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

1. ## 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.

2. 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.

$\displaystyle x=0, 2$.

3. Originally Posted by luobo $\displaystyle x=0, 2$.
Thanks, but how did you get to this solution?

4. 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

5. 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.

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