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Math Help - u-substitution

  1. #1
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    u-substitution

     \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 )<br />
\mbox{answer I get is 2080/3 \ is this correct?}

    also - is there a way to write limits on the integral line?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by weezie23 View Post
     \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 )<br />
\mbox{answer I get is 2080/3 \ is this correct?}
    nope, sorry. show your work

    also - is there a way to write limits on the integral line?
    yes, type \int_{lower limit goes here}^{upper limit goes here} to get the limits on the integral sign

    so you would type [tex]\int_{-1}^{1}(3x + 1)^5~dx[/tex] to get \int_{-1}^{1}(3x + 1)^5~dx
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  3. #3
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    Quote Originally Posted by weezie23 View Post
     \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 )<br />
\mbox{answer I get is 2080/3 \ is this correct?}

    also - is there a way to write limits on the integral line?
    To write subscript, use "_{script}", or, if the script is only one character long, "_s". To write superscript, use "^{script}", or "^s" for a single character. Please note that superscript and subscript can be written on top of each other, and in any place. For example, we have:

    n_0^1

    ...yields...

    n_0^1

    So, for an integral, we just make the integral sign and then the limits...

    \int_0^5dx

    \int_0^5dx

    For super/subscript longer than one character:

    \int_{genesis}^{deuteronomy}inerrancy\;d(pentateuc h)=mosaic\;law

    \int_{genesis}^{deuteronomy}inerrancy\;d(pentateuc  h)=mosaic\;law
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  4. #4
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    Quote Originally Posted by weezie23 View Post
     \int (3x+1)^5\ dx \ \mbox{limits - lower} (-1), \mbox{upper -} \ (1 )<br />
\mbox{answer I get is 2080/3 \ is this correct?}

    also - is there a way to write limits on the integral line?
    \displaystyle\int^{1}_{-1} (3x+1)^5  \ \mathrm{d}x

    Subsitute u=3x+1.

    This integral works out nicely and you should get a whole number as the answer when you apply the limits.
    Last edited by Simplicity; August 27th 2008 at 02:21 AM.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    nope, sorry. show your work

    yes, type \int_{lower limit goes here}^{upper limit goes here} to get the limits on the integral sign

    so you would type [tex]\int_{-1}^{1}(3x + 1)^5~dx[/tex] to get \int_{-1}^{1}(3x + 1)^5~dx

    \int_{-1}^{1}(3x + 1)^5~dx u=(3x+1) du=(3dx)

    Lower limits x=-1 (3(-1)+1)=-2 Upper limits x=1 u=(3(1)+1)=4

    \frac{1}{3} \int_{-2}^{4} u^5 du

    I get 4,842,434
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  6. #6
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    Quote Originally Posted by weezie23 View Post
    \int_{-1}^{1}(3x + 1)^5~dx u=(3x+1) du=(3dx)

    Lower limits x=-1 (3(-1)+1)=-2 Upper limits x=1 u=(3(1)+1)=4

    \frac{1}{3} \int_{-2}^{4} u^5 du

    I get 4,842,434
    No, it's incorrect.

    \frac{1}{3} \int_{-2}^{4} u^5 \ \mathrm{d}u

    =\frac13 \left[ \frac{u^6}{6}\right]_{-2}^{4}
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