# Find the area of the region bounded by the curves...

• Aug 13th 2009, 04:44 PM
sun1
Find 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 PM
luobo
Quote:

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.

$x=0, 2$.
• Aug 13th 2009, 04:58 PM
sun1
Quote:

Originally Posted by luobo
$x=0, 2$.

Thanks, but how did you get to this solution?
• Aug 13th 2009, 05:17 PM
pickslides
luobo did not solve $\sin(\frac{\pi}{2}x) = x^2-2x$

The fact that both these functions have common zeros at 0 and 2 was used.

You should consider the fact that $\sin(\frac{\pi}{2}x)$ has period of 4 to help you understand
• Aug 13th 2009, 05:27 PM
luobo
Quote:

Originally Posted by pickslides
luobo did not solve $\sin(\frac{\pi}{2}x) = x^2-2x$

The fact that both these functions have common zeros at 0 and 2 was used.

You should consider the fact that $\sin(\frac{\pi}{2}x)$ has period of 4 to help you understand

Right.