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Evaluate the definite integral

  1. Apr 27, 2013 #1
    1. [itex]\int^{a}_{0} x\sqrt{x^{2}+a^{2}}[/itex] a > 0



    2. u = x[itex]^{2}[/itex] + a[itex]^{2}[/itex], du = 2x



    3. [itex]\frac{1}{2}\int^{a}_{0}\frac{2u^{3/2}}{3} = \frac{1}{3}\int ^{a}_{0}(x^{2}+a^{2})^{3/2}[/itex] How do I solve this? Any hints?
     
    Last edited by a moderator: Apr 27, 2013
  2. jcsd
  3. Apr 27, 2013 #2

    Mark44

    Staff: Mentor

    You have omitted dx. You might think this to be an unimportant appendage that can be ignored, but as you learn additional integration techniques, it becomes very important.

    In your ordinary substitution, u = x2 + a2, and du = 2xdx.

    Making the substitution, you get
    $$ \frac{1}{2}\int_0^a u^{1/2}du = \frac{2}{2 \cdot 3}\left. u^{3/2} \right|_0^a$$
    $$ =\frac{1}{3} \left. (\sqrt{x^2 + a^2})^3 \right|_0^a$$

    Evaluate the quantity at x = a, and then at x = 0, and subtract the two values.
     
  4. Apr 27, 2013 #3
    So we let a = 1 and 1+1 = 2 and it would be sqrt 2^3 which would be 2 sqrt 2 - (0+1)^3 = 1 all to a^3 * 1/3.
     
  5. Apr 27, 2013 #4

    Mark44

    Staff: Mentor

    No, you don't assign a value to a.

    Instead of writing stuff like "which would be 2 sqrt 2 - (0+1)^3 = 1 all to a^3 * 1/3" that makes little sense, try writing some actual mathematics, with = between equal expressions.
     
  6. Apr 27, 2013 #5
    So [itex]\left|x\right| = a[/itex] and -a.
     
  7. Apr 27, 2013 #6
    Is there an equation I can use to compute the values of x?
     
  8. Apr 27, 2013 #7
    I got it when a is evaluated x = +/- 1 and since x=a +/- 1^2 + +/- 1^2 = 2 and when the lower limit is evaluated x = 0, but a > 0 so that makes the expression equal to 1.
     
  9. Apr 27, 2013 #8

    SammyS

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    "a" is simply "a", an arbitrary positive real number. Your answer should have "a" in it.
     
  10. Apr 27, 2013 #9
    I have [itex]\frac{1}{3}(2\sqrt{2}-1)a^{3}[/itex]
     
  11. Apr 27, 2013 #10

    SammyS

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    Looks good.
     
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