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Math Help - Antiderivative via FTC and Intergration by parts

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    Antiderivative via FTC and Intergration by parts

    Could anyone please help me with this?

    Define the function g(x) = anti derivative of t^2 * cos(t) dt
    top value = 1/x^2 bottom value = 0

    a) solve using FTC
    b) solve using intergration by parts
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  2. #2
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    Quote Originally Posted by Lupus7874 View Post
    Could anyone please help me with this?

    Define the function g(x) = anti derivative of t^2 * cos(t) dt
    top value = 1/x^2 bottom value = 0

    a) solve using FTC
    b) solve using intergration by parts
    a) You want \frac{dg}{dx} = \frac{d}{dx} \int_0^{1/x^2} t^2 \cos t \, dt. Let  u = \frac{1}{x^2}. Then from the chain rule:

    \frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du} \int_0^{u} t^2 \cos t \, dt \cdot \frac{du}{dx}

    and you should be able to do the necessary calculations.

    b) Use integration by parts twice. It should be clear what choices to make (Hint: You want to end up integrating only a trig function).
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    I'm rather confused on the whole methodology of this problem, any chance someone could walk me through it?

    I mean assuming I substitute what you said, I'm still lost on how to take the antiderivative of a product
    Last edited by Lupus7874; December 7th 2009 at 06:04 AM. Reason: Insufficient Data
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    Quote Originally Posted by Lupus7874 View Post
    I'm rather confused on the whole methodology of this problem, any chance someone could walk me through it?

    I mean assuming I substitute what you said, I'm still lost on how to take the antiderivative of a product
    I told you exactly how to get the solution: \frac{dg}{dx} = \frac{dg}{du} \cdot \frac{du}{dx} = \frac{d}{du} \int_0^{u} t^2 \cos t \, dt \cdot \frac{du}{dx}.

    \frac{d}{du} \int_0^{u} t^2 \cos t \, dt is found using the Fundamental Theorem of Calculus (then substitute back u = \frac{1}{x^2}).

    Getting \frac{du}{dx} from u = \frac{1}{x^2} should be trivial at the level you're apparently studying.

    I don't see where the trouble can be in this ....
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