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Adding two integrals with different limits of integration

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Interpret the integrals (from 0 to 4)∫ (3x/4) dx + (from 4 to 5)∫ (sqrt(25-x^2)) dx as areas and use the result to express the sum above as one definite integral. Evaluate the new integral.


    2. Relevant equations



    3. The attempt at a solution
    I see that I could integrate them separately and add the two values to find the total area. The issue I am having is writing them as one integral. I suspect that one way would be to re-write one with the equivilant function but using the limits of integration of the other. However, the only way I've ever changed limits of integration (and all I can find using searches) is when you're doing a u-substition and I'm not sure if the same methodology applies here.

    Can I simply add '4' to the first function, 3x/4, to move it over 4 units to the right and then stick that whole thing, 3x/4 + 4, into the other integral between 4 and 5 (along with sqrt(25-x^2)? That would look something like:

    the integral from 4 to 5 of (sqrt(25-x^2)+ 3x/4 + 4) dx

    Thank you,
    Odd
     
    Last edited: Dec 11, 2011
  2. jcsd
  3. Dec 11, 2011 #2

    Mark44

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    No. Adding 4 to the formula of a function shifts the graph up, not right.
    I think that you are missing the point of this exercise. The first part of this problem asks you to interpret the two integrals as areas, which implies that you should sketch a graph of each function being integrated. It doesn't seem to me that you have done that.

    To express the sum as a single integral is easy to do if you are working with polar integrals.
     
  4. Dec 11, 2011 #3
    I did graph them and it's easy to visualize the sum. I believe you when you suggest using polar integrals but I have -no idea- how to go about that. I just watched a video on double integrals using polar coordinates to see if it shed some light but I think I'm more confused than before now. I'll keep looking for more on that topic but would greatly appreciate any hints on how to get started.

    I suspect the solution is simple but I feel that I'm simply missing a basic idea here.
     
    Last edited: Dec 11, 2011
  5. Dec 11, 2011 #4

    Dick

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    What does the region you are finding the total area of look like? Can you describe it geometrically?
     
  6. Dec 11, 2011 #5
    Sure, a straight line starting at the origin up and to the right (described by 3x/4) and when x=4 the function curves downwards and ends up at (5,0) (described by sqrt(25-x^2). I did this by graphing them both separately and visualizing the combination. Maybe that's skipping a step you're looking for me to take?

    EDIT: Sorry, I described the line. The region would be the area between that line and the x-axis.
     
  7. Dec 11, 2011 #6

    Dick

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    What does the curve y=sqrt(25-x^2) look like geometrically? Answer that one and if you are still confused, I'll tell you the answer I'm fishing for.
     
  8. Dec 11, 2011 #7
    Half of an oval. More specifically, the 'upper' half of an oval with the longest axis parallel with to y-axis. Similar to a parabola.

    I suspect that you'd want me to express this in a polar system. I don't know how to go about this :(
     
  9. Dec 11, 2011 #8

    Dick

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    I'd call your 'oval' a 'circle'. The region whose area you are finding is a sector of a circle. Hence, Mark44's hint about polar coordinates. And it's not a double integral. Look up 'area in polar coordinates'.
     
  10. Dec 11, 2011 #9
    The end point of the first segment is (5,arcsin(3/5)). Working on the second now.
     
  11. Dec 11, 2011 #10

    Dick

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    You can probably stop working on it, because I don't know what you are talking about. The region is a region sliced out a circle that looks like a slice of pie. It has fixed radius and a beginning angle and an ending angle. Work on those instead.
     
  12. Dec 11, 2011 #11
    Haha, this is getting pretty silly. I appreciate all the help so far. I think the confusion comes from the image I have in my mind versus the image I should have and the image you are visualizing.

    MiA8i.png

    That's what I was thinking. You add the two areas of the lines given the different limits of integration.

    What I think you're looking for now is something like this: (extracted from a youtube video from patrickJMT titled 'Finding Areas in Polar Coordinates').

    rp72N.png

    What I'm having a hard time doing is visualizing the -addition- of the two functions while they sit on top of one another finding two different values for theta. That idea makes me think subtraction, not addition. Thoughts?


    The point where the two lines meet in the first picture I posted is described in polar form as (5, arcsin(3/5)). arcsin(3/5) is the angle theta that the first function makes with the x-axis and 5 is the radius.
     
  13. Dec 11, 2011 #12

    Dick

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    You've got exactly the right polar coordinate reference. But forget about adding anything. Now you have a single simple region. Find it's area with a single simple integral in polar coordinates.
     
  14. Dec 11, 2011 #13
    That's the only issue I'm having. I've had the area of this function down about one minute into the problem by simply adding the areas of each respective integral.

    The issue I'm having is writing out -one integral- to represent them both. I'm working on understanding how to write these two as one integral using polar equations but am not yet there.
     
  15. Dec 11, 2011 #14

    Dick

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    Stop saying "these two areas". In polar coordinates you can write the integral expressing the area of the region as a single integral. That's the whole point. Don't do them separately! What's r as a function of theta in the combined region? Then what are the limits for theta?
     
  16. Dec 11, 2011 #15
    Like I said, that's exactly what I am trying to do... I see that as theta changes r changes as it follows the curve down. Then I see that if I integrate that function with respect to theta (multiplied by a half because it's a triangle not a rectangle) it will give me the overall area from theta at point 4 to theta at point 5.

    Thanks
     
  17. Dec 11, 2011 #16

    Dick

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    How does r change? Do you still think it's 'oval'? It's not. It's a circle. Even if it weren't you could still write the area as a single integral. It would just be more complicated.
     
  18. Dec 11, 2011 #17
    When I plot that function it looks like an oval.

    http://www.wolframalpha.com/input/?i=plot+y=sqrt(25-x^2)+x=0..5

    I'm not saying you're wrong. I just don't know how else to think about it.

    At this point I don't know how r changes but I'm sure I can find that out using one of the polar formulas. Finals + missing concepts = frustration. I will have to come back to this problem because it's driving me nuts.

    Thanks again and I will report back here after I get some sleep.
     
  19. Dec 11, 2011 #18

    Dick

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  20. Dec 11, 2011 #19

    Mark44

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    The only reason the graph looks like part of an ellipse is that the x and y axes are scaled differently. Seriously, if you are working with integrals, you should know that the graph of y = [itex]\sqrt{25 - x^2}[/itex] is the upper half of a circle.
     
  21. Dec 11, 2011 #20
    One thing that happens when the human brain is on very little sleep, is stressed, and is overworked is it begins to overlook things. I defaulted to the easiest method I know to understand the shape of something and that is to graph it. Yes, I know the equation of a circle. I'm sorry that I missed the swap from x^2+y^2=r^2 to y=sqrt(25-x^2) but you don't have to be rude about it. It really doesn't help solve anything...

    Why I missed it probably has something to do with the fact that up until one post before the mention of shape it didn't matter if it was a circle or oval in the way I was visualizing the information.
     
    Last edited: Dec 12, 2011
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