Discussion
If the quadrature formula
\[ \begin{aligned} &\int_{-1}^{1} f(x)\,dx \\ &\approx \frac{1}{9}\Big(c_{1} f(-1) + c_{2} f\big(\tfrac{1}{2}\big) \\ &\qquad + c_{3} f(1)\Big) \end{aligned} \]
is exact for all polynomials of degree less than or equal to 2, then
Answer: A
The relation that holds is \(c_{1} + \frac{c_{2}}{4} + c_{3} = 6\).
Exactness for degree up to 2 means the rule must integrate \(1\), \(x\) and \(x^{2}\) exactly over \([-1, 1]\), where the true integrals are 2, 0 and \(\frac{2}{3}\). Using \(f(x) = x^{2}\), with \(f(-1) = 1\), \(f(\frac{1}{2}) = \frac{1}{4}\), \(f(1) = 1\):
\[ \begin{aligned} \tfrac{1}{9}\left(c_{1} + \tfrac{c_{2}}{4} + c_{3}\right) &= \tfrac{2}{3} \\ c_{1} + \tfrac{c_{2}}{4} + c_{3} &= 6 \end{aligned} \]
The other two conditions fix the weights: \(f = 1\) gives \(c_{1} + c_{2} + c_{3} = 18\), and \(f = x\) gives \(-c_{1} + \frac{c_{2}}{2} + c_{3} = 0\). Together with the result above they give \(c_{1} = 5\), \(c_{2} = 16\), \(c_{3} = -3\).
Check the rest against these weights: \(c_{1} + \frac{c_{2}}{3} + c_{3} = \frac{22}{3}\), not 4; \(c_{1} + \frac{c_{2}}{2} + c_{3} = 10\), not 2; and \(c_{1} + c_{2} + c_{3} = 18\), not 5.
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