Consider the following numerical integration rule: $$I = \int_0^1 \sqrt x f(x)\ dx = w_1f(x_1)$$
At part (a) I showed that $w_1=2/3$ and $x_1 = 3/5$ to make the rule exact for linear polynomials.
Now, I was asked to show that:
$$I - w_1f(x_1) = cf^{''}(\xi), \ \xi \in [0,1]$$
I think I Lagrange's interpolation is needed but not sure how to apply it here.
I'd be glad for help!
Find the Taylor expansion of $f(x)$ at $3/5$ and plug it into the integral:
$$I = \int_0^1 \sqrt x f(x)\ dx = \int_0^1\sqrt{x}\Big(f(\frac{3}{5})+f'(\frac{3}{5})(x-\frac{3}{5})+f''(\xi)\frac{(x-\frac{3}{5})^2}{2}\Big)dx$$
Integrating this gives you
$$\frac{2}{3}f\Big(\frac{3}{5}\Big)+cf''(\xi)$$ for some constant $c$. This proves the error.