Approximate a definite integral to three decimal places: $\int_0^2 \frac{dx}{\sqrt[3]{64+x^3}}$. I try to expand function $$\frac1{\sqrt[3]{64+x^3}}$$ using Maclaurin series. So, $f(x) = 64{(1+ \frac{x^3}{64})}^{-1/3}$. I expand it and I get $$64\sum_0^n(-1)^n\frac{\left(\frac13\right)\ldots\left(\frac23-n\right)}{(3n+1)}{\left(\frac1{64}\right)}^nx^{3n}$$.
So, even assuming that I expanded $f(x)$ correctly, I am unable to evaluate the answer.
The result I am getting seems to be way too big, assuming that I have not been mislead by online definite integral calculators...
With that said, I think the answer should be around $0.495$, however I failed to dig up to it.
Any ideas how do I do it?
 A: Just perform the integral using some quadrature rule, say a three point quadrature (a.k.a Simpson's rule):
$$\int_{a}^{b} f(x) \, dx \approx \frac{b-a}{6}\left[f(a) + 4f\left(\tfrac{a+b}{2}\right)+f(b)\right]$$
Then subdivide until you reach the desired accuracy.
In this case, the basic rule is enough with no subdivisions.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{2}{\dd x \over \root[3]{64 + x^{3}}}}$

\begin{align}
&\int_{0}^{2}{\dd x \over \root[3]{64 + x^{3}}}=
\int_{0}^{1/2}{\dd x \over \root[3]{x^{3} + 1}}
=\sum_{n = 0}^{\infty}a_{n}\int_{0}^{1/2}x^{3n}\,\dd x
=\sum_{n = 0}^{\infty}a_{n}\,{1 \over 2^{3n + 1}\pars{3n + 1}}
\end{align}
  Find the first $n$ such that $\ds{a_{n}}$ satisfies
  $\ds{\verts{a_{n}\,{1 \over 2^{3n + 1}\pars{3n + 1}}} < 10^{-4}\quad\imp\quad
\verts{a_{n}} < 2^{3n + 1}\pars{3n + 1}\,10^{-4}}$

It turns out that $\ds{n \leq 3}$:
$$
\int_{0}^{2}{\dd x \over \root[3]{64 + x^{3}}}
\approx \sum_{n = 0}^{3}{a_{n} \over 2^{3n + 1}\pars{3n + 1}}\,,
\qquad
\left\lbrace
\begin{array}{rcr}
a_{0} & = & 1
\\
a_{1} & = & -\,{1 \over 3}
\\
a_{2} & = & {2 \over 9}
\\
a_{3} & = & -\,{14 \over 81}
\end{array}\right.
$$

$$
\int_{0}^{2}{\dd x \over \root[3]{64 + x^{3}}}
\approx \half - {1 \over 192} + {1 \over 4032} - {7 \over 414720}
={1437071 \over 2903040} \approx \color{#c00000}{\large 0.495}0228037
$$

Result: $\ds{\color{#00f}{\large\approx 0.495}}$
