Given the integration rule $Q(x) = \alpha_1f(0)+\alpha_2f(1)+\alpha_3f'(0)$ for interpolating the integral $\int_0^1f(x) dx$ , I need to find $\alpha_1,\alpha_2,\alpha_3$ values s.t Q has exact degree of n=2.
For exact degree of n=2 , I know that we must have : $\int_0^1 1 = 1,\int_0^1 x = 0.5,\int_0^1 x^2 = 1/3$ , but I cannot find out how to find $\alpha_1,\alpha_2,\alpha_3$ that answer these constraints.
I'd be grateful if anyone could point me to the solution,thanks in advance.
For any function $f$, let $$I(f) = \int^{1}_{0} f(x) \, dx$$ provided the integral exists. Let $$Q(f) = \alpha_1f(0) + \alpha_2f(1) + \alpha_3f'(0)$$ be a quadrature that approximates $I(f)$. The error in the approximation is given by $$E(f) = I(f) - Q(f).$$ If the quadrature has degree of precision $2$, then $E(1) = E(x) = E(x^2) = 0$.
$$E(1) = \int^{1}_{0} dx - (\alpha_1 + \alpha_2) = 1 - \alpha_1 - \alpha_2 = 0.$$ $$E(x) = \int^{1}_{0} x \, dx - (\alpha_2 + \alpha_3) = \frac{1}{2} - \alpha_2 - \alpha_3 = 0.$$ $$E(x^2) = \int^{1}_{0} x^2 \, dx - \alpha_2 = \frac{1}{3} - \alpha_2 = 0.$$ The solution to this linear system is $$\alpha_1 = \frac{2}{3}, \quad \alpha_2 = \frac{1}{3}, \quad \alpha_3 = \frac{1}{6}.$$