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Integral 'Riddles'

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that ∫g(x) dx = ∫g(x^2) 2xdx with a=0 and b=1

    The followup question asks me to prove that the above integrals are NOT equal, by providing two functions that disprove it (one where the former integral is larger, and one where the latter integral is larger).

    The last question with this set asks which of the previous two integrals is larger if g(x) is increasing on (0,1)?

    2. Relevant equations



    3. The attempt at a solution

    I used g(x)= x and I got u= x^2
    du= 2x dx

    1 | x^2/2
    0 |

    1 | u^2/2
    0 | x^4/2

    1/2 = 1/2

    So for the former integral being larger, my example is g(x)= 3x + 6. For the latter integral being larger, my example is g(x)= -(x^2) + 4.

    For the last question in the set, I have no clue what to do, because only 3x+ 6 is increasing on that interval. In fact, -(x^2) + 4 is decreasing, so I'm stumped. Help please? Thank you.
     
  2. jcsd
  3. May 28, 2013 #2

    Mentz114

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    The question seems incomplete - where do a and b appear ? are they the limits of integration ?
     
  4. May 28, 2013 #3
    Yes they are the limits of integration. Sorry about that!
     
  5. May 28, 2013 #4

    LCKurtz

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    Do you mean ##\int_0^1g(x)\, dx = \int_0^1 g(x^2)2x\, dx##?

    So you are asked to prove it is true and then also asked to give counterexamples to its truth??? I suggest you need to quote the exact and complete statement of the problem.
     
  6. May 28, 2013 #5
    Yes that's what the second question asks.

    Show that the integral of g(x) dx is not equal to the integral of g(x^2) dx when a =0 and b=1 by finding two examples, one where the former integral is larger, and the other where the latter integral is larger. Give explicit formulae for your functions.
     
  7. May 28, 2013 #6

    Mentz114

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    No problem. This a bit better
    abg(x) dx = ∫abg(x^2) 2x dx

    I still don't get how you can be asked to prove the equality, and then to *disprove* it.
     
  8. May 28, 2013 #7
    Well they are two parts to the same question, so I guess that's what it is.
     
  9. May 28, 2013 #8

    LCKurtz

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    Are you sure you aren't supposed to show it is true when a = 0 and b = 1 but false for other values of a and b? Like I asked earlier, please give a word-for-word statement of the problem. Otherwise you are just wasting our time.
     
  10. May 28, 2013 #9
    This is the EXACT wording of the question. Sorry if I'm wasting your time...

    And no it is not asking for me to change the upper and lower limits of the integral, but it's asking me to select different functions altogether.
     
  11. May 28, 2013 #10
    $$\int_0^1 g(x) \, dx = \int_0^1 g(x^2) 2x \, dx = \int_0^1 g(u) \frac{du}{dx} \, dx = \int_0^1 g(u) \, du$$

    What went wrong with that? :confused:
     
  12. May 28, 2013 #11
    The bolded part is very important. It is not the same integral as in the original question.
     
  13. May 28, 2013 #12
    Yes slider142. That's exactly what made me rethink if I had even done this correctly. Thank you for bolding that.
     
  14. May 28, 2013 #13

    HallsofIvy

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    slider142's point is that first problem asks you to show that
    [itex]\int_0^1 g(x)dx= \int_0^1 g(x^2) 2x dx[/itex]
    while the second asks you to give examples to show that [itex]\int_0^1 g(x)dx[/itex] is NOT the same as [itex]\int_0^1 g(x^2)dx[/itex].

    Do you see the difference? There is no "2x" in the second.
     
  15. May 28, 2013 #14

    LCKurtz

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    And that's why I asked for the exact statement of the problem. You didn't have it stated correctly.
     
  16. May 28, 2013 #15
    I realized that the problem is stated that way LCKurtz, and that everyone wondered why it was so, which is why I retyped the whole thing out to clarify that that is exactly how it was worded. I solved it despite that, and I'm almost 1000% certain it's wrong, which is why I decided to ask the experts on here.
     
  17. May 28, 2013 #16

    haruspex

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    Not sure I understand what you're saying there. With the exact wording, the question makes sense. What is it that you are almost a thousand percent sure is wrong?
     
  18. May 28, 2013 #17

    LCKurtz

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    That the problem was stated what way?

    This confuses me too. I can't tell if you understand yet that your original post did not state the problem correctly.
     
  19. May 29, 2013 #18
    I was a 1000% percent sure that MY solution to the problem is incorrect. The problem was stated in the way slider142 presented it: that the integral in the second part of the problem is NOT the same as the integral in the first part of the problem. This is where I'm stuck. That's why I wanted to ask all of you experts, if my solution is right or not.
     
  20. May 29, 2013 #19

    LCKurtz

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    Looking at your solution above, for part a you have shown that it works for ##g(x) = x##. But you were to show it for all ##g(x)##. So no, your solution for that one isn't correct.

    For your solutions to one or the other being larger, instead of just giving answers, show us what you did. I don't know whether they are right or not without working them myself. You shouldn't expect us to work them out for ourselves to see if we agree. Just show us what you did.
     
  21. May 29, 2013 #20
    Okay, so I would have to choose another function for part a, and show the work for that.

    For part B,

    g(x) = 3x + 6
    integral g(x) = 3/2x^2 + 6x (dx)
    integral g(1)= 7.5

    integral g(x^2)= x^3 + 6x
    integral g(1^2)= 7

    Other function:

    g(x) = -(x^2) + 4
    integral of g(x)= =1/3x^3 + 4x
    g(1)= 11/3

    integral of g(x^2)= -1/5x^5 + 4x
    g(1^2)= 19/5

    As for part C, I'm clueless. Thank you.
     
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