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.

Printable View

- Aug 13th 2009, 04:44 PMsun1Find the area of the region bounded by the curves...
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.

- Aug 13th 2009, 04:52 PMluobo
- Aug 13th 2009, 04:58 PMsun1
- Aug 13th 2009, 05:17 PMpickslides
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 - Aug 13th 2009, 05:27 PMluobo