Evaluate the integral $$\int _{0}^{1}\sqrt [3] {2x^{3}-3x^{2}-x+1}~\mathrm dx.$$
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Christian Ivicevic
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FMath
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2Is it 0? This is based on a very crude guess. – Student Dec 08 '15 at 19:13
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I doubt that this can be expressed in terms of simple constants: Mathematica 7 gives a very ugly result in terms of the Appell hypergeometric function $F_1$. – Alex M. Dec 08 '15 at 19:13
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We seperated lines from 0 to (1/2) and from (1/2) to 1 ,because at (1/2) point second derivative is 0 – FMath Dec 08 '15 at 19:19
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i have got a complex value of this integral – Dr. Sonnhard Graubner Dec 08 '15 at 19:31
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Substitute $u=x +1/2$ to see that this function is odd hence the integral equals zero. – Michael Hoppe Dec 08 '15 at 19:39
1 Answers
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Note firstly that:
$(2x^{3} -3x^{2}-x+1)^{1/3} =[(2(x-\frac{1}{2})((x-\frac{1}{2})^{2} -\frac{5}{4})]^{1/3}$
Thus let $u=x-\frac{1}{2}$.
The upper limit becomes $\frac{1}{2} $ and the lower limit becomes $\frac{-1}{2}$.
And so we have:
$\int^{1/2}_{-1/2} (2u^{3} -\frac{5u}{2}) du$
Note that the expression inside the integral is odd and passes through the origin, and since the integral is the sum of the signed areas bounded by the curve and the $x$ axis, the sum of these areas must be $0$. Consequently the integral is $0$.
J.Gudal
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