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anonymous

  • 5 years ago

Consider the area between the graphs x+3y=1 and x+9=y2. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals; Alternatively this area can be computed as a single integral

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  1. amistre64
    • 5 years ago
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    and the question for this statement?

  2. amistre64
    • 5 years ago
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    we can integrate them both at the same time, or seperately and add the sums...

  3. anonymous
    • 5 years ago
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    I need both please!

  4. anonymous
    • 5 years ago
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    And I need a and b, as well as c

  5. amistre64
    • 5 years ago
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    the common line between them is the sum of there parts./2 right? gets an averge?

  6. anonymous
    • 5 years ago
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    I also need f(x) and g(x) separately

  7. amistre64
    • 5 years ago
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    f(x)=-x/3 +1/3; g(x) = sqrt(x+9)

  8. amistre64
    • 5 years ago
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    [S] f(x) dx - [S] g(x) dx [S] f(x) - g(x) dx

  9. anonymous
    • 5 years ago
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    The fist thing you have to do is to find the intersection of the two lines. Secondly, you have to find the intersection of the two lines with the x axis. As the third step, you have to integrate the frist integral from the x-coordinate of the first line until the x-cordinate of the intersection point. You have to do the same with the seconde line

  10. amistre64
    • 5 years ago
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    it doesnt provide an interval, and so assuming that it just wants the intersections is presumptuous; prolly true, but not accurate

  11. amistre64
    • 5 years ago
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    it could want the interval [0,root to the right]

  12. anonymous
    • 5 years ago
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    Im not sure how to do that. Sorry

  13. amistre64
    • 5 years ago
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    you get something that looks like this

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  14. amistre64
    • 5 years ago
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    without a stated interval, you are prolly correct in assuming the left and right intercepts

  15. anonymous
    • 5 years ago
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    ahhh ok. give me one second to thing about it

  16. amistre64
    • 5 years ago
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    its got issues of being on the other side of the y axis; and youd have to split it into intervals from the common point to the left and right

  17. amistre64
    • 5 years ago
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    maybe +9 to get it all on the right side of the y axis?

  18. amistre64
    • 5 years ago
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    and my line parts to steep , but its just an artist interpretation lol

  19. amistre64
    • 5 years ago
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    kinda like this eh

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  20. anonymous
    • 5 years ago
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    Anyway, you have to split the total area in two sides.Carry out integration from intersection of the parable and the x-axis until the intersection of the parable and the line. They you have to integrate from the intersection point to the intersection of the line with the x-axis. That is how I would do that my friend

  21. anonymous
    • 5 years ago
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    I got x=1 for c, but it keeps saying that it's wrong.

  22. anonymous
    • 5 years ago
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    which c?

  23. anonymous
    • 5 years ago
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    the end point on the right

  24. anonymous
    • 5 years ago
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    I dont know my friend. I got a little bit confused. Sorry :(

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