# Integral / Complex Nos.

1. Feb 17, 2005

### 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.

Last edited: Feb 17, 2005
2. Feb 17, 2005

### 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 $$\frac{1}{\sqrt[3]{-1}} = e^{-i(2\pi k+\pi)/3}$$ $$\int _1 ^9 \frac{1}{\sqrt[3]{9-x}} dx = 6$$ as you suggest therefore, the answer is : $$6e^{-i(2\pi k+\pi)/3}$$, k=0,1,2 or: $$6 (cos(\pi/3)-isin(\pi/3))= 3 ( 1-i\sqrt{3} )$$ $$6 (cos(3\pi/3)-isin(3\pi/3))=-6$$ $$6 (cos(5\pi/3)-isin(5\pi/3))=3 ( 1+i\sqrt{3} )$$ 3. Feb 17, 2005 ### 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 4. Feb 17, 2005 ### dextercioby Make the substitution [itex] 9-x=u$ and solve the integral.

Daniel.

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