Approximate a definite integral to three decimal places: $\int_0^2 \frac{dx}{\sqrt[3]{64+x^3}}$.

Question

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?

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{2}{\dd x \over \root[3]{64 + x^{3}}}}$

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

Answer 2

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.