What does this integral mean and how can it be solved?

[DivGradCurl]

\int _1 ^9 \frac{1}{\sqrt[3]{x-9}}\: dx = -6\cos \left( \frac{2k\pi}{3} \right) - i \mbox{ } 6\sin \left( \frac{2k\pi}{3} \right) \quad k \in \mathbb{N}

How do I interpret this integral? The only real number that I can get out of it is -6.

Thanks

PS: I mean, maybe the integral can't be interpreted as the area under the graph anymore. What happens in this case? I'm a bit confused.

 

[vincentchan]

\int _1 ^9 \frac{1}{\sqrt[3]{x-9}} dx = \int _1 ^9 \frac{1}{\sqrt[3]{(-1)(9-x)}} dx = \frac{1}{\sqrt[3]{-1}} \int _1 ^9 \frac{1}{\sqrt[3]{9-x}} dx

now, it become a real integral multiply by a complex number..

since \sqrt[3]{-1} = e^{i(2\pi k+\pi)/3} [/tex], k=0,1,2<br /> <br /> \frac{1}{\sqrt[3]{-1}} = e^{-i(2\pi k+\pi)/3}<br /> <br /> \int _1 ^9 \frac{1}{\sqrt[3]{9-x}} dx = 6 as you suggest<br /> <br /> therefore, the answer is :<br /> 6e^{-i(2\pi k+\pi)/3}, k=0,1,2 <br /> or:<br /> 6 (cos(\pi/3)-isin(\pi/3))= 3 ( 1-i\sqrt{3} )<br /> 6 (cos(3\pi/3)-isin(3\pi/3))=-6<br /> 6 (cos(5\pi/3)-isin(5\pi/3))=3 ( 1+i\sqrt{3} )

 

[DivGradCurl]

Oh, I see what you mean. The given integral can still be interpreted as an area (after some rearrangements) that is multiplied by a complex number, making the result complex.

Thanks

 

[dextercioby]

Make the substitution 9-x=u and solve the integral.

Daniel.