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How can I solve the integral below?

$$\int_0^1 (x^4+x^3+x^2+x^1+1)^{1/2} \mathrm dx $$

Jean-Sébastien
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Mahsa Ghaffari
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  • I've edited the question and made the title more explicit. Make sure it still reflected what you wanted. – Jean-Sébastien Sep 11 '13 at 18:03
  • Where did you encounter this integral? – dfeuer Sep 11 '13 at 18:06
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    Either way, it can't be done in terms of elementary functions. I have seen this question on a different forum. In a book I have seen it where students had to apply Simpson Rule – imranfat Sep 11 '13 at 18:13
  • The Maple code $$with(Student[Calculus1]):evalf(ApproximateInt(sqrt(x^4+x^3+x^2+x+1), x = 0 .. 1, method = simpson)) $$ produces $1.468637500$ and $$int(sqrt(x^4+x^3+x^2+x+1), x = 0 .. 1, numeric, epsilon = 0.1e-1) $$ gives $ 1.468637565.$ – user64494 Sep 11 '13 at 18:33
  • Unfortunately, I don't have access to Maple, is any Matlab code for it?? Is it possible for you to let me know how is it solved, with which algorithm? – Mahsa Ghaffari Sep 11 '13 at 19:02
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    When you want to extract arc length from the a cubic Bezier curve you will face such an Integral....... – Mahsa Ghaffari Sep 11 '13 at 18:30
  • Unfortunately, I don't have access to Maple, is any Matlab code for it?? Is it possible for you to let me know how is it solved, with which algorithm? – Mahsa Ghaffari Sep 11 '13 at 19:01
  • @MahsaGhaffari Any numerical quadrature algorithm will work, for example Simpson's rule. In Matlab, just use integral: http://www.mathworks.com/help/matlab/ref/integral.html – user7530 Sep 11 '13 at 19:08
  • @Mahsa Ghaffari: Also one can use the WolframAlpha to this end. – user64494 Sep 11 '13 at 19:15
  • @MahsaGhaffari : you don't solve definite integrals, you evaluate them. I would recommend trying a computing algebra system first to see if it is even possible. wolframalpha.com is is free and online. All you need is an Internet connection and a Web browser. It seems to accept Mathematica or Maple syntax, at least. In Maplese, you enter int(sqrt(x^4+x^3+x^2+x+1),x=0..1) in the box. If you don't get an exact answer, it probably cannot be done exactly using named functions. wolframalpha will probably give you an approximate answer whether or not it can do it exactly. Enjoy! – Stefan Smith Sep 12 '13 at 01:30

1 Answers1

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You can use the horrible result of calculating $\int(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$ by Wolfram Integrator, but I have the following relatively simpler approach:

$\int_0^1(x^4+x^3+x^2+x+1)^\frac{1}{2}~dx$

$=\int_0^1\left(\dfrac{1-x^5}{1-x}\right)^{\frac{1}{2}}~dx$

$=\int_0^1\dfrac{(1-x^5)^\frac{1}{2}}{(1-x)^\frac{1}{2}}~dx$

$=\int_0^1\sum\limits_{n=0}^\infty\dfrac{(2n)!x^{5n}}{4^n(n!)^2(1-2n)(1-x)^\frac{1}{2}}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!B\left(5n+1,\dfrac{1}{2}\right)}{4^n(n!)^2(1-2n)}$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!(5n)!\left(-\dfrac{1}{2}\right)!}{4^n(n!)^2\left(5n+\dfrac{1}{2}\right)!(1-2n)}$

$=\sum\limits_{n=0}^\infty\dfrac{(2n)!(5n)!\sqrt\pi}{4^n(n!)^2\dfrac{(10n)!\sqrt\pi}{4^{5n}(5n)!}(1-2n)}$

$=\sum\limits_{n=0}^\infty\dfrac{256^n(2n)!((5n)!)^2}{(n!)^2(10n)!(1-2n)}$

Harry Peter
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  • (+1) This result/method can also be expressed in terms of the generalised hypergeometric function: $$\int _{0}^{1}!\sqrt {{\frac {{x}^{5}-1}{x-1}}}{dx}=2, {\mbox{$_6$F$_5$}(-\frac {1}{2},\frac {1}{5},\frac {2}{5},\frac {3}{10},\frac {4}{5},1;,\frac {3}{10},\frac {1}{2},{\frac {7}{10}},{\frac {9}{10}},{\frac {11}{10}};,1)}$$ not that that neccessarily adds much. – Graham Hesketh Sep 12 '13 at 16:10
  • @ Harry Peter: Up to Wiki, this is not a closed-form expression – user64494 Sep 12 '13 at 16:11
  • @ Graham Heskesh: Could you base your claim? – user64494 Sep 12 '13 at 16:15
  • @user64494, the series can be written as a single summation, the summand can be expressed in terms of the pochhammer symbol and that series can be compared to the series for the generalised hypergeometric function (see e.g. Maple help "hypergeom") some algebra is involved to get a recognisable comparison, this might be of academic interest but I don't think it adds much in this context. – Graham Hesketh Sep 13 '13 at 10:33