# calculus

posted by on .

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

• calculus - ,

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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