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Math Help - WolframAlpha wrong?

  1. #1
    No one in Particular VonNemo19's Avatar
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    WolframAlpha wrong?

    Hi.

    I usually check my work with WolframAlpha and for this problem there seems to be a discrepency...

    Here's what I've done:

    \int_1^5\frac{dx}{(x-2)^{1/3}}

    Let u=x-2\Rightarrow{du}=dx

    This means that

    \int_1^5\frac{dx}{(x-2)^{1/3}}=\int_{x=1}^{x=5}\frac{du}{u^{1/3}} .

    As for the limits with respect to u, we observe that

    x=1\Rightarrow{u=-1} and x=5\Rightarrow{u=3}.

    Finally, evaluating:

    \int_{-1}^3u^{-1/3}du=\frac{3}{2}u^{2/3}|_{-1}^3=\frac{3}{2}[(3)^{2/3}-(-1)^{2/3}]

    Therefore, \int_1^5\frac{dx}{(x-2)^{1/3}}=\frac{3}{2}(\sqrt[3]{9}-1)

    But WAlpha gets something different.
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post

    Finally, evaluating:

    \int_{-1}^3u^{-1/3}du=\frac{3}{2}u^{2/3}|_{-1}^3=\frac{3}{2}[(3)^{2/3}-(-1)^{2/3}]
    \displaystyle (-1)^{2/3}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i
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  3. #3
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    There is a discontinuity in the graph of that function at x=2
    and you are integrating across it.
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    Hi.

    I usually check my work with WolframAlpha and for this problem there seems to be a discrepency...

    Here's what I've done:

    \int_1^5\frac{dx}{(x-2)^{1/3}}

    Let u=x-2\Rightarrow{du}=dx

    This means that

    \int_1^5\frac{dx}{(x-2)^{1/3}}=\int_{x=1}^{x=5}\frac{du}{u^{1/3}} .

    As for the limits with respect to u, we observe that

    x=1\Rightarrow{u=-1} and x=5\Rightarrow{u=3}.

    Finally, evaluating:

    \int_{-1}^3u^{-1/3}du=\frac{3}{2}u^{2/3}|_{-1}^3=\frac{3}{2}[(3)^{2/3}-(-1)^{2/3}]

    Therefore, \int_1^5\frac{dx}{(x-2)^{1/3}}=\frac{3}{2}(\sqrt[3]{9}-1)

    But WAlpha gets something different.
    The function you are integrating has a discontinuity over the interval of integration (at x = 2) so you have to take greater care in dealing with the resulting improper integrals.

    Further, WolframAlpha does not give a real value to (-1)^(2/3) (nor does a CAS calculator. There are technical reasons for this). There are three complex values, namely \displaystyle cis\left(- \frac{2\pi}{3}\right), \displaystyle cis\left(\frac{2\pi}{3}\right) and cis(2\pi) = 1. Since your integral is real (why?) ......


    Edit: Fixed a careless typo. Misread cis(-pi/2) = -1 instead of cis(-pi) = -1.
    Last edited by mr fantastic; January 13th 2011 at 02:48 PM.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle (-1)^{2/3}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i
    Isn't true that -1^{2/3}=(\sqrt[3]{-1})^2=-1^2=1

    or equivalently

    -1^{2/3}=\sqrt[3]{-1^2}=\sqrt[3]1=1 ?
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    Isn't true that -1^{2/3}=(\sqrt[3]{-1})^2=-1^2=1

    or equivalently

    -1^{2/3}=\sqrt[3]{-1^2}=\sqrt[3]1=1 ?
    Polar form.
    r=1 \ \ \theta=\tan^{-1}{(-\sqrt{3})}

    \displaystyle\exp{\left(\frac{2\pi}{3}i\right)}=\c  os{\left(\frac{2\pi}{3}\right)}+i\sin{\left(\frac{  2\pi}{3}\right)}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by mr fantastic View Post
    ... the function you are integrating has a discontinuity over the interval of integration (at x = 2) so you have to take greater care in dealing with the resulting improper integrals...
    Of course that is true and I suggest to start with the indefinite integral...

