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Math Help - integrals

  1. #1
    Junior Member
    Joined
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    integrals

    I am having a hard time understanding this...you all do a much better job of explaining things than my textbook...hopefully between both, I can get a better understanding...

    Integral (3square root y) - (2/y^3) + (1/y) dy
    =3y^1/2 - 2y^-3 + y^-1 dy
    =3(y3/2dy) - 2(y^-2dy) + y^0) dy
    =2y^3/2 = 1/y^2 + y + C


    and on this one I am even more lost:
    integral (1/2y - 2/y^2 + 3/square root y)
    = 2y^-1 - 2y^-2 + 3y^-1/2 dy
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  2. #2
    Super Member

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    Hello, Becky!

    You're expected to know that: . \int \frac{1}{u}\,du \;=\;\ln|u| + C


    \int\left(3\sqrt{y} \:-\: \frac{2}{y^3} \:+\: \frac{1}{y}\right)dy

    We have: . \int\left(3y^{\frac{1}{2}} \:-\: 2y^{-3} \:+\: \frac{1}{y}\right)dy

    . . = \;3\!\cdot\!\frac{y^{\frac{3}{2}}}{\frac{3}{2}} \:-\: 2\!\cdot\!\frac{y^{-2}}{-2} \:+\: \ln|y| \:+\: C

    . . = \;2y^{\frac{3}{2}} \:+\: y^{-2} \:+\: \ln|y| \:+\: C



    \int \left(\frac{1}{2y} \:-\: \frac{2}{y^2} \:+\: \frac{3}{\sqrt{y}}\right)dy

    We have: . \int\left(\frac{1}{2}\!\cdot\!\frac{1}{y} \:-\: 2y^{-2} \:+\: 3y^{-\frac{1}{2}}\right)dy

    . . = \;\frac{1}{2}\ln|y| \:-\: 2\!\cdot\!\frac{y^{-1}}{-1} \:+\: 3\!\cdot\!\frac{y^{\frac{1}{2}}}{\frac{1}{2}} \:+\: C

    . . = \;\frac{1}{2}\ln|y| \:+\: 2y^{-1} \:+\: 6y^{\frac{1}{2}} \:+\: C

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