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Integration problem

  1. Oct 2, 2004 #1
    Hello everybody
    I'm working on a section for integration by substitution and I came across an integral that I don't know how to do a substitution for

    [tex]\int_{1}^{2} x\sqrt{x-1} \,dx [/tex]

    How can I do this problem?
     
  2. jcsd
  3. Oct 2, 2004 #2
    This isn't simple substitution. You can multiply the equation fancily by 1 and then use substitution. Tell me if this is too vague.

    Edit: It seems someone else found it just by substitution. Oh well, I like my way better :p
     
    Last edited by a moderator: Oct 2, 2004
  4. Oct 2, 2004 #3

    Clausius2

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    [tex] t=\sqrt{x-1}[/tex]
    [tex] x=t^2+1[/tex]
    [tex] dx=2tdt[/tex]

    Your integral is:

    [tex]\int_{make it you}^{make it you} (t^2+1)t^2 2dt[/tex]
     
  5. Oct 2, 2004 #4

    arildno

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    Another choice is:
    [tex]\int_{1}^{2}x\sqrt{x-1}dx=\int_{1}^{2}(x-1)^{\frac{3}{2}}dx+\int_{1}^{2}\sqrt{x-1}dx[/tex]
     
  6. Oct 2, 2004 #5
    Thanks, everybody. Would you all care to check my work please?

    [itex]\int_{1}^{2} x\sqrt{x-1} \,dx [/itex]
    [itex]u = \sqrt{x-1} [/itex]
    [itex]x = u^2 + 1 [/itex]
    [itex]\,du = \frac{1}{2\sqrt{x-1}}\,dx [/itex]
    [itex]\,dx = 2\sqrt{x-1} \,du = 2u\,du [/itex]
    [itex]\int_{1}^{2} x\sqrt{x-1} \,dx = \int_{0}^{1} (u^2 + 1)u 2u\,du = \int_{0}^{1} 2u^4 + 2u^2 \,du = 2\int_{0}^{1} u^4 \,du + 2\int_{0}^{1} u^2 \,du = 2 + 2 = 4[/itex]

    Is this correct?
     
  7. Oct 2, 2004 #6

    arildno

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    Do the following integrals once more:
    [tex]\int_{0}^{1}u^{4}du,\int_{0}^{1}u^{2}du[/tex]
     
  8. Oct 2, 2004 #7
    Oh, whoops. Thanks for pointing that out.

    [tex]\int u^4 \,du = \frac{u^5}{5} , \int u^2 \,du = \frac{u^3}{3} [/tex]
    So then it becomes
    [tex]\frac{2}{5} + \frac{2}{3} = \frac{16}{15}[/tex]
     
  9. Oct 2, 2004 #8

    arildno

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    Seems much better! :biggrin:
     
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