My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.

Solution (a). Firstly, we set \(f(x) = 1\), we have \[ \int_a^b f(x) \, dx = b - a = A_0 + A_1. \] Given that \(A_0\) and \(A_1\) are the same, we have \(A_0 = A_1 = \frac{b - a}{2}\). Now, set \(f(x) = x\), we have \[ \int_a^b x \, dx = \frac{b^2}{2} - \frac{a^2}{2} = \frac{(b-a)\cdot (b + a)}{2}= \frac{b - a}{2} x_0 + \frac{b - a}{2} x_1. \] Thus, we get \(x_0 + x_1 = b+a\), which implies that \(x_0 = b + a - x_1\). Then, we set \(f(x) = x^2\), we have \[ \int_a^b x^2 \, dx = \frac{b^3}{3} - \frac{a^3}{3} = \frac{(b-a)\cdot (b^2 + ab + a^2)}{3} = \frac{b - a}{2} x_0^2 + \frac{b - a}{2} x_1^2. \] It shows that \[ \frac{2\cdot (b^2 + ab + a^2)}{3} = x_0^2 + x_1^2. \] We substitute \(x_0 = b + a - x_1\) into the equation above, we get \[ \begin{align} \frac{2\cdot (b^2 + ab + a^2)}{3} &= (b + a - x_1)^2 + x_1^2 \\ \frac{2\cdot ((a + b)^2 - ab)}{3}&= (b + a)^2 - 2x_1(b+a) + 2x_1^2 \\ 2x_1^2 - 2x_1(b+a) + (b + a)^2 &- \frac{2\cdot ((a + b)^2 - ab)}{3} = 0 \\ 2x_1^2 - 2(b+a)x_1 + \frac{(a + b)^2}{3} + \frac{2ab}{3} &= 0 \\ \end{align} \] Now, using the quadratic formula, we have \[ x_1 = \frac{2(b+a) \pm \sqrt{4(b+a)^2 - 4\cdot 2\cdot \left(\frac{(a + b)^2}{3} + \frac{2ab}{3}\right)}}{4} = \frac{b+a \pm \sqrt{\frac13b^2 - \frac23ab + \frac13a^2}}{2} = \frac{b+a \pm \frac{\sqrt{3}}{3}(b-a)}{2}. \] Hence, we have \(x_0 = \frac{b+a}{2} - \frac{\sqrt{3}(b-a)}{6}\), and \(x_1 = \frac{b+a}{2} + \frac{\sqrt{3}(b-a)}{6}\). Therefore, we have \[ \int_a^b f(x) \, dx \approx \frac{b - a}{2} f\left(\frac{b+a}{2} - \frac{\sqrt{3}(b-a)}{6}\right) + \frac{b - a}{2} f\left(\frac{b+a}{2} + \frac{\sqrt{3}(b-a)}{6}\right). \]