Posted by **Liz** on Wednesday, January 30, 2013 at 7:21pm.

Find the area of region bounded by the curves y=sin(pi/2*x)and y=x^2-2x.

- calculus -
**Reiny**, Wednesday, January 30, 2013 at 8:20pm
At first it might look difficult to find the intersection by setting

sin(πx/2) = x^2 - 2x , but a quick sketch makes it quite easy.

the period of the sin curve is 2π/(π/2) = 4

making x-intercepts of 0, 2, and 4

the parabola y - x^2 - 2x has x-intercepts of

0 and 2

How convenient

make a rough sketch

Area = ∫(upper y - lower y) dx from 0 to 2

=∫(sin (πx/2) - x^2 + 2x) dx from 0 to 2

= [ (-2/π)cos (πx/2) - x^3/3 + x^2] from 0 to 2

= (-2/π)(-1) - 8/3 + 4 - ( (-2/π)(1) - 0 + 0)

= 2/π - 8/3 + 4 + 2/π

= 4/3

check my arithmetic

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