calculus
posted by Liz .
Find the area of region bounded by the curves y=sin(pi/2*x)and y=x^22x.

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 xintercepts of 0, 2, and 4
the parabola y  x^2  2x has xintercepts 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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