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Area under the curve

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the area under the curve of the function y=5x^2+3 from x=0 to x=2


    2. Relevant equations

    -

    3. The attempt at a solution

    First I sketched it out on paper, finding that at x=2,y=28. I created a rectangle with l*w of 6 (3*2), would the remaining area not included in the rectangle be 23/3 ? I guess this because it asks from 0 to 2 and the value of 0 is x=3 and value of 2 is x=28.

    So my final answer would be 6+23/3 ?
     
  2. jcsd
  3. Jun 18, 2013 #2

    CompuChip

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    Why 23/3? The area of a triangle with vertices (0, 3), (2, 3) and (2, 28) would be 1/2 * (28 - 3) * 2 = 25, which is different from 23/3 = 7 + 2/3.

    To be honest, I'm a bit lost at the moment as to how you would solve this question without resorting to integration...
     
  4. Jun 18, 2013 #3
    Welcome to my precal teacher's logic, she never taught us integration.. So I'm pretty lost with this
     
  5. Jun 18, 2013 #4

    CAF123

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    Are you being taught approximation methods at the moment? You'll get onto integration later.
     
  6. Jun 18, 2013 #5
    Honestly I don't even know what she's doing. Our exam is tomorrow I'm just reviewing now and it didn't make sense how she got her answer of 6+8/3.

    She literally gives us constants for values above rectangles such as x^2 being 8/3, then she gets us to create a rectangle from the x to x values given, and it becomes simply l*w then you add the constant.
     
  7. Jun 18, 2013 #6

    CompuChip

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    Ah, so you have been given that the area below x² between x = 0 and 2 is 8/3?

    In that case: look how the graph for the given formula 5x² + 3 can be made from the standard graph of x².
    There are two steps, can you describe them geometrically (so in terms of scaling and shifting)?
     
  8. Jun 18, 2013 #7
    As in a horizontal translation of 3 units upwards? And the hs of 1/5. But the 5x wouldn't effect the area in this particular case would it?
     
  9. Jun 18, 2013 #8

    CompuChip

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    Yep, except upwards is called vertical. And note that you have to do them in the correct order: first a horizontal scaling, then the translation; otherwise you would get 5(x + 3) = 5x + 15, instead of 5x + 3.

    So the translation is what gives you the rectangle of 2 by 3 units which you had already found. Now let's look at the scaling, which actually does affect the area. Try looking at a straight line y = x between x = 0 and 2. What happens if I apply the scaling there (i.e. switch to y = 5x)?
    You could consider such a straight line as a (crude) approximation to the actual curve, which should make it more intuitive to you that it doesn't only hold for straight lines.
     
  10. Jun 18, 2013 #9
    your slope will be changed to 1/5 once you apply the 5x
     
  11. Jun 18, 2013 #10

    CompuChip

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    Yep, but I was not so much interested in the slope, but more in the area when I do that rescaling.
    How does the area change if you scale y = x by 1/5?
     
  12. Jun 18, 2013 #11
    It would become 5 times greater?
     
  13. Jun 18, 2013 #12

    CompuChip

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    Exactly.
    It's basically a triangle with the same base but 5 times the height.

    Now, as I hinted, the same happens to any function. So if you know the area under x² from x = 0 to 2, you can also find the area of 5x².
     
  14. Jun 18, 2013 #13
    But I don't know how to find the area of x^2.. :( haha
     
  15. Jun 18, 2013 #14

    CAF123

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    I think the method provided by CompuChip was suggested on the basis of you already knowing what the area under y=x2 from 0 to 2 was. (i.e your teacher had provided that for you). Is this the case?
     
  16. Jun 18, 2013 #15

    CompuChip

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    At least it is what you said :)
     
  17. Jun 18, 2013 #16
    Oh okay I see, but how can I actually figure out that it is 8/3 myself?
     
  18. Jun 18, 2013 #17

    CAF123

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    You would have to compute the integral $$\int_0^2 x^2 dx,$$ but I think you said you have not done this. If you have done approximation methods (I.e something like approximating the area as the sum of finitely many rectangles) then you can certainly obtain an approximation to that 8/3.
     
  19. Jun 18, 2013 #18
    Okay, so she gave me the constant as the area under the curve from x=0 to x=2 to be 8/3, this is for y=x^2. So If I had y=5x^2+3 .. Does that mean my area is 5*(8/3) ? Seem right? I know I've been taking long intervals with replying but I have been studying very hard.. so if anyone could answer that would be awesome. And thank you so much for hanging along with me you're really helping a lot.
     
  20. Jun 18, 2013 #19

    Mark44

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    No, because you're not taking into account the "+3" part. If y = 5x2, then the area under this curve, from x = 0 to x = 2, is 5 * 8/3.

    Adding 3 has the effect of shifting the graph of y = 5x2 up by 3 units, so what will that do to the area beneath y = 5x2 + 3?
     
  21. Jun 18, 2013 #20
    Well I already solved for the area from x=0 to x=2 by creating a rectangle with a width of 2 and height/vertical length of 3 (due to my shift). So l*w = 6. But I still have my area above this undetermined.. So I know that x^2 has an remainder area above the rectangle of 8/3 as my teacher gave us.. but I don't get how the 5x^2 will effect this
     
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