Find the area of the region bounded by the curves y = sin (π x/2), y = x²-2x. I'm stuck with the point of intersection: sin (π x/2) = x²-2x.
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Originally Posted by sun1 Find the area of the region bounded by the curves y = sin (π x/2), y = x²-2x. I'm stuck with the point of intersection: sin (π x/2) = x²-2x. $\displaystyle x=0, 2$.
Last edited by mr fantastic; Sep 19th 2009 at 12:04 AM. Reason: Restored original reply
Originally Posted by luobo $\displaystyle x=0, 2$. Thanks, but how did you get to this solution?
luobo did not solve $\displaystyle \sin(\frac{\pi}{2}x) = x^2-2x$ The fact that both these functions have common zeros at 0 and 2 was used. You should consider the fact that $\displaystyle \sin(\frac{\pi}{2}x) $ has period of 4 to help you understand
Originally Posted by pickslides luobo did not solve $\displaystyle \sin(\frac{\pi}{2}x) = x^2-2x$ The fact that both these functions have common zeros at 0 and 2 was used. You should consider the fact that $\displaystyle \sin(\frac{\pi}{2}x) $ has period of 4 to help you understand Right.
Last edited by mr fantastic; Sep 19th 2009 at 12:05 AM. Reason: Restored original reply
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