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Connection between Improper Integrals

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Explain geometrically why the integral of dx/sqrt(x) for (x, 0,1)= [the integral of dx/x^2 for (x, 1, infinity)] + 1
    Including graphs as well.


    2. Relevant equations



    3. The attempt at a solution
    I graphed both functions and they met at a point (1,1) but i don't really understand how to show it geometrically.
     
  2. jcsd
  3. Jul 11, 2010 #2

    Mark44

    Staff: Mentor

    Graph the regions whose areas are represented by the two integrals. That might get you going in the right direction.

    Also, there is a relationship between y = 1/sqrt(x) and y = 1/x^2. Can you figure out what it is?
     
  4. Jul 12, 2010 #3
    I know the two functions are both inverse of each other. I graphed the areas of each region. But i still dont know what to write to answer the question.
     
  5. Jul 12, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Do you know that if two functions are inverses of each other then their graphs are reflections of each other across the line x=y?
     
  6. Jul 12, 2010 #5
    How does this affect their area on the intervals i wrote in the initial question i asked.Why do need to add 1 to make both areas equal
     
  7. Jul 12, 2010 #6

    Mark44

    Staff: Mentor

    The area represented by the integral [tex]\int_0^1 \frac{dx}{\sqrt{x}}[/tex] is a bit larger (1 unit) than the area represented by the integral [tex]\int_1^{\infty}\frac{dx}{x^2}}[/tex].

    For the first integral, the area is calculated using vertical strips of width dx and height 1/sqrt(x). What would the integral be if you used horizontal strips?
     
  8. Jul 13, 2010 #7
    Yeah, finally solve it. I used the inverses and reflection of the two graphs at x=y.
     
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