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Integrating the area under a curve

  1. Jun 22, 2012 #1
    Hey everyone, I had a recent post similar to this one, and everyone may have not understood it because I didn't use LaTeX, so here it is.

    1. The problem statement, all variables and given/known data
    Integrate the area under [itex]\frac{x}{3x}[/itex] and above [itex]\frac{x}{3x^.5}[/itex] between x=1 and x=4.
    Same as: [itex]\int[/itex] [itex]\frac{x}{3x}[/itex] - [itex]\frac{x}{3x^.5}[/itex] dx between x=1 and x=4.

    2. Relevant equations
    See above

    3. The attempt at a solution
    I would integrate each individual fraction by raising each by 1 power, and dividing the coefficient by that new exponent. Each time that I try this, no matter how I simplify it, the answer turns out to be way too small.

    What should I try next?
     
  2. jcsd
  3. Jun 22, 2012 #2

    Mark44

    Staff: Mentor

    Before integrating, you really should simplify the fractions. Both fractions have a factor of 1/3, so that should be factored out and brought out front of your integral.

    After simplification, what does your integral look like?

    Tip: itex is used for inline expressions, but it makes fractions and integrals very small. For those kinds of things use tex instead of itex.
     
  4. Jun 22, 2012 #3
    Factoring's not my strong topic:

    [itex]\frac{1}{3}[/itex][itex]\int[/itex] 1 - [itex]\frac{x}{x^.5}[/itex] dx between 1 and 4
     
  5. Jun 22, 2012 #4

    Mark44

    Staff: Mentor

    [tex]\frac{x^1}{x^{.5}} = x^{.5}[/tex]

    So your integral should look like this:
    $$ \frac{1}{3}\int_1^4 1 - x^{1/2}~dx$$
     
  6. Jun 22, 2012 #5
    And from there, just regularly integrate?
     
  7. Jun 22, 2012 #6
    Thank you so much Mark
     
  8. Jun 22, 2012 #7

    Mark44

    Staff: Mentor

    Yes.
     
  9. Jun 22, 2012 #8
    Actually, one last question. Why do when I compute it, it is a negative number? Is there any rules I should be aware of... I just started learning about Calculus and the definite integral.
     
  10. Jun 22, 2012 #9
    Specifically, -1.667
     
  11. Jun 22, 2012 #10

    could be a problem regarding the area below 1/3 and above √x/3, but as x varies from 1 to 4, √x/3 varies from 1/3 to 2/3, so the setup doesn't make much sense.
     
  12. Jun 23, 2012 #11

    Mark44

    Staff: Mentor

    The integral should be
    ## \frac{1}{3}\int_1^4 x^{1/2} - 1 ~dx##

    The graph of y = (1/3)x1/2 lies above the graph of y = 1/3 on the interval [1, 4]. When you measure a distance, you need to subtract the smaller number from the larger number, otherwise you'll get a negative value for the length of the typical area element.
     
  13. Jun 23, 2012 #12
    Thank you very much Mark!
     
  14. Jun 23, 2012 #13
    My final answer came out to be 1/3, does that sound right?
     
  15. Jun 23, 2012 #14

    Mark44

    Staff: Mentor

    No, you have made a mistake. Show us what you did to get 1/3 for your answer.
     
  16. Jun 23, 2012 #15
    I see what I did. Let me try it again:

    [itex]\frac{1}{3} [/itex][itex]\int[/itex] (x^.5 - 1)dx

    = [itex]\frac{1}{3}[/itex] (([itex]\frac{x}{1.5}[/itex] ^1.5) - x)) for 1 through 4

    = [itex]\frac{1}{3}[/itex] ([itex]\frac{4}{1.5}[/itex] ^1.5) - 4) - ([itex]\frac{1}{1.5}[/itex] ^1.5 - 1))

    =[itex]\frac{1}{3}[/itex] (((2.66^1.5) - 4) - ((.66^1.5) - 1))

    =1/3((4.338-4)- (.536 - 1)) = 1/3(.338 + .464) = 1/3(.802) = .267

    ???
     
    Last edited: Jun 23, 2012
  17. Jun 23, 2012 #16

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Here is your error. The integral of [itex]x^{.5}[/itex] is [itex]x^{1.5}/1.5[/itex] NOT [itex](x/1.5)^{1.5}[/itex] so that, at x= 4, it is [itex]4^{1.5}/1.5= 8/1.5= 16/3[/itex], NOT [itex](4/1.5)^{1.5}= (8/3)^{1.5}[/itex].

    By the way, it is a lot easier to put entire formulas in LaTeX, not just "bits and pieces".
     
  18. Jun 23, 2012 #17
    Thanks once again.
     
  19. Jun 23, 2012 #18
    Now I get about .55. This doesn't seem right, again.
     
  20. Jun 23, 2012 #19

    Mark44

    Staff: Mentor

    The actual answer is 5/9, which is approximately .55.
     
  21. Jun 23, 2012 #20
    When I graph the original two lines it seems to have a much larger area, but an answer is an answer. Thanks.
     
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