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Analysis: Integration

  1. Nov 23, 2011 #1
    I'm not sure about #4. I know probably #5 and definitely #6 are wrong. For some reason I'm stumped on #6 especially.

    http://i111.photobucket.com/albums/n149/camarolt4z28/4-1.png [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/5.jpg [Broken]

    Question for #6: http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

    http://i111.photobucket.com/albums/n149/camarolt4z28/6a.jpg [Broken]
    http://i111.photobucket.com/albums/n149/camarolt4z28/6b.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 23, 2011 #2

    Mark44

    Staff: Mentor

    Speaking for myself, I would really appreciate seeing the problem directly in the post, instead of having to go to another site to see what the problem is and what you have done.
     
  4. Nov 23, 2011 #3
    All you're doing is clicking a link to a picture of my work. That's the fastest, easiest, and clearest way for me to express my work.
     
  5. Nov 23, 2011 #4

    Mark44

    Staff: Mentor

    OTOH, I have to have two windows open - one to see what you did and another to make comments about what you did. And I can't insert a comment at a particular point if there's something wrong. Instead, I have to give some context, like "in the line where blah, blah, you have xyz and it should be abc."

    It may be fast, easy, and clear for you, but it's a hassle for me that I don't need. When you're asking for help from someone, the key is to make it as easy as possible for them, not necessarily for yourself.
     
  6. Nov 23, 2011 #5
    I understand your point, but since I don't know latex, in problems like these it's more clear to simply take a picture. Fortunately, there aren't too many lines to worry about.
     
  7. Nov 23, 2011 #6
    Latex is not so difficult. One can try and learn how to use it. The 'preview' facility is a great help.
     
  8. Nov 23, 2011 #7

    Mark44

    Staff: Mentor

  9. Nov 23, 2011 #8
    Number 4 is not correct. How do you know the point c is the same for both functions? You need to use MVT for integrals but to show a "new" function is 0.

    Number 5 looks ok, idk why you need to say F is uniformly cont.

    I don't see what problem 6 is?
     
  10. Nov 23, 2011 #9
    #4

    Suppose ∫abf(x)dx = ∫abg(x)dx
    Then ∫ab[f(x)-g(x)]dx = 0

    Now you can use continuity to complete the proof. I don't think the mean value theorem can help you in this problem. The mean value theorem is used to prove that a point can be a certain slope, which has little to do with your problem.

    #5

    What's the lower bound of integration in the last line? Hint: c∈[a,b]

    #6

    What is the question?
     
    Last edited: Nov 23, 2011
  11. Nov 23, 2011 #10
    For #4, I can set h(x) = f(x) - g(x) and take it from there, right?

    For #5, I probably don't need to say it's uniformly continuous.

    Oops. Sorry, I forgot to include the question.
     
  12. Nov 23, 2011 #11
    For #5, I forgot the b at the bottom. I just swapped x and b.
     
  13. Nov 23, 2011 #12

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I heartily agree with these sentiments. It is also a pain to follow the replies. That is why I didn't respond to this thread when I first saw it. Well, that and the fact that looked like some kind of take-home assignment.
     
  14. Nov 23, 2011 #13
    I reworked #4 and cleaned up #5. I'm going take another shot at #6.
     
  15. Nov 26, 2011 #14
    Any ideas on #6? I'm sure it's an easy problem. I just keep running into dead ends.
     
  16. Nov 26, 2011 #15
    You are going to have to get rid of that integral sign, and no, don't evaluate the integral would be my hint.
     
  17. Nov 26, 2011 #16
    Okay. I think I'm over-complicating this or approaching from the wrong side of the FTOC. Haha.

    And if f is continuous at a point c in the interval

    F'(c) = f(c)

    How about?

    [tex]\int_a^x f(t)dt = \int_a^b f(t)dt -\int_x^b f(t)dt[/tex]

    [tex]\int_a^x f(t)dt = F(x) = \int_x^b f(t)dt[/tex]

    [tex]F(x) = \alpha - F(x)[/tex]

    [tex]F'(x) = 0 - F'(x)[/tex]

    [tex]f(x) = -f(x)[/tex]

    [tex]\Rightarrow f(x) = 0[/tex]
     
    Last edited: Nov 26, 2011
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