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Math Help - Check for me?

  1. #1
    Member integral's Avatar
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    Check for me?

    I was just playing with integrals a little (because I really do not understand them fully yet)

    what I got first:
    \int_{0}^{t}xdx=\frac{t^2}{2}
    and what I got next

    \int_{0}^{t}x^2dx=\frac{t^3}{3}

    Which lead me to believe:

    \int_{0}^{t}x^ndx=\frac{t^{n+1}}{n+1}

    edit:
    My last:
    \int_{0}^{t}ax^ndx=\frac{t^{n+1}a}{n+1}
    Last edited by integral; January 30th 2010 at 07:01 AM.
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  2. #2
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    You are correct, good work.

    You can always check your work here:

    Wolfram|Alpha
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  3. #3
    Member integral's Avatar
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    Thanks
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  4. #4
    Member integral's Avatar
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    One more thing:

    If you have
    \int_{0}^{t}ax^n+bx^{n-1}dx
    and you take two individual integrals like this:

    \int_{0}^{t}ax^n=\frac{at^{n+1}}{n+1}
    and
    \int_{0}^{t} bx^z dx=\frac {bt^{z+1}}{{z+1}}
    will the answer be :

    \int_{0}^{t}ax^n+bx^{n-1}dx=\int_{0}^{t}ax^ndx+\int_{0}^{t}bx^{n-1}dx=\frac{at^{n+1}}{n+1}+\frac{bt^n}{n}

    And, hopefully this would be true:
    \int_{0}^{t}ax^n+bx^{n-1}+cx^{n-2} \cdots dx dx=

    \int_{0}^{t}ax^n+\int_{0}^{t}bx^{n-1}+\int_{0}^{t}cx^{n-2} \cdots +\int_{0}^{t}(dx)dx=

    \frac{at^{n+1}}{n+1}+\frac{bt^n}{n}+\frac{ct^n-1} {n-1} \cdots +dt

    right?




    >.> I hope this makes sense because i always get in trouble when mods don't understand my stuff (all through my falt of course... but still)
    and I tried wolffram, I could not get it to work.
    edit:
    (sry for dble posting ...)
    double edit: general.... thanks, does that mean there is something wrong with the equation above?
    Last edited by integral; January 30th 2010 at 11:21 AM.
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  5. #5
    Super Member General's Avatar
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    Quote Originally Posted by integral View Post
    I was just playing with integrals a little (because I really do not understand them fully yet)

    what I got first:
    \int_{0}^{t}xdx=\frac{t^2}{2}
    and what I got next

    \int_{0}^{t}x^2dx=\frac{t^3}{3}

    Which lead me to believe:

    \int_{0}^{t}x^ndx=\frac{t^{n+1}}{n+1}

    edit:
    My last:
    \int_{0}^{t}ax^ndx=\frac{t^{n+1}a}{n+1}
    For the last two, n \neq -1
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  6. #6
    Member integral's Avatar
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    ok so the equation only works if
    n is greater than the number of terms
    or
    if n is less than -1
    ??
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  7. #7
    Super Member General's Avatar
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    Quote Originally Posted by integral View Post
    ok so the equation only works if n>number of terms-2
    No, It works for all n \neq -1.
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  8. #8
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    Quote Originally Posted by integral View Post
    One more thing:

    If you have
    \int_{0}^{t}ax^n+bx^{n-1}dx
    and you take two individual integrals like this:

    \int_{0}^{t}ax^n=\frac{at^{n+1}}{n+1}
    and
    \int_{0}^{t} bx^z dx=\frac {bt^{z+1}}{{z+1}}
    will the answer be :

    \int_{0}^{t}ax^n+bx^{n-1}dx=\int_{0}^{t}ax^ndx+\int_{0}^{t}bx^{n-1}dx=\frac{at^{n+1}}{n+1}+\frac{bt^n}{n}

    And, hopefully this would be true:
    \int_{0}^{t}ax^n+bx^{n-1}+cx^{n-2} \cdots dx dx=

    \int_{0}^{t}ax^n+\int_{0}^{t}bx^{n-1}+\int_{0}^{t}cx^{n-2} \cdots +\int_{0}^{t}(dx)dx=

    \frac{at^{n+1}}{n+1}+\frac{bt^n}{n}+\frac{ct^n-1} {n-1} \cdots +dt

    right?




    >.> I hope this makes sense because i always get in trouble when mods don't understand my stuff (all through my falt of course... but still)
    and I tried wolffram, I could not get it to work.
    edit:
    (sry for dble posting ...)
    double edit: general.... thanks, does that mean there is something wrong with the equation above?
    This is generally correct, but as General remarked, make sure to add the stipulation that the denominator is nonzero. So if the denominator is n+1, we can not have n be -1. Also, when you write the integral that is written after "And hopefully this will be true", there is no reason to write dx all those times. One dx for the entire integral is sufficient, and then when you break the integral up by linearity, then each integral gets its own "dx" so to speak.
    Also, your integral of (dx)dx doesn't really make sense, we can't really make an integral like that. The other parts of it are okay though.
    As for Wolfram not working, I suppose it is because of all the constants you have floating around...it works with ease when you just have one variable, but can be a pain otherwise.
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  9. #9
    Member integral's Avatar
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    Yes I see that I should have placed a dx behind each integral...sorry.
    And the (dx)dx... it was just a continuation ax,bx,cx,dx
    thanks
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