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Math Help - Intergration Question (fairly basic)

  1. #1
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    Intergration Question (fairly basic)

    \int(e^x - 3)^2 dx

    I think it goes as following, here is my working.

    \int (e^{2x} - 6e^x + 9) dx

    = \frac{e^{2x}}{2} - e^x + 9x


    Can anyone confirm or tell me where I've gone wrong? Thank you very much.
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  2. #2
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    Hi.

    Quote Originally Posted by Peleus View Post
    \int(e^x - 3)^2 dx

    I think it goes as following, here is my working.

    \int (e^{2x} - 6e^x + 9) dx

    = \frac{e^{2x}}{2} - e^x + 9x


    Can anyone confirm or tell me where I've gone wrong? Thank you very much.
    Yea, the work seems to be correct, but you made a mistake

    \int - 6e^x dx \not= -e^x

    But I guess you know that

    And maybe you should add a constant to your solution

    = \frac{e^{2x}}{2} - 6e^x + 9x + const

    Yours
    Rapha
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  3. #3
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    Yep, sorry missed that.

    \int 6e^x

    which turns into

    6 \int \frac{1}{6}e^x

    6 * e^x

    6e^x

    Thanks, forgot to type in the const as well.
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  4. #4
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    Quote Originally Posted by Peleus View Post
    Yep, sorry missed that.

    \int 6e^x

    which turns into

    6 \int \frac{1}{6}e^x
    No!!!!

    Quote Originally Posted by Peleus View Post
    6 * e^x
    Yes

    Quote Originally Posted by Peleus View Post
    6e^x
    Yes

    Actually it is - \int 6e^x dx = -6*\int e^xdx = -6*e^x

    Quote Originally Posted by Peleus View Post
    Thanks
    You're welcome
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  5. #5
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    I have absolutely no idea what I was thinking there, even the working doesn't work.

    Yes, you're right and I understand how you got it

    Cheers.
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