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Homework Help: Integration questions

  1. Oct 23, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm wondering if anyone can check these integrations for me, or suggest alternative answers if they're not quite right, or can be simplified?

    1) [tex]\int x^{\sqrt{2}}[/tex]

    2) [tex]\int x . \sqrt{x}[/tex]

    3) [tex]\int \frac{1}{x^\pi}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    1) [tex]\int x^{1.4} = \frac{x^{2.4}}{2.4} + C [/tex]

    2) [tex]\int x . x^{\frac{1}{2}} = \frac{x^2}{2} . \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{x^{\frac{3}{2}}}{3} . x^2 + C [/tex]

    3) [tex]\int \frac{1}{x^\pi} = \int x^{-\pi} = \frac{x^{-\pi+1}}{-\pi+1} + C [/tex]
  2. jcsd
  3. Oct 23, 2008 #2
    The first and the third look okay to me, although in the first one I would leave the [tex]\sqrt{2}[/tex] rather than putting 2.4.

    In the second your almost there. When doing this sort of integral it is often easier if you simplify the integrand. By combining [tex]x*\sqrt{x}[/tex] into [tex]\sqrt{x*x^2}[/tex] to get [tex]{x^{3/2}}[/tex] it should be easier to integrate.
  4. Oct 23, 2008 #3


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    [itex]\sqrt{2}[/itex] is NOT equal to 1.4! There is no reason to change [itex]x^{\sqrt{2}}[/itex] to [itex]x^{1.4}[/itex].

    An unfortunately common mistake: just as you cannot differentiate a product by just differentiating each part, you cannot integrate a product that way either. As Vuldoraq said, [itex]x\sqrt{x}= x(x^{1/2})= x^{3/2}[/itex]. Integrate that.

    Yes, this is correct.
  5. Oct 23, 2008 #4
    Thanks for the replies Vuldoraq, HallsofIvy.

    So, my answer should be [tex]\int x^{\frac{3}{2}} = \frac{2x^{\frac{3}{2}}}{3} [/tex]
  6. Oct 23, 2008 #5
    You forgot to increase the exponent by one before dividing!
  7. Oct 23, 2008 #6
    Are you aware of the fact that you're leaving out the dx in the integrals?
  8. Oct 23, 2008 #7

    Second attempt.

    [tex] \int x^{\frac{3}{2}} = \frac{2x^{\frac{5}{2}}}{5} [/tex]
  9. Oct 23, 2008 #8
    Thats it, much better. :smile:

    To repeat what HallsofIvy said: you can't integrate products of the function being integrated in the usual manner. So you have to simplify and when you can't simplify you have to use a different method (like integration by parts or substitution, you'll come accross these later on).

    May your Math prosper
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