# Math Help - WolframAlpha wrong?

1. ## 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.

2. Originally Posted by VonNemo19

Finally, evaluating:

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

3. There is a discontinuity in the graph of that function at x=2
and you are integrating across it.

4. Originally Posted by VonNemo19
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.

5. Originally Posted by dwsmith
$\displaystyle (-1)^{2/3}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i$
Isn't true that $-1^{2/3}=(\sqrt[3]{-1})^2=-1^2=1$

or equivalently

$-1^{2/3}=\sqrt[3]{-1^2}=\sqrt[3]1=1$ ?

6. Originally Posted by VonNemo19
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.
$r=1 \ \ \theta=\tan^{-1}{(-\sqrt{3})}$

$\displaystyle\exp{\left(\frac{2\pi}{3}i\right)}=\c os{\left(\frac{2\pi}{3}\right)}+i\sin{\left(\frac{ 2\pi}{3}\right)}=\frac{-1}{2}+\frac{\sqrt{3}}{2}i$

7. Originally Posted by mr fantastic
... 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

$\chi$ $\sigma$

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

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

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

am I on the right track?

9. Originally Posted by VonNemo19
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.

10. Originally Posted by mr fantastic
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?

11. Originally Posted by VonNemo19
OK, but if I continue with my last post, I will acheive the correct resukt?
Yes. Do it for $(2,5]$

12. Originally Posted by mr fantastic
[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.