Find the area of the region bounded by the given curves. y = sin2(x), y = sin3(x), 0 ≤ x ≤ π?

1 Answer

  • Does that mean sin^2(x) and sin^3(x) ??

    Or does it mean sin(2x) and sin(3x) ?

    In either case, your notation is not very good.

    I’m going to assume sin^2(x) and sin^3(x), because the intersection problem is then very simple. The functions sin^2(x) and sin^3(x) intersect at x = pi/2, but on both sides of that intersection, sin^2(x) > sin^3(x), so you’d want the integral of

    [sin^2(x) – sin^3(x)] dx

    from x = 0 to x = pi.

    The indefinite integral of sin^2(x)dx is the integral of

    [1 – cos(2x)] dx/2

    = x/2 – (1/4)sin(2x);

    from 0 to pi, it’s pi/2.

    The indefinite integral of sin^3(x)dx is the integral of

    [sin(x)]*[1 – cos^2(x)] dx

    = -cos(x) + (1/3)cos^3(x).

    From 0 to pi, it’s

    -(-1) + (1/3)(-1) – (-1) – (1/3)(1) = 4/3.

    Putting this all together, you get pi/2 – 4/3

    which is about 0.237.

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