calculus

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evaluate the following integrals by any means possible:


b. integral from -1 to 1 (w/ (w^2+1)) (dw)


c. integral from -3 to 0 sqrt(9-x^2)dx

  • calculus -

    b) zero because the function is odd.

    c) substitute x = 3 sin(t)

    Integral from -pi/2 to 0 of

    9 cos^2(t) dt

    If you were to replace cos^2 by sin^2 in this integral, the answer would be the same because

    cos(x) = -sin(-pi/2-x) ------->

    cos^2(x) = sin^(-pi/2 - x)

    So, the values attained by cos^2 are also attained by sin^2 in the inteval from -pi/2 to 0. If we call the integral I, then:

    2 I = 9 Integral of
    [cos^2(t) + sin^2(t)] dt from -pi/2 to 0 =

    9* pi/2 --------->

    I = 9/4 pi

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