Post a New Question

calculus

posted by .

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

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question