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Special integral

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data

    im taking a course on methods including gamma beta ,bessel functions and related stuff, i met this question in the library on old exam papers for the course, [tex] \int^{3}_{0} \frac{t^3}{\sqrt{3-t}}dt [/tex]




    2. Relevant equations

    im not sure how to do it but did think of the beta function
    [tex] B(m,n)= \int^{1}_{0} t^{m-1}(1-t)^{n-1} dt [/tex]


    3. The attempt at a solution
    [tex] \int^{3}_{0} t^{4-1}(1-t)^{\frac{1}{2} -1}} dt [/tex]
    but the limits are wrong and im probably heading in the wrong direction
     
  2. jcsd
  3. Feb 26, 2010 #2

    vela

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    Good try. You're going in the right direction, but you need to be a little more careful. What happened to the three in the radical?
     
  4. Feb 26, 2010 #3
    thanks vela it should have remained since i did not do anything special to it, let me edit that.
     
  5. Feb 26, 2010 #4
    Correction: this is what the attempt at a solution should be
    [tex] \int^{3}_{0} t^{4-1}(3-t)^{\frac{1}{2} -1}} dt [/tex]
     
  6. Feb 26, 2010 #5
    Now if i make substitution [itex] u= \frac{t}{3} [/itex] at t=3, u=1
    and t=0, u = 0. [tex] du=\frac {dt}{3} [/tex]
    now [tex] \int ^{1}_{0} 3^{4-1} u^{4-1} (1-u)^{\frac{1}{2}-1} 3 du [/tex]
    which is, [itex] 3^4 B(4, \frac{1}{2}) [/itex]
    im i ok?
    And finaly [tex]
    = 3^4 \frac{gamma (4) gamma(\frac{1}{2} ) } {gamma(4+\frac{1}{2} ) } [/tex]
     
  7. Feb 26, 2010 #6
    Oops i fogot to take 1/sqrt(3) out when i divided throug the stuff in the radical by 3
    correction
    Now if i make substitution [itex] u= \frac{t}{3} [/itex] at t=3, u=1
    and t=0, u = 0. [tex] du=\frac {dt}{3} [/tex]
    now [tex] \frac{1}{\sqrt{3}} \int ^{1}_{0} 3^{4-1} u^{4-1} (1-u)^{\frac{1}{2}-1} 3 du [/tex]
    which is, [itex]\frac{1}{\sqrt{3}} 3^4 B(4, \frac{1}{2}) [/itex]
    im i ok?
    And finaly [tex]
    = \frac{1}{\sqrt{3}} 3^4 \frac{gamma (4) gamma(\frac{1}{2} ) } {gamma(4+\frac{1}{2} ) } [/tex]
     
  8. Feb 26, 2010 #7

    Dick

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    You're ok. You can also evaluate it without the beta functions. Substitute u=3-t and expand t^3 in terms of u. You'll get the same thing.
     
  9. Feb 26, 2010 #8
    thanks, it would work(i tried it) but since its a course on special methods, odinary integral calculus would not earn a mark.
     
  10. Feb 26, 2010 #9

    Dick

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    Sure. But it's an easy way to check your solution.
     
  11. Feb 26, 2010 #10
    :blushing: thanks, il keep that in mind.
     
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