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Math Help - need help with "simple" integration

  1. #1
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    need help with "simple" integration

    Hey everybody, first time using the forum because im taking cal II this semester and Im sure ill need help : )

    I dont know how to integrate when an exponent is in the denominator. For ex:
    4/y^2

    my book does this:

    4/y^2 = 4(-1/y) when integrated and i dont know how they did it, im sure its a combination of some algebraic trick that im just not catching.

    For reference the book is: Thomas Calculus early transcendentals pg. 431 ex. 7

    thanks for any help you can offer
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cknicker View Post
    Hey everybody, first time using the forum because im taking cal II this semester and Im sure ill need help : )

    I dont know how to integrate when an exponent is in the denominator. For ex:
    4/y^2

    my book does this:

    4/y^2 = 4(-1/y) when integrated and i dont know how they did it, im sure its a combination of some algebraic trick that im just not catching.

    For reference the book is: Thomas Calculus early transcendentals pg. 431 ex. 7

    thanks for any help you can offer
    note that \frac{1}{y^a}=y^{-a}.

    So in your case \frac{4}{y^2}=4y^{-2}.

    When you integrate, you apply power rule to get \int 4y^{-2}\,dy=\frac{4y^{-2+1}}{-2+1}+C=-4y^{-1}+C=-\frac{4}{y}+C

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    note that \frac{1}{y^a}=y^{-a}.

    So in your case \frac{4}{y^2}=4y^{-2}.

    When you integrate, you apply power rule to get \int 4y^{-2}\,dy=\frac{4y^{-2+1}}{-2+1}+C=-4y^{-1}+C=-\frac{4}{y}+C

    Does this make sense?

    yes it does now, thank you very much for your help.
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