Homework Help: Integral / Complex Nos.

1. Feb 17, 2005

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

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..

Daniel.