Using series find $\int_0^1 \sqrt{1+x^4}\hspace{1mm} dx$ up to $2$ decimal places

Question

I cannot figure out an aesthetic way to do this.

Can someone give a beautiful solution to this ugly question?

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This is what I have tried yet.

I used the fact that $$x = \dfrac{1}{1-\left(1-\dfrac{1}{x}\right)}$$

Hence $$x = \sum_{n=0}^{\infty} \left(1-\dfrac{1}{x}\right)^n$$

Now replacing $x$ with $\sqrt{1+x^4}$, to get $$\sqrt{1+x^4} = \sum_{n=0}^{\infty} \left(1-\dfrac{1}{\sqrt{1+x^4}}\right)^n$$

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Just wondering if there are better alternatives available.

Answer 1

Hint :

Maclaurin series for Binomial series is $$ (1+y)^n=\sum_{k=0}^\infty\binom{n}{k} y^n, $$ for $|y|<1$ and all $n$. Also use identities binomial coefficients: $$ \binom{n}{0}=1 $$ and $$ \binom{n}{k+1}=\binom{n}{k}\frac{n-k}{k+1}. $$ Now, replace $y=x^4$ and then integrate it term by term.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \int_{0}^{1}\root{1 + x^{4}}\,\dd x&= \int_{0}^{1}\sum_{n = 0}^{\infty}{1/2 \choose n}x^{4n}\,\dd x =\sum_{n = 0}^{\infty}{1/2 \choose n}\int_{0}^{1}x^{4n}\,\dd x =\sum_{n = 0}^{\infty}{1/2 \choose n}{1 \over 4n + 1} \end{align}

Now we 'cut' the series once the absolute value of the sum 'general term' $\ds{{1/2 \choose n}{1 \over 4n + 1}}$ becomes smaller than $\ds{10^{-2}}$: $$ \begin{array}{rclrcl} \verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 0} & = & 1\,,&\qquad \verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 1} & = & 0.1 \\[3mm] \verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 2} & \approx &-0.0139\,,&\qquad \verts{{1/2 \choose n}{1 \over 4n + 1}}_{\,n\ =\ 3} & \approx & 0.0048 \end{array} $$

Then, I'll use three terms $\ds{\pars{~n = 0,1,2}}$: \begin{align} \color{#66f}{\large\int_{0}^{1}\root{1 + x^{4}}\,\dd x} &\approx {1/2 \choose 0}\,{1 \over 4\times 0 + 1} +{1/2 \choose 1}\,{1 \over 4\times 1 + 1} + {1/2 \choose 2}\,{1 \over 4\times 2 + 1} \\[3mm]&=1 + 0.1 - 0.0139 = \color{#66f}{\large 1.08\color{#c00000}{61}} \end{align}

A numerical calculation of the integral yields $\ds{\approx \color{#66f}{\large 1.08}9429413}$.