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Math Help - Integrals: #1 Help with fraction #2 Moment of inertia

  1. #1
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    Integrals: #1 Help with fraction #2 Moment of inertia

    Problems:
    25-2-EX9
    The time rate of change of the displacement (velocity) of a robot arm is ds/dt = 8t/(t^2 + 4)^2. Find the expression for the displacement as a function of time if s = -1 m when t = 0 s.
    26-5-9
    Find the moment of inertia of a plate covering the first-quadrant region bounded by y^2 = x, x = 9 and the x-axis with respect to x-axis.

    Attempt
    25-2-EX9
    This is a book example, so the solution is here. I am hung up on one step though.
    ds = (int)/(8tdt)((t^2 + 4)^2) = 4 (int) (t^2 + 4)^-2(2t dt)

    This next step throws me off. What happened to (2t dt)??

    it just disappears. Can someone explain? Thanks.

    s = 4(1)(-1))(t^2 + 4)^-1 + C

    26-5-9
    Ix =k (int) y^2(x2 - x1)dy

    9ky^2 dx = 9kx = 9k1/2x^2

    4.5x^2 = 4.5 * 3^2

    4.5*9 = 40.5

    However, the answer is 162/5k = 32.4

    so where'd I go wrong? thanks.
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  2. #2
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    Re: Integrals: #1 Help with fraction #2 Moment of inertia

    Quote Originally Posted by togo View Post
    ...

    Attempt
    25-2-EX9
    This is a book example, so the solution is here. I am hung up on one step though.
    ds = (int)/(8tdt)((t^2 + 4)^2) = 4 (int) (t^2 + 4)^-2(2t dt)

    This next step throws me off. What happened to (2t dt)??

    it just disappears. Can someone explain? Thanks.

    ...
    If you have
    \int(t^2+4)^{-2} \cdot 2t dt
    Then use integration by substitution:

    u = t^2+4~\implies~\frac{du}{dt} = 2t~\implies~du = 2t dt

    Now your integral becomes:
    \int u^{-2} du = -u^{-1}+c
    that means:
    \int(t^2+4)^{-2} \cdot 2t dt = -(t^2+4)^{-1}+c
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  3. #3
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    Re: Integrals: #1 Help with fraction #2 Moment of inertia

    any thoughts on question #2?
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