I'm going through the "Calculating a general row in the matrix" section of this tutorial right now and there's one step with the following integral
$b_i = \int_{x_{i-1}}^{x_i}\frac{x-x_{i-1}}{h}f(x)dx+\int_{x_{i}}^{x_{i+1}}(1-\frac{x-x_i}{h})f(x)dx $
where $f(x) = x(1-x)$ and $h=x_{i+1}-x_i$ (or $x_i-x_{i-1}$). They end up with a solution for $b_1$ of:
$b_1 = \frac{h^2}{12}(12-14h)$
How do they do this? After substituting in the $f(x)$ expression, I get a jumble of terms I can't seem to simplify, and I can't reproduce this answer. I broke it apart into:
$b_i = \int\frac{x}{h}f(x) - \int \frac{x_{i-1}}{h}f(x) + \int f(x) - \int \frac{x f(x)}{h} - \frac{x_i}{h} \int f(x)$
and evaluated each, but I end up with $x^4$ terms from the fourth term. I'm not sure how this reduces into only a $h^3$ term like in the answer. Thanks!
The highest order terms in the result will be $\frac{x_{i+1}^4 - x_i^4}{h} = \frac{(x_i + h)^4 - x_i^4}{h} = 4x_i^3 + 6x_i^2h + 4x_ih^2 + h^3$.