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True/false question on integration

  1. Feb 6, 2012 #1
    Let f(x) > 0 be a continuous function. Then F(x) = ∫f(t) dt ≥ 0 [0, x] for all values of x.

    I would have chosen true, but the answer is false. If f(x) is always positive, then how can the integral ever be negative?

    Thanks.
     
  2. jcsd
  3. Feb 6, 2012 #2

    jedishrfu

    Staff: Mentor

    is this answer in a book or the prof said so?

    is the t going from 0 to x or x to 0? that could be the difference.
     
  4. Feb 6, 2012 #3
    It is part of an old exam, the solutions say that the answer is false.
     
  5. Feb 6, 2012 #4

    jedishrfu

    Staff: Mentor

    then its probably due to whether its evaluated from 0..x or x..0
     
  6. Feb 6, 2012 #5
    The bottom limit on the integral is 0 and the top limit is x.
     
  7. Feb 6, 2012 #6

    jedishrfu

    Staff: Mentor

    okay so is f(x) real or complex?
     
  8. Feb 6, 2012 #7
    We haven't discussed the differences between real and complex integrals, so it's real.
     
  9. Feb 6, 2012 #8

    jedishrfu

    Staff: Mentor

    I can't think of any other trick so I think you're right and I'd go with that. You could ask a teacher about it.

    For f(x)>0 then conceptually sum of the vertical strip areas from 0 to x would make the answer positive. The trick is looking for some way to have ( F(x) - F(0) ) < 0 but ( F(x) - F(0) ) is the area vertical strip areas from 0 to x.

    So given that f(x) is real then the only way is for the integral to be negative is if its evaluated from x to 0, Since it wasn't explicitly shown at least in your example.
     
  10. Feb 6, 2012 #9
    I think I may have gotten it. Is it because x can be any value, so that the integral can go from 0 to -3, for example, where we would receive a negative value?
     
  11. Feb 6, 2012 #10

    jedishrfu

    Staff: Mentor

    perhaps but wouldn't the closed interval be written as [x,0] to imply that x<0
     
  12. Feb 6, 2012 #11

    Bacle2

    User Avatar
    Science Advisor

    If you're going in the standard direction [0,x] , then your F(x) is positive:

    F(x):=∫x0 f(t)dt

    So F(0)=0 ; F'(x)=f(x) , by FTCalculus, and f(x)>0 by assumption.

    If x<0 , then F(x)<0 by same argument.
     
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