Hi,
Me and my friend tried to solve:
integral from 1 to 9 of -2/(x-9)^(1/3) dx
So we set it as a lim t --> 9- and got -12 but the homework system says it's wrong. I checked wolfram and it gives the answer as 12(-1)^(2/3) so it's a bit confusing. Not to mention that the graph's area is positive from 1 to 9. Any clarification is appreciated
Well, you've run into a weird part of Mathematics. The integral comes out to beOriginally Posted by DannyMath
. You might think that, since -2 is the cube root of -8 and then you square it that
. But you'd be wrong. We must take that negative out and leave it on its own:
.
This is a similar rule to why you have to factor -1s out of the following:. This expression is not
. This is
.
-Dan
In case it makes you feel better, playing around in the complex plane for a bit gives
^{2/3} = -\frac{1}{2} + i~\frac{ \sqrt{3}}{2})
I've not learned about imaginary numbers but I plan on it in the summer. From your post I get that you also say that it's -12? It just seemed odd to me that it would be negative since the function is above the x-axis on the given limits of integration.
here is a link to some discussion over Wolfram alpha's response to roots of negative values ...
Wolfram|Alpha Community • View topic - Odd Roots of Negative Numbers
since you're dealing with real values, you'd be safe in saying the value of the improper integral is +4.
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