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Math Help - [SOLVED] Integrals with fractional powers

  1. #1
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    Cool [SOLVED] Integrals with fractional powers

    In integral questions including exponents of variable (x^5) we just collect like terms and add them. But what if there is no variables involved how would you solve the expression.

    Example how would you solve the following:

    a) 2 - 3 / 2 + 2 - 1 / 2 + 2 1 / 2 + 2 3 / 2


    (2 raise to power negative 3 over/divide by 2) plus ( 2 raise to power negative 1 over/divide by 2) plus ( 2 raise to power 1 over/divide by 2) plus (2 raise to power 3 over/divide by 2)


    Thanks!
    Last edited by unstopabl3; September 26th 2009 at 03:59 AM.
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  2. #2
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    Hello unstopabl3
    Quote Originally Posted by unstopabl3 View Post
    In integral questions including exponents of variable (x^5) we just collect like terms and add them. But what if there is no variables involved how would you solve the expression.

    Example how would you solve the following:

    a) 2 - 3 / 2 + 2 - 1 / 2 + 2 1 / 2 + 2 3 / 2


    (2 raise to power negative 3 over/divide by 2) plus ( 2 raise to power negative 1 over/divide by 2) plus ( 2 raise to power 1 over/divide by 2) plus (2 raise to power 3 over/divide by 2)


    Thanks!
    There are no hard and fast rules here, but one thing always to look out for is: Can I factorise the expression?

    All of the terms in 2^{-\frac32}+2^{-\frac12}+2^{\frac12} + 2^{\frac32} have (essentially) a 2^{\frac12} in common. So look at what happens if we 'split off' 2^{\frac12} from each term and use the rule of indices x^a\times x^b=x^{a+b}:

    2^{-\frac32}+2^{-\frac12}+2^{\frac12} + 2^{\frac32}=(2^{\frac12}\times2^{-2})+(2^{\frac12}\times2^{-1})+(2^{\frac12}\times1) + (2^{\frac12}\times2^{1})

    We now have a common factor 2^{\frac12}, which we can take outside a bracket:

     =2^{\frac12}(2^{-2}+2^{-1}+1 +2^{1})

    = 2^{\frac12}(\tfrac14+\tfrac12+1+2)

    =2^\frac12(3\tfrac34)

    = \frac{15\sqrt2}{4}

    Grandad
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  3. #3
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    Cool

    I don't want to crap your awesome post but I am so very happy that I have to say this ... after I made the post, I started trying different variations on my own and I finally figured it out.

    But your post has explained it completely

    Thanks a bunch!
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    Another question in similar category is bugging me. I've almost got it but can't get the final answer right.



    Thanks for your help in advance!
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  5. #5
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    Hello unstopabl3
    Quote Originally Posted by unstopabl3 View Post
    Another question in similar category is bugging me. I've almost got it but can't get the final answer right.



    Thanks for your help in advance!
    Writing \sqrt3 as 3^{\tfrac12}, we get

    \Big(3^{\tfrac12}\Big)^{-3}+\Big(3^{\tfrac12}\Big)^{-2}+\Big(3^{\tfrac12}\Big)^{-1}+\Big(3^{\tfrac12}\Big)^{0}+\Big(3^{\tfrac12}\Bi  g)^{1}+\Big(3^{\tfrac12}\Big)^{2}+\Big(3^{\tfrac12  }\Big)^{3}

    =3^{-\tfrac32}+3^{-1}+3^{-\tfrac12}+3^0+3^{\tfrac12}+3^1+3^{\tfrac32}

    =3^{-\tfrac32}+3^{-\tfrac12}+3^{\tfrac12}+3^{\tfrac32}+3^{-1}+3^0+3^1

    =3^{\tfrac12}(3^{-2}+3^{-1}+1+3^{1})+\tfrac13+1+3

    =3^{\tfrac12}(\tfrac19+\tfrac13+1+3) + 4\tfrac13

    =3^{\tfrac12}(4\tfrac49) + 4\tfrac13

    = \frac{40\sqrt3}{9}+\frac{13}{3}

    =\frac{40\sqrt3+39}{9}

    Grandad
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    Cool

    Big thanks! You are awesome, are you a teacher by any chance?
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  7. #7
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    Quote Originally Posted by unstopabl3 View Post
    Big thanks! You are awesome, are you a teacher by any chance?
    Used to be - retired now. Busier than ever!

    Grandad
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    Cool

    I thought so How are you busier than ever when you have retired ? Hehe just curious.

    Thanks again sir
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