    \displaystyle \int \frac{dx}{(x-2)^{\frac{1}{3}}} = \frac{3}{2}\ (x-2)^{\frac{2}{3}} + c (1)

    Now we can devide the definite integral in two parts...

    \displaystyle \int_{1}^{5} \frac{dx}{(x-2)^{\frac{1}{3}}} = \int_{1}^{2} \frac{dx}{(x-2)^{\frac{1}{3}}} + \int_{2}^{5} \frac{dx}{(x-2)^{\frac{1}{3}}} (2)

    But both the integrals in (2) converge so that is...

    \displaystyle \int_{1}^{5} \frac{dx}{(x-2)^{\frac{1}{3}}} =  \frac{3}{2}\ (0 - 1 + 3^{\frac{2}{3}} - 0)= \frac{3}{2}\ (3^{\frac{2}{3}} -1) (3)

    So I agree with VonNemo and the result of WolframAlpha is difficult to undestand... at least for me...

    Kind regards

    \chi \sigma
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  8. #8
    No one in Particular VonNemo19's Avatar
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    So, should I proceed like so...

    \displaystyle\int_1^5\frac{1}{(x-2)^{1/3}}dx=\int_1^2\frac{1}{(x-2)^{1/3}}dx+\int_2^5\frac{1}{(x-2)^{1/3}}dx.

    Now, Let u=x-2\Rightarrow{du}=dx.

    So, \displaystyle\int_1^2\frac{1}{(x-2)^{1/3}}dx=\int_{x=1}^{x=2}u^{-1/3}du

    =\displaystyle\lim_{b\to2^-}\int_{x=1}^{x=b}u^{-1/3}du=\displaystyle\lim_{b\to2^-}<br />
\frac{3}{2}u^{2/3}\Bigg|_{x=1}^{x=b}

    =\displaystyle\lim_{b\to2^-}\frac{3}{2}(\sqrt[3]{x-2})^2\Bigg|_{1}^{b}=\frac{3}{2}(-1)

    am I on the right track?
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  9. #9
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    Quote Originally Posted by VonNemo19 View Post
    Isn't true that -1^{2/3}=(\sqrt[3]{-1})^2=-1^2=1

    or equivalently

    -1^{2/3}=\sqrt[3]{-1^2}=\sqrt[3]1=1 ?
    Yes, I misread something I wrote down. See my edit. What I said is still true but some details needed correction.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Yes, I misread something I wrote down. See my edit. What I said is still true but some details needed correction.
    OK, but if I continue with my last post, I will acheive the correct resukt?
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  11. #11
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    Quote Originally Posted by VonNemo19 View Post
    OK, but if I continue with my last post, I will acheive the correct resukt?
    Yes. Do it for (2,5]
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    [snip]

    Further, WolframAlpha does not give a real value to (-1)^(2/3) (nor does a CAS calculator. There are technical reasons for this). There are three complex values, namely \displaystyle cis\left(- \frac{2\pi}{3}\right), \displaystyle cis\left(\frac{2\pi}{3}\right) and cis(2\pi) = 1.

    [snip]
    I will elaborate on this statement.

    When the TI-89 CAS calculator is operating in real mode it uses the real branch (when it exists) for fractional powers that have a reduced exponent with odd denominator. So for (-1)^(2/3) it returns a value of 1.

    However, when operating in complex mode, or when the real branch does not exist, the TI-89 uses the principle branch. Thus, for (-1)^(2/3) it returns a value of \displaystyle - \frac{1}{2} + i\, \frac{\sqrt{3}}{2} (which is the rectangular form of cis\left( \frac{2\pi}{3}\right), by the way). No doubt something similar is happening with Mathematica: &#40;-1&#41;&#94;&#40;2&#47;3&#41; - Wolfram|Alpha

    and this is what's causing the trouble with regards to the output for the definite integral that initiated this thread.
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