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Calculus

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1. Find the area of the region between the curves y=sin(x pi/2) and y=x.

2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.

  • Calculus - ,

    I will do the harder of the two
    2.

    If you make a sketch you will see that the curves intersect in your domain 0 ≤ x ≤ π/2

    sin2x = sinx
    2sinxcosx - sinx = 0
    sinx(2cosx - 1) = 0
    sinx = 0 or cosx = 1/2
    x = 0 , your left domain, or
    x = π/3

    so we have to do this in two parts

    area = ∫(sin2x - sinx) dx from 0 to π/3 + ∫(sinx - sin2x) dx from π/3 to π/2

    = [(-1/2)cos2x + cosx] form 0 to π/3 + [-cosx + 1/2)cos2x ] from π/3 to π/2

    = (1/4 + 1/2 - (-1/2) + 1)) + (0 + 0 -(-1/2 +(1/2)(-1/2))
    = .... you do the arithmetic
    and please check my arithmetic above, should have written it out on paper first.

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