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Antiderivative and Intergrals

  1. Apr 3, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex] (X-1) [tex]\sqrt{X}[/tex] dx

    3. The attempt at a solution
    It is multiple choice, I believe the answer is either (2/5)x^(5/2)-(2-3)x^(3/2)+c or

    I have tried to find the derivatives of both these answers yet neither of them gave me the correct antidrivatives. I must be doing something wrong. To do it the right way I have done the u's but I just can't figure it out.
  2. jcsd
  3. Apr 3, 2008 #2
    [tex]\int(x-1)\sqrt xdx[/tex]

    What was your first step?

    Hint: Distribute the [tex]\sqrt x[/tex]
  4. Apr 3, 2008 #3
    I distributed the [tex]\sqrt{x}[/tex] and replaced with u's I then got

    [tex]\int[/tex] u-u^(1/2) This lets see where the -2/3x^(3/2) comes from but I still don't understand where the 2/5x^(5/2) comes from.
  5. Apr 3, 2008 #4
    You don't need to replace with u's. What are the rules for multiplying variables with the same base with exponents?
  6. Apr 3, 2008 #5
    Thank you, I forgot how to distribute for a second.
  7. Apr 4, 2008 #6
    This seems a fairly straightforward case of multiplying out the brackets and solving using the sum rule of integrals:

    [tex]\int \left(f \pm g\right) \,dx = \int f \,dx \pm \int g \,dx\rightarrow[/tex]

    [tex]\int (x-1)\sqrt{x}\,dx\rightarrow \int x^{\frac{3}{2}}-x^{\frac{1}{2}}\,dx=\frac{2}{5}x^{\frac{5}{2}}-\frac{2}{3}x^{\frac{3}{2}}+c[/tex]

    No need to use the u unless the question asks you to? Or am I missing something here?
    Last edited: Apr 4, 2008
  8. Apr 4, 2008 #7

    Gib Z

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    Homework Helper

    I'm quite surprised they have multiple-choice anti derivative questions! I mean, if one doesn't really know how to integrate it they can just differentiate every option and see which one matches.
  9. Apr 4, 2008 #8
    It sounds like calculus for dummies. :smile: I've never heard of multiple choice exams either? Not in A' Level or anywhere else?
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