Such a rookie mistake now I see it in front of me. Think I over complicated things in my head. Thank you! – IanWalter Apr 08 '18 at 18:51
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Using the two point Gauss Quadrature formula (here), there is a change of interval formula. This will yield an exact answer for polynomials of degree (2n-1) = 3 in our case (and therefore degree 2 as well).
Jokus
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There is only one sampling point free, $x_0$, the other is fixed as $x_1=1$. Gauß would apply if both sampling points were free. – Lutz Lehmann Apr 08 '18 at 16:18