I've been working out questions regarding Newton Raphson and Simpson's Rule, whilst they're fairly easy to execute, the latter seems to boggle my mind a little bit more in terms of what the examiner's usually ask. Below is a part of a question (which is an amallgemation of the previous two).
Evaluate the integral $\int_{0}^{0.4} \ln(1+x^3)dx$ by means of (i) Simpson's Rule with an interval width of $h = 0.1$ and (ii) Integrating the first two terms of the series expansion of $\ln(1+x^3)$. Give answer to 4 d.p
So working the first one out went quite well, my answer boiled down to $$\int_{0}^{0.4} \ln(1+x^3)dx \approx 0.0063$$
However the second part is where my mind is blown as I'm not sure what he's asking for. I do remember something to do with the expansion of $\ln$ being something like $(-1)^{r-1} \frac{x^r}{r}$ however am not sure how this applies to Simpson's rule or anything similar to it.
I presume that this question is to prove the accuracy of the above approximation.
The second part is a different method of approximation, unrelated to Simpson's rule: By the geometric series formula $1/(1 + t) = 1 - t + t^{2} - \cdots$ for $|t| < 1$, and the fact that a convergent power series may be integrated term by term, $$ \ln(1 + u) = \int_{0}^{u} \frac{1}{1 + t}\, dt = \int_{0}^{u} [1 - t + \cdots]\, dt = u - \frac{u^{2}}{2} + \cdots\quad\text{for $|u| < 1$.} $$ Substituting $u = x^{3}$, $\ln(1 + x^{3}) \approx x^{3} - \frac{1}{2} x^{6}$ for $|x| < 1$. So, you can integrate this polynomial approximation using the fundamental theorem of calculus, and use the value as a numerical approximation of $$ \int_{0}^{0.4} \ln(1 + x^{3})\, dx. $